# Eight to Late

Sensemaking and Analytics for Organizations

## A gentle introduction to Monte Carlo simulation for project managers

This article covers the why, what and how of Monte Carlo simulation using a canonical example from project management –  estimating the duration of a small project. Before starting, however, I’d like say a few words about the tool I’m going to use.

In keeping with the format of the tutorials on this blog, I’ve assumed very little prior knowledge about probability, let alone Monte Carlo simulation. Consequently, the article is verbose and the tone somewhat didactic.

### Introduction

Estimation is key part of a project manager’s role. The most frequent (and consequential) estimates they are asked deliver relate to time and cost.  Often these are calculated and presented as point estimates: i.e. single numbers – as in, this task will take 3 days. Or, a little better, as two-point ranges – as in, this task will take between 2 and 5 days.  Better still, many use a PERT-like approach wherein estimates are based on 3 points: best, most likely and worst case scenarios – as in, this task will take between 2 and 5 days, but it’s most likely that we’ll finish on day 3.  We’ll use three-point estimates as a starting point for Monte Carlo simulation, but first, some relevant background.

It is a truism, well borne out by experience, that it is easier to estimate small, simple tasks than large, complex ones. Indeed, this is why one of the early to-dos in a project is the construction of a work breakdown structure. However, a problem arises when one combines the estimates for individual elements into an overall estimate for a project or a phase thereof. It is that a straightforward addition of individual estimates or bounds will almost always lead to a grossly incorrect estimation of overall time or cost. The reason for this is simple: estimates are necessarily based on probabilities and probabilities do not combine additively. Monte Carlo simulation provides a principled and intuitive way to obtain probabilistic estimates at the level of an entire project based on estimates of the individual tasks that comprise it.

### The problem

The best way to explain Monte Carlo is through a simple worked example. So, let’s consider a 4 task project shown in Figure 1. In the project, the second task is dependent on the first, and third and fourth are dependent on the second but not on each other. The upshot of this is that the first two tasks have to be performed sequentially and the last two can be done at the same time, but can only be started after the second task is completed.

To summarise: the first two tasks must be done in series and the last two can be done in parallel.

Figure 1; A project with 4 tasks.

Figure 1 also shows the three point estimates for each task – that is the minimum, maximum and most likely completion times. For completeness I’ve listed them below:

• Task 1 – Min: 2 days; Most Likely: 4 days; Max: 8 days
• Task 2 – Min: 3 days; Most Likely: 5 days; Max: 10 days
• Task 3 – Min: 3 days; Most Likely: 6 days; Max: 9 days
• Task 4 – Min: 2 days; Most Likely: 4 days; Max: 7 days

OK, so that’s the situation as it is given to us. The first step to  developing  an estimate is to formulate the problem in a way that it can be tackled using Monte Carlo simulation. This bring us to the important topic of the shape of uncertainty aka probability distributions.

### The shape of uncertainty

Consider the data for Task 1. You have been told that it most often finishes on day 4.  However, if things go well, it could take as little as 2 days; but if things go badly it could take as long as 8 days.  Therefore, your range of possible finish times (outcomes) is between 2 to 8 days.

Clearly, each of these outcomes is not equally likely.  The most likely outcome is that you will finish the task in 4 days (from what your team member has told you). Moreover, the likelihood of finishing in less than 2 days or more than 8 days is zero. If we plot the likelihood of completion against completion time, it would look something like Figure 2.

Figure 2: Likelihood of finishing on day 2, day 4 and day 8.

Figure 2 begs a couple of questions:

1. What are the relative likelihoods of completion for all intermediate times – i.e. those between 2 to 4 days and 4 to 8 days?
2. How can one quantify the likelihood of intermediate times? In other words, how can one get a numerical value of the likelihood for all times between 2 to 8 days?  Note that we know from the earlier discussion that this must be zero for any time less than 2 or greater than 8 days.

The two questions are actually related. As we shall soon see, once we know the relative likelihood of completion at all times (compared to the maximum), we can work out its numerical value.

Since we don’t know anything about intermediate times (I’m assuming there is no other historical data available), the simplest thing to do is to assume that the likelihood increases linearly (as a straight line) from 2 to 4 days and decreases in the same way from 4 to 8 days as shown in Figure 3. This gives us the well-known triangular distribution.

Jargon Buster: The term distribution is simply a fancy word for a plot of likelihood vs. time.

Figure 3: Triangular distribution fitted to points in Figure 1

Of course, this isn’t the only possibility; there are an infinite number of others. Figure 4 is another (admittedly weird) example.

Figure 4: Another distribution that fits the points in Figure 2.

Further, it is quite possible that the upper limit (8 days) is not a hard one. It may be that in exceptional cases the task could take much longer (for example, if your team member calls in sick for two weeks) or even not be completed at all (for example, if she then leaves for that mythical greener pasture).  Catering for the latter possibility, the shape of the likelihood might resemble Figure 5.

Figure 5: A distribution that allows for a very long (potentially) infinite completion time

The main takeaway from the above is that uncertainties should be expressed as shapes rather than numbers, a notion popularised by Sam Savage in his book, The Flaw of Averages.

[Aside:  you may have noticed that all the distributions shown above are skewed to the right – that  is they have a long tail. This is a general feature of distributions that describe time (or cost) of project tasks. It would take me too far afield to discuss why this is so, but if you’re interested you may want to check out my post on the inherent uncertainty of project task estimates.

### From likelihood to probability

Thus far, I have used the word “likelihood” without bothering to define it.  It’s time to make the notion more precise.  I’ll begin by asking the question: what common sense properties do we expect a quantitative measure of likelihood to have?

Consider the following:

1. If an event is impossible, its likelihood should be zero.
2. The sum of likelihoods of all possible events should equal complete certainty. That is, it should be a constant. As this constant can be anything, let us define it to be 1.

In terms of the example above, if we denote time by $t$ and the likelihood by $P(t)$  then:

$P(t) = 0$ for $t< 2$ and  $t> 8$

And

$\sum_{t}P(t) = 1$ where $2\leq t< 8$

Where $\sum_{t}$ denotes the sum of all non-zero likelihoods – i.e. those that lie between 2 and 8 days. In simple terms this is the area enclosed by the likelihood curves and the x axis in figures 2 to 5.  (Technical Note:  Since $t$ is a continuous variable, this should be denoted by an integral rather than a simple sum, but this is a technicality that need not concern us here)

$P(t)$ is , in fact, what mathematicians call probability– which explains why I have used the symbol $P$ rather than $L$. Now that I’ve explained what it  is, I’ll use the word “probability” instead of ” likelihood” in the remainder of this article.

With these assumptions in hand, we can now obtain numerical values for the probability of completion for all times between 2 and 8 days. This can be figured out by noting that the area under the probability curve (the triangle in figure 3 and the weird shape in figure 4) must equal 1, and we’ll do this next.  Indeed, for the problem at hand, we’ll assume that all four task durations can be fitted to triangular distributions. This is primarily to keep things  simple. However, I should emphasise that you can use any shape so long as you can express it mathematically, and I’ll say more about this towards the end of this article.

### The triangular distribution

Let’s look at the estimate for Task 1. We have three numbers corresponding to a minimummost likely and maximum time.  To keep the discussion general, we’ll call these $t_{min}$, $t_{ml}$ and $t_{max}$ respectively, (we’ll get back to our estimator’s specific numbers later).

Now, what about the probabilities associated with each of these times?

Since $t_{min}$ and $t_{max}$ correspond to the minimum and maximum times,  the probability associated with these is zero. Why?  Because if it wasn’t zero, then there would be a non-zero probability of completion for a time less than $t_{min}$ or greater than $t_{max}$ – which isn’t possible [Note: this is a consequence of the assumption that the probability varies continuously –  so if it takes on non-zero value, $p_{0}$,  at $t_{min}$ then it must take on a value slightly less than $p_{0}$ – but greater than 0 –  at $t$ slightly smaller than $t_{min}$ ] .   As far as  the most likely time,  $t_{ml}$,  is concerned:  by definition, the probability attains its highest value at time $t_{ml}$.    So, assuming the probability can be described by a triangular function, the distribution must have the form shown in Figure 6 below.

Figure 6: Triangular distribution redux.

For the simulation, we need to know the equation describing the above distribution.  Although Wikipedia will tell us the answer in a mouse-click, it is instructive to figure it out for ourselves. First, note that the area under the triangle must be equal to  1 because the task must finish at some time between $t_{min}$ and $t_{max}$.   As a consequence we have:

$\frac{1}{2}\times{base}\times{altitude}=\frac{1}{2}\times{(t_{max}-t_{min})}\times{p(t_{ml})}=1\ldots\ldots{(1)}$

where $p(t_{ml})$ is the probability corresponding to time $t_{ml}$.  With a bit of rearranging we get,

$p(t_{ml})=\frac{2}{(t_{max}-t_{min})}\ldots\ldots(2)$

To derive the probability for any time $t$ lying between $t_{min}$ and $t_{ml}$, we note that:

$\frac{(t-t_{min})}{p(t)}=\frac{(t_{ml}-t_{min})}{p(t_{ml})}\ldots\ldots(3)$

This is a consequence of the fact that the ratios on either side of equation (3)  are  equal to the slope of the line joining the points $(t_{min},0)$ and $(t_{ml}, p(t_{ml}))$.

Figure 7

Substituting (2) in (3) and simplifying a bit, we obtain:

$p(t)=\frac{2(t-t_{min})}{(t_{ml}-t_{min})(t_{max}-t_{min})}\dots\ldots(4)$ for $t_{min}\leq t \leq t_{ml}$

In a similar fashion one can show that the probability for times lying between $t_{ml}$ and $t_{max}$ is given by:

$p(t)=\frac{2(t_{max}-t)}{(t_{max}-t_{ml})(t_{max}-t_{min})}\dots\ldots(5)$ for $t_{ml}\leq t \leq t_{max}$

Equations 4 and 5 together describe the probability distribution function (or PDF)  for all times between $t_{min}$ and $t_{max}$.

As it turns out, in Monte Carlo simulations, we don’t directly work with the probability distribution function. Instead we work with the cumulative distribution function (or CDF) which is the probability, $P$,  that the task is completed by time $t$. To reiterate, the PDF, $p(t)$, is the probability of the task finishing at time $t$ whereas the CDF, $P(t)$, is the probability of the task completing by time $t$. The CDF, $P(t)$,  is essentially a sum of all probabilities between $t_{min}$ and $t$. For $t < t_{min}$ this is the area under the triangle with apexes at   ($t_{min}$, 0), (t, 0) and (t, p(t)).  Using the formula for the area of a triangle (1/2 base times height) and equation (4) we get:

$P(t)=\frac{(t-t_{min})^2}{(t_{ml}-t_{min})(t_{max}-t_{min})}\ldots\ldots(6)$ for $t_{min}\leq t \leq t_{ml}$

Noting that for $t \geq t_{ml}$, the area under the curve equals the total area minus the area enclosed by the triangle with base between t and $t_{max}$, we have:

$P(t)=1- \frac{(t_{max}-t)^2}{(t_{max}-t_{ml})(t_{max}-t_{min})}\ldots\ldots(7)$ for $t_{ml}\leq t \leq t_{max}$

As expected,  $P(t)$  starts out with a value 0 at $t_{min}$ and then increases monotonically, attaining a value of 1 at $t_{max}$.

To end this section let’s plug in the numbers quoted by our estimator at the start of this section: $t_{min}=2$, $t_{ml}=4$ and $t_{max}=8$.  The resulting PDF and CDF are shown in figures 8 and 9.

Figure 8: PDF for triangular distribution (tmin=2, tml=4, tmax=8)

Figure 9 – CDF for triangular distribution (tmin=2, tml=4, tmax=8)

### Monte Carlo in a minute

Now with all that conceptual work done, we can get to the main topic of this post:  Monte Carlo estimation. The basic idea behind Monte Carlo is to simulate the entire project (all 4 tasks in this case) a large number N (say 10,000) times and thus obtain N overall completion times.  In each of the N trials, we simulate each of the tasks in the project and add them up appropriately to give us an overall project completion time for the trial.  The resulting N overall completion times will all be different, ranging from the sum of the minimum completion times to the sum of the maximum completion times.  In other words, we will obtain the PDF and CDF for the overall completion time, which will enable us to answer questions such as:

• How likely is it that the project will be completed within 17 days?
• What’s the estimated time for which I can be 90% certain that the project will be completed? For brevity, I’ll call this the 90% completion time in the rest of this piece.

“OK, that sounds great”, you say, “but how exactly do we simulate a single task”?

Good question, and I was just about to get to that…

### Simulating a single task using the CDF

As we saw earlier, the CDF for the triangular has a S shape and ranges from 0 to 1 in value. It turns out that the S shape is characteristic of all CDFs, regardless of the details underlying PDF. Why? Because, the cumulative probability must lie between 0 and 1 (remember, probabilities can never exceed 1, nor can they be negative).

OK, so to simulate a task, we:

• generate a random number between 0 and 1, this corresponds to the probability that the task will finish at time t.
• find the time, t, that this corresponds to this value of probability. This is the completion time for the task for this trial.

Incidentally, this method is called inverse transform sampling.

An example might help clarify how inverse transform sampling works.  Assume that the random number generated is 0.4905. From the CDF for the first task, we see that this value of probability corresponds to a completion time of 4.503 days, which is the completion for this trial (see Figure 10). Simple!

Figure 10: Illustrating inverse transform sampling

In this case we found the time directly from the computed CDF. That’s not too convenient when you’re simulating the project 10,000 times. Instead, we need a programmable math expression that gives us the time corresponding to the probability directly. This can be obtained by solving equations (6) and (7) for $t$. Some straightforward algebra, yields the following two expressions for $t$:

$t = t_{min} + \sqrt{P(t)(t_{ml} - t_{min})(t_{max} - t_{min})} \ldots\ldots(8)$ for $t_{min}\leq t \leq t_{ml}$

And

$t = t_{max} - \sqrt{[1-P(t)](t_{max} - t_{ml})(t_{max} - t_{min})} \ldots\ldots(9)$ for $t_{ml}\leq t \leq t_{max}$

These can be easily combined in a single Excel formula using an IF function, and I’ll show you exactly how in a minute. Yes, we can now finally get down to the Excel simulation proper and you may want to download the workbook if you haven’t done so already.

### The simulation

Open up the workbook and focus on the first three columns of the first sheet to begin with. These simulate the first task in Figure 1, which also happens to be the task we have used to illustrate the construction of the triangular distribution as well as the mechanics of Monte Carlo.

Rows 2 to 4 in columns A and B list the min, most likely and max completion times while the same rows in column C list the probabilities associated with each of the times. For $t_{min}$ the probability is 0 and for $t_{max}$ it is 1.  The probability at $t_{ml}$ can be calculated using equation (6) which, for $t=t_{max}$, reduces to

$P(t_{ml}) =\frac{(t_{ml}-t_{min})}{t_{max}-t_{min}}\ldots\ldots(10)$

Rows 6 through 10005 in column A are simulated probabilities of completion for Task 1. These are obtained via the Excel RAND() function, which generates uniformly distributed random numbers lying between 0 and 1.  This gives us a list of probabilities corresponding to 10,000 independent simulations of Task 1.

The 10,000 probabilities need to be translated into completion times for the task. This is done using equations (8) or (9) depending on whether the simulated probability is less or greater than $P(t_{ml})$, which is in cell C3 (and given by Equation (10) above). The conditional statement can be coded in an Excel formula using the IF() function.

Tasks 2-4 are coded in exactly the same way, with distribution parameters in rows 2 through 4 and simulation details in rows 6 through 10005 in the columns listed below:

• Task 2 – probabilities in column D; times in column F
• Task 3 – probabilities in column H; times in column I
• Task 4 – probabilities in column K; times in column L

That’s basically it for the simulation of individual tasks. Now let’s see how to combine them.

For tasks in series (Tasks 1 and 2), we simply sum the completion times for each task to get the overall completion times for the two tasks.  This is what’s shown in rows 6 through 10005 of column G.

For tasks in parallel (Tasks 3 and 4), the overall completion time is the maximum of the completion times for the two tasks. This is computed and stored in rows 6 through 10005 of column N.

Finally, the overall project completion time for each simulation is then simply the sum of columns G and N (shown in column O)

Sheets 2 and 3 are plots of the probability and cumulative probability distributions for overall project completion times. I’ll cover these in the next section.

### Discussion – probabilities and estimates

The figure on Sheet 2 of the Excel workbook (reproduced in Figure 11 below) is the probability distribution function (PDF) of completion times. The x-axis shows the elapsed time in days and the y-axis the number of Monte Carlo trials that have a completion time that lie in the relevant time bin (of width 0.5 days). As an example, for the simulation shown in the Figure 11, there were 882 trials (out of 10,000) that had a completion time that lie between 16.25 and 16.75 days. Your numbers will vary, of course, but you should have a maximum in the 16 to 17 day range and a trial number that is reasonably close to the one I got.

Figure 11: Probability distribution of completion times (N=10,000)

I’ll say a bit more about Figure 11 in the next section. For now, let’s move on to Sheet 3 of workbook which shows the cumulative probability of completion by a particular day (Figure 12 below).  The figure shows the cumulative probability function (CDF), which is the sum of all completion times from the earliest possible completion day to the particular day.

Figure 12: Probability of completion by a particular day (N=10,000)

To reiterate a point made earlier,  the reason we work with the CDF  rather than the PDF is that we are interested in knowing the probability of completion by a particular date (e.g. it is 90% likely that we will finish by April 20th) rather than the probability of completion on a particular date (e.g. there’s a 10% chance we’ll finish on April 17th). We can now answer the two questions we posed earlier. As a reminder, they are:

• How likely is it that the project will be completed within 17 days?
• What’s the 90% likely completion time?

Both questions are easily answered by using the cumulative distribution chart on Sheet 3 (or Fig 12).  Reading the relevant numbers from the chart, I see that:

• There’s a 60% chance that the project will be completed in 17 days.
• The 90% likely completion time is 19.5 days.

How does the latter compare to the sum of the 90% likely completion times for the individual tasks? The 90% likely completion time for a given task can be calculated by solving Equation 9 for $t$, with appropriate values for the parameters $t_{min}$, $t_{max}$ and $t_{ml}$ plugged in, and $P(t)$ set to 0.9. This gives the following values for the 90% likely completion times:

• Task 1 – 6.5 days
• Task 2 – 8.1 days
• Task 3 – 7.7 days
• Task 4 – 5.8 days

Summing up the first three tasks (remember, Tasks 3 and 4 are in parallel) we get a total of 22.3 days, which is clearly an overestimation. Now, with the benefit of having gone through the simulation, it is easy to see that the sum of 90% likely completion times for individual tasks does not equal the 90% likely completion time for the sum of the relevant individual tasks – the first three tasks in this particular case. Why? Essentially because a Monte Carlo run in which the first three tasks tasks take as long as their (individual) 90% likely completion times is highly unlikely. Exercise:  use the worksheet to estimate how likely this is.

There’s much more that can be learnt from the CDF. For example, it also tells us that the greatest uncertainty in the estimate is in the 5 day period from ~14 to 19 days because that’s the region in which the probability changes most rapidly as a function of elapsed time. Of course, the exact numbers are dependent on the assumed form of the distribution. I’ll say more about this in the final section.

To close this section, I’d like to reprise a point I mentioned earlier: that uncertainty is a shape, not a number. Monte Carlo simulations make the uncertainty in estimates explicit and can help you frame your estimates in the language of probability…and using a tool like Excel can help you explain these to non-technical people like your manager.

### Closing remarks

We’ve covered a fair bit of ground: starting from general observations about how long a task takes, saw how to construct simple probability distributions and then combine these using Monte Carlo simulation.  Before I close, there are a few general points I should mention for completeness…and as warning.

First up, it should be clear that the estimates one obtains from a simulation depend critically on the form and parameters of the distribution used. The parameters are essentially an empirical matter; they should be determined using historical data. The form of the function, is another matter altogether: as pointed out in an earlier section, one cannot determine the shape of a function from a finite number of data points. Instead, one has to focus on the properties that are important. For example, is there a small but finite chance that a task can take an unreasonably long time? If so, you may want to use a lognormal distribution…but remember, you will need to find a sensible way to estimate the distribution parameters from your historical data.

Second, you may have noted from the probability distribution curve (Figure 11)  that despite the skewed distributions of the individual tasks, the distribution of the overall completion time is somewhat symmetric with a minimum of ~9 days, most likely time of ~16 days and maximum of 24 days.  It turns out that this is a general property of distributions that are generated by adding a large number of independent probabilistic variables. As the number of variables increases, the overall distribution will tend to the ubiquitous Normal distribution.

The assumption of independence merits a closer look.  In the case it hand,  it implies that the completion times for each task are independent of each other. As most project managers will know from experience, this is rarely the case: in real life,  a task that is delayed will usually have knock-on effects on subsequent tasks. One can easily incorporate such dependencies in a Monte Carlo simulation. A formal way to do this is to introduce a non-zero correlation coefficient between tasks as I have done here. A simpler and more realistic approach is to introduce conditional inter-task dependencies As an example, one could have an inter-task delay that kicks in only if the predecessor task takes more than 80%  of its maximum time.

Thirdly, you may have wondered why I used 10,000 trials: why not 100, or 1000 or 20,000. This has to do with the tricky issue of convergence. In a nutshell, the estimates we obtain should not depend on the number of trials used.  Why? Because if they did, they’d be meaningless.

Operationally, convergence means that any predicted quantity based on aggregates should not vary with number of trials.  So, if our Monte Carlo simulation has converged, our prediction of 19.5 days for the 90% likely completion time should not change substantially if I increase the number of trials from ten to twenty thousand. I did this and obtained almost the same value of 19.5 days. The average and median completion times (shown in cell Q3 and Q4 of Sheet 1) also remained much the same (16.8 days). If you wish to repeat the calculation, be sure to change the formulas on all three sheets appropriately. I was lazy and hardcoded the number of trials. Sorry!

Finally, I should mention that simulations can be usefully performed at a higher level than individual tasks. In their highly-readable book,  Waltzing With Bears: Managing Risk on Software Projects, Tom De Marco and Timothy Lister show how Monte Carlo methods can be used for variables such as  velocity, time, cost etc.  at the project level as opposed to the task level. I believe it is better to perform simulations at the lowest possible level, the main reason being that it is easier, and less error-prone, to estimate individual tasks than entire projects. Nevertheless, high level simulations can be very useful if one has reliable data to base these on.

There are a few more things I could say about the usefulness of the generated distribution functions and Monte Carlo in general, but they are best relegated to a future article. This one is much too long already and I think I’ve tested your patience enough. Thanks so much for reading, I really do appreciate it and hope that you found it useful.

Acknowledgement: My thanks to Peter Holberton for pointing out a few typographical and coding errors in an earlier version of this article. These have now been fixed. I’d be grateful if readers could bring any errors they find to my attention.

Written by K

March 27, 2018 at 4:11 pm

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## A gentle introduction to logistic regression and lasso regularisation using R

In this day and age of artificial intelligence and deep learning, it is easy to forget that simple algorithms can work well for a surprisingly large range of practical business problems.  And the simplest place to start is with the granddaddy of data science algorithms: linear regression and its close cousin, logistic regression. Indeed, in his acclaimed MOOC and accompanying textbook, Yaser Abu-Mostafa spends a good portion of his time talking about linear methods, and with good reason too: linear methods are not only a good way to learn the key principles of machine learning, they can also be remarkably helpful in zeroing in on the most important predictors.

My main aim in this post is to provide a beginner level introduction to logistic regression using R and also introduce LASSO (Least Absolute Shrinkage and Selection Operator), a powerful feature selection technique that is very useful for regression problems. Lasso is essentially a regularization method. If you’re unfamiliar with the term, think of it as a way to reduce overfitting using less complicated functions (and if that means nothing to you, check out my prelude to machine learning).  One way to do this is to toss out less important variables, after checking that they aren’t important.  As we’ll discuss later, this can be done manually by examining p-values of coefficients and discarding those variables whose coefficients are not significant. However, this can become tedious for classification problems with many independent variables.  In such situations, lasso offers a neat way to model the dependent variable while automagically selecting significant variables by shrinking the coefficients of unimportant predictors to zero.  All this without having to mess around with p-values or obscure information criteria. How good is that?

### Why not linear regression?

In linear regression one attempts to model a dependent variable (i.e. the one being predicted) using the best straight line fit to a set of predictor variables.  The best fit is usually taken to be one that minimises the root mean square error,  which is the sum of square of the differences between the actual and predicted values of the dependent variable. One can think of logistic regression as the equivalent of linear regression for a classification problem.  In what follows we’ll look at binary classification – i.e. a situation where the dependent variable takes on one of two possible values (Yes/No, True/False, 0/1 etc.).

First up, you might be wondering why one can’t use linear regression for such problems. The main reason is that classification problems are about determining class membership rather than predicting variable values, and linear regression is more naturally suited to the latter than the former. One could, in principle, use linear regression for situations where there is a natural ordering of categories like High, Medium and Low for example. However, one then has to map sub-ranges of the predicted values to categories. Moreover, since predicted values are potentially unbounded (in data as yet unseen) there remains a degree of arbitrariness associated with such a mapping.

Logistic regression sidesteps the aforementioned issues by modelling class probabilities instead.  Any input to the model yields a number lying between 0 and 1, representing the probability of class membership. One is still left with the problem of determining the threshold probability, i.e. the probability at which the category flips from one to the other.  By default this is set to p=0.5, but in reality it should be settled based on how the model will be used.  For example, for a marketing model that identifies potentially responsive customers, the threshold for a positive event might be set low (much less than 0.5) because the client does not really care about mailouts going to a non-responsive customer (the negative event). Indeed they may be more than OK with it as there’s always a chance – however small – that a non-responsive customer will actually respond.  As an opposing example, the cost of a false positive would be high in a machine learning application that grants access to sensitive information. In this case, one might want to set the threshold probability to a value closer to 1, say 0.9 or even higher. The point is, the setting an appropriate threshold probability is a business issue, not a technical one.

### Logistic regression in brief

So how does logistic regression work?

For the discussion let’s assume that the outcome (predicted variable) and predictors are denoted by Y and X respectively and the two classes of interest are denoted by + and – respectively.  We wish to model the conditional probability that the outcome Y is +, given that the input variables (predictors) are X. The conditional probability is denoted by p(Y=+|X)   which we’ll abbreviate as p(X) since we know we are referring to the positive outcome Y=+.

As mentioned earlier, we are after the probability of class membership so we must ensure that the hypothesis function (a fancy word for the model) always lies between 0 and 1. The function assumed in logistic regression is:

$p(X) = \dfrac{\exp^{\beta_0+\beta_1 X}}{1+\exp^{\beta_0 + \beta_1 X}} .....(1)$

You can verify that $p(X)$ does indeed lie between 0 and  1 as $X$ varies from $-\infty$ to $\infty$.  Typically, however, the values of $X$ that make sense are bounded as shown in the example (stolen from Wikipedia) shown in Figure 1. The figure also illustrates the typical S-shaped  curve characteristic of logistic regression.

Figure 1: Logistic function

As an aside, you might be wondering where the name logistic comes from. An equivalent way of expressing the above equation is:

$\log(\dfrac{p(X)}{1-p(X)}) = \beta_0+\beta_1 X .....(2)$

The quantity on the left is the logarithm of the odds. So, the model is a linear regression of the log-odds, sometimes called logit, and hence the name logistic.

The problem is to find the values of $\beta_0$  and $\beta_1$ that results in a $p(X)$ that most accurately classifies all the observed data points – that is, those that belong to the positive class have a probability as close as possible to 1 and those that belong to the negative class have a probability as close as possible to 0. One way to frame this problem is to say that we wish to maximise the product of these probabilities, often referred to as the likelihood:

$\displaystyle\log ( {\prod_{i:Y_i=+} p(X_{i}) \prod_{j:Y_j=-}(1-p(X_{j}))})$

Where $\prod$ represents the products over i and j, which run over the +ve and –ve classed points respectively. This approach, called maximum likelihood estimation, is quite common in many machine learning settings, especially those involving probabilities.

It should be noted that in practice one works with the log likelihood because it is easier to work with mathematically. Moreover, one minimises the negative  log likelihood which, of course, is the same as maximising the log likelihood.  The quantity one minimises is thus:

$L = - \displaystyle\log ( {\prod_{i:Y_i=+} p(X_{i}) \prod_{j:Y_j=-}(1-p(X_{j}))}).....(3)$

However, these are technical details that I mention only for completeness. As you will see next, they have little bearing on the practical use of logistic regression.

### Logistic regression in R – an example

In this example, we’ll use the logistic regression option implemented within the glm function that comes with the base R installation. This function fits a class of models collectively known as generalized linear models. We’ll apply the function to the Pima Indian Diabetes dataset that comes with the mlbench package. The code is quite straightforward – particularly if you’ve read earlier articles in my “gentle introduction” series – so I’ll just list the code below  noting that the logistic regression option is invoked by setting family=”binomial”  in the glm function call.

Here we go:

#set working directory if needed (modify path as needed)
#setwd(“C:/Users/Kailash/Documents/logistic”)
library(mlbench)
data(“PimaIndiansDiabetes”)
#set seed to ensure reproducible results
set.seed(42)
#split into training and test sets
PimaIndiansDiabetes[,”train”] <- ifelse(runif(nrow(PimaIndiansDiabetes))<0.8,1,0)
#separate training and test sets
trainset <- PimaIndiansDiabetes[PimaIndiansDiabetes$train==1,] testset <- PimaIndiansDiabetes[PimaIndiansDiabetes$train==0,]
#get column index of train flag
trainColNum <- grep(“train”,names(trainset))
#remove train flag column from train and test sets
trainset <- trainset[,-trainColNum]
testset <- testset[,-trainColNum]
#get column index of predicted variable in dataset
typeColNum <- grep(“diabetes”,names(PimaIndiansDiabetes))
#build model
glm_model <- glm(diabetes~.,data = trainset, family = binomial)
summary(glm_model)
Call:
glm(formula = diabetes ~ ., family = binomial, data = trainset)
<<output edited>>
Coefficients:
Estimate  Std. Error z value Pr(>|z|)
(Intercept)-8.1485021 0.7835869 -10.399  < 2e-16 ***
pregnant    0.1200493 0.0355617   3.376  0.000736 ***
glucose     0.0348440 0.0040744   8.552  < 2e-16 ***
pressure   -0.0118977 0.0057685  -2.063  0.039158 *
triceps     0.0053380 0.0076523   0.698  0.485449
insulin    -0.0010892 0.0009789  -1.113  0.265872
mass        0.0775352 0.0161255   4.808  1.52e-06 ***
pedigree    1.2143139 0.3368454   3.605  0.000312 ***
age         0.0117270 0.0103418   1.134  0.256816
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#predict probabilities on testset
#type=”response” gives probabilities, type=”class” gives class
glm_prob <- predict.glm(glm_model,testset[,-typeColNum],type=”response”)
#which classes do these probabilities refer to? What are 1 and 0?
contrasts(PimaIndiansDiabetes$diabetes) pos neg 0 pos 1 #make predictions ##…first create vector to hold predictions (we know 0 refers to neg now) glm_predict <- rep(“neg”,nrow(testset)) glm_predict[glm_prob>.5] <- “pos” #confusion matrix table(pred=glm_predict,true=testset$diabetes)
glm_predict  neg pos
neg    90 22
pos     8 33
#accuracy
mean(glm_predict==testset$diabetes) [1] 0.8039216 Although this seems pretty good, we aren’t quite done because there is an issue that is lurking under the hood. To see this, let’s examine the information output from the model summary, in particular the coefficient estimates (i.e. estimates for $\beta$) and their significance. Here’s a summary of the information contained in the table: • Column 2 in the table lists coefficient estimates. • Column 3 list s the standard error of the estimates (the larger the standard error, the less confident we are about the estimate) • Column 4 the z statistic (which is the coefficient estimate (column 2) divided by the standard error of the estimate (column 3)) and • The last column (Pr(>|z|) lists the p-value, which is the probability of getting the listed estimate assuming the predictor has no effect. In essence, the smaller the p-value, the more significant the estimate is likely to be. From the table we can conclude that only 4 predictors are significant – pregnant, glucose, mass and pedigree (and possibly a fifth – pressure). The other variables have little predictive power and worse, may contribute to overfitting. They should, therefore, be eliminated and we’ll do that in a minute. However, there’s an important point to note before we do so… In this case we have only 9 variables, so are able to identify the significant ones by a manual inspection of p-values. As you can well imagine, such a process will quickly become tedious as the number of predictors increases. Wouldn’t it be be nice if there were an algorithm that could somehow automatically shrink the coefficients of these variables or (better!) set them to zero altogether? It turns out that this is precisely what lasso and its close cousin, ridge regression, do. ### Ridge and Lasso Recall that the values of the logistic regression coefficients $\beta_0$ and $\beta_1$ are found by minimising the negative log likelihood described in equation (3). Ridge and lasso regularization work by adding a penalty term to the log likelihood function. In the case of ridge regression, the penalty term is $\beta_1^2$ and in the case of lasso, it is $|\beta_1|$ (Remember, $\beta_1$ is a vector, with as many components as there are predictors). The quantity to be minimised in the two cases is thus: $L +\lambda \sum \beta_1^2.....(4)$ – for ridge regression, and $L +\lambda \sum |\beta_1|.....(5)$ – for lasso regression. Where $\lambda$ is a free parameter which is usually selected in such a way that the resulting model minimises the out of sample error. Typically, the optimal value of $\lambda$ is found using grid search with cross-validation, a process akin to the one described in my discussion on cost-complexity parameter estimation in decision trees. Most canned algorithms provide methods to do this; the one we’ll use in the next section is no exception. In the case of ridge regression, the effect of the penalty term is to shrink the coefficients that contribute most to the error. Put another way, it reduces the magnitude of the coefficients that contribute to increasing $L$. In contrast, in the case of lasso regression, the effect of the penalty term is to set the these coefficients exactly to zero! This is cool because what it mean that lasso regression works like a feature selector that picks out the most important coefficients, i.e. those that are most predictive (and have the lowest p-values). Let’s illustrate this through an example. We’ll use the glmnet package which implements a combined version of ridge and lasso (called elastic net). Instead of minimising (4) or (5) above, glmnet minimises: $L +\lambda[ (1-\alpha)\sum [\beta_1^2 + \alpha\sum|\beta_1|]....(6)$ where $\alpha$ controls the “mix” of ridge and lasso regularisation, with $\alpha=0$ being “pure” ridge and $\alpha=1$ being “pure” lasso. ### Lasso regularisation using glmnet Let’s reanalyse the Pima Indian Diabetes dataset using glmnet with $\alpha=1$ (pure lasso). Before diving into code, it is worth noting that glmnet: • does not have a formula interface, so one has to input the predictors as a matrix and the class labels as a vector. • does not accept categorical predictors, so one has to convert these to numeric values before passing them to glmnet. The glmnet function model.matrix creates the matrix and also converts categorical predictors to appropriate dummy variables. Another important point to note is that we’ll use the function cv.glmnet, which automatically performs a grid search to find the optimal value of $\lambda$. OK, enough said, here we go: #load required library library(glmnet) #convert training data to matrix format x <- model.matrix(diabetes~.,trainset) #convert class to numerical variable y <- ifelse(trainset$diabetes==”pos”,1,0)
#perform grid search to find optimal value of lambda
#family= binomial => logistic regression, alpha=1 => lasso
# check docs to explore other type.measure options
cv.out <- cv.glmnet(x,y,alpha=1,family=”binomial”,type.measure = “mse” )
#plot result
plot(cv.out)

The plot is shown in Figure 2 below:

Figure 2: Error as a function of lambda (select lambda that minimises error)

The plot shows that the log of the optimal value of lambda (i.e. the one that minimises the root mean square error) is approximately -5. The exact value can be viewed by examining the variable lambda_min in the code below. In general though, the objective of regularisation is to balance accuracy and simplicity. In the present context, this means a model with the smallest number of coefficients that also gives a good accuracy.  To this end, the cv.glmnet function  finds the value of lambda that gives the simplest model but also lies within one standard error of the optimal value of lambda. This value of lambda (lambda.1se) is what we’ll use in the rest of the computation. Interested readers should have a look at this article for more on lambda.1se vs lambda.min.

#min value of lambda
lambda_min <- cv.out$lambda.min #best value of lambda lambda_1se <- cv.out$lambda.1se
#regression coefficients
coef(cv.out,s=lambda_1se)
10 x 1 sparse Matrix of class “dgCMatrix”
1
(Intercept) -4.61706681
(Intercept)  .
pregnant     0.03077434
glucose      0.02314107
pressure     .
triceps      .
insulin      .
mass         0.02779252
pedigree     0.20999511
age          .

The output shows that only those variables that we had determined to be significant on the basis of p-values have non-zero coefficients. The coefficients of all other variables have been set to zero by the algorithm! Lasso has reduced the complexity of the fitting function massively…and you are no doubt wondering what effect this  has on accuracy. Let’s see by running the model against our test data:

#get test data
x_test <- model.matrix(diabetes~.,testset)
#predict class, type=”class”
lasso_prob <- predict(cv.out,newx = x_test,s=lambda_1se,type=”response”)
#translate probabilities to predictions
lasso_predict <- rep(“neg”,nrow(testset))
lasso_predict[lasso_prob>.5] <- “pos”
#confusion matrix
table(pred=lasso_predict,true=testset$diabetes) pred neg pos neg 94 28 pos 4 27 #accuracy mean(lasso_predict==testset$diabetes)
[1] 0.7908497

Which is a bit less than what we got with the more complex model. So, we get  a similar out-of-sample accuracy as we did before, and we do so using a way simpler function (4 non-zero coefficients) than the original one (9  nonzero coefficients). What this means is that the simpler function does at least as good a job fitting the signal in the data as the more complicated one.  The bias-variance tradeoff tells us that the simpler function should be preferred because it is less likely to overfit the training data.

Paraphrasing William of Ockhamall other things being equal, a simple hypothesis should be preferred over a complex one.

### Wrapping up

In this post I have tried to provide a detailed introduction to logistic regression, one of the simplest (and oldest) classification techniques in the machine learning practitioners arsenal. Despite it’s simplicity (or I should say, because of it!) logistic regression works well for many business applications which often have a simple decision boundary. Moreover, because of its simplicity it is less prone to overfitting than flexible methods such as decision trees. Further, as we have shown, variables that contribute to overfitting can be eliminated using lasso (or ridge) regularisation, without compromising out-of-sample accuracy. Given these advantages and its inherent simplicity, it isn’t surprising that logistic regression remains a workhorse for data scientists.

Written by K

July 11, 2017 at 10:00 pm

## A gentle introduction to support vector machines using R

### Introduction

Most machine learning algorithms involve minimising an error measure of some kind (this measure is often called an objective function or loss function).  For example, the error measure in linear regression problems is the famous mean squared error – i.e. the averaged sum of the squared differences between the predicted and actual values. Like the mean squared error, most objective functions depend on all points in the training dataset.  In this post, I describe the support vector machine (SVM) approach which focuses instead on finding the optimal separation boundary between datapoints that have different classifications.  I’ll elaborate on what this means in the next section.

Here’s the plan in brief. I’ll begin with the rationale behind SVMs using a simple case of a binary (two class) dataset with a simple separation boundary (I’ll clarify what “simple” means in a minute).  Following that, I’ll describe how this can be generalised to datasets with more complex boundaries. Finally, I’ll work through a couple of examples in R, illustrating the principles behind SVMs. In line with the general philosophy of my “Gentle Introduction to Data Science Using R” series, the focus is on developing an intuitive understanding of the algorithm along with a practical demonstration of its use through a toy example.

### The rationale

The basic idea behind SVMs is best illustrated by considering a simple case:  a set of data points that belong to one of two classes, red and blue, as illustrated in figure 1 below. To make things simpler still, I have assumed that the boundary separating the two classes is a straight line, represented by the solid green line in the diagram.  In the technical literature, such datasets are called linearly separable.

Figure 1: Linearly separable data

In the linearly separable case, there is usually a fair amount of freedom in the way a separating line can be drawn. Figure 2 illustrates this point: the two broken green lines are also valid separation boundaries. Indeed, because there is a non-zero distance between the two closest points between categories, there are an infinite number of possible separation lines. This, quite naturally, raises the question as to whether it is possible to choose a separation boundary that is optimal.

Figure 2: Illustrating multiple separation boundaries

The short answer is, yes there is. One way to do this is to select a boundary line that maximises the margin, i.e. the distance between the separation boundary and the points that are closest to it.  Such an optimal boundary is illustrated by the black brace in Figure 3.  The really cool thing about this criterion is that the location of the separation boundary depends only on the points that are closest to it. This means, unlike other classification methods, the classifier does not depend on any other points in dataset. The directed lines between the boundary and the closest points on either side are called support vectors (these are the solid black lines in figure 3). A direct implication of this is that the fewer the support vectors, the better the generalizability of the boundary.

Figure 3: Optimal separation boundary in linearly separable case

Although the above sounds great, it is of limited practical value because real data sets are seldom (if ever) linearly separable.

So, what can we do when dealing with real (i.e. non linearly separable) data sets?

A simple approach to tackle small deviations from linear separability is to allow a small number of points (those that are close to the boundary) to be misclassified.  The number of possible misclassifications is governed by a free parameter C, which is called the cost.  The cost is essentially the penalty associated with making an error: the higher the value of C, the less likely it is that the algorithm will misclassify a point.

This approach – which is called soft margin classification – is illustrated in Figure 4. Note the points on the wrong side of the separation boundary.  We will demonstrate soft margin SVMs in the next section.  (Note:  At the risk of belabouring the obvious, the purely linearly separable case discussed in the previous para is simply is a special case of the soft margin classifier.)

Figure 4: Soft margin classifier (linearly separable data)

Real life situations are much more complex and cannot be dealt with using soft margin classifiers. For example, as shown in Figure 5, one could have widely separated clusters of points that belong to the same classes. Such situations, which require the use of multiple (and nonlinear) boundaries, can sometimes be dealt with using a clever approach called the kernel trick.

Figure 5: Non-linearly separable data

### The kernel trick

Recall that in the linearly separable (or soft margin) case, the SVM algorithm works by finding a separation boundary that maximises the margin, which is the distance between the boundary and the points closest to it. The distance here is the usual straight line distance between the boundary and the closest point(s). This is called the Euclidean distance in honour of the great geometer of antiquity. The point to note is that this process results in a separation boundary that is a straight line, which as Figure 5 illustrates, does not always work. In fact in most cases it won’t.

So what can we do? To answer this question, we have to take a bit of a detour…

What if we were able to generalize the notion of distance in a way that generates nonlinear separation boundaries? It turns out that this is possible. To see how, one has to first understand how the notion of distance can be generalized.

The key properties that any measure of distance must satisfy are:

1. Non-negativity – a distance cannot be negative, a point that needs no further explanation I reckon 🙂
2. Symmetry – that is, the distance between point A and point B is the same as the distance between point B and point A.
3. Identity– the distance between a point and itself is zero.
4. Triangle inequality – that is the sum of distances between point A and B and points B and C must be less than or equal to the distance between A and C (equality holds only if all three points lie along the same line).

Any mathematical object that displays the above properties is akin to a distance. Such generalized distances are called metrics and the mathematical space in which they live is called a metric space. Metrics are defined using special mathematical functions designed to satisfy the above conditions. These functions are known as kernels.

The essence of the kernel trick lies in mapping the classification problem to a  metric space in which the problem is rendered separable via a separation boundary that is simple in the new space, but complex – as it has to be – in the original one. Generally, the transformed space has a higher dimensionality, with each of the dimensions being (possibly complex) combinations of the original problem variables. However, this is not necessarily a problem because in practice one doesn’t actually mess around with transformations, one just tries different kernels (the transformation being implicit in the kernel) and sees which one does the job. The check is simple: we simply test the predictions resulting from using different kernels against a held out subset of the data (as one would for any machine learning algorithm).

It turns out that a particular function – called the radial basis function kernel  (RBF kernel) – is very effective in many cases.  The RBF kernel is essentially a Gaussian (or Normal) function with the Euclidean distance between pairs of points as the variable (see equation 1 below).   The basic rationale behind the RBF kernel is that it creates separation boundaries that it tends to classify points close together (in the Euclidean sense) in the original space in the same way. This is reflected in the fact that the kernel decays (i.e. drops off to zero) as the Euclidean distance between points increases.

$\exp (-\gamma |\mathbf{x-y}|)....(1)$

The rate at which a kernel decays is governed by the parameter $\gamma$ – the higher the value of $\gamma$, the more rapid the decay.  This serves to illustrate that the RBF kernel is extremely flexible….but the flexibility comes at a price – the danger of overfitting for large values of $\gamma$ .  One should choose appropriate values of C and $\gamma$ so as to ensure that the resulting kernel represents the best possible balance between flexibility and accuracy. We’ll discuss how this is done in practice later in this article.

Finally, though it is probably obvious, it is worth mentioning that the separation boundaries for arbitrary kernels are also defined through support vectors as in Figure 3.  To reiterate a point made earlier, this means that a solution that has fewer support vectors is likely to be more robust than one with many. Why? Because the data points defining support vectors are ones that are most sensitive to noise- therefore the fewer, the better.

There are many other types of kernels, each with their own pros and cons. However, I’ll leave these for adventurous readers to explore by themselves.  Finally, for a much more detailed….and dare I say, better… explanation of the kernel trick, I highly recommend this article by Eric Kim.

### Support vector machines in R

In this demo we’ll use the svm interface that is implemented in the e1071 R package. This interface provides R programmers access to the comprehensive libsvm library written by Chang and Lin. I’ll use two toy datasets: the famous iris dataset available with the base R package and the sonar dataset from the mlbench package. I won’t describe details of the datasets as they are discussed at length in the documentation that I have linked to. However, it is worth mentioning the reasons why I chose these datasets:

1. As mentioned earlier, no real life dataset is linearly separable, but the iris dataset is almost so. Consequently, it is a good illustration of using linear SVMs. Although one almost never uses these in practice, I have illustrated their use primarily for pedagogical reasons.
2. The sonar dataset is a good illustration of the benefits of using RBF kernels in cases where the dataset is hard to visualise (60 variables in this case!). In general, one would almost always use RBF (or other nonlinear) kernels in practice.

With that said, let’s get right to it. I assume you have R and RStudio installed. For instructions on how to do this, have a look at the first article in this series. The processing preliminaries – loading libraries, data and creating training and test datasets are much the same as in my previous articles so I won’t dwell on these here. For completeness, however, I’ll list all the code so you can run it directly in R or R studio (a complete listing of the code can be found here):

#set working directory if needed (modify path as needed)
setwd(“C:/Users/Kailash/Documents/svm”)
library(e1071)
data(iris)
#set seed to ensure reproducible results
set.seed(42)
#split into training and test sets
iris[,”train”] <- ifelse(runif(nrow(iris))<0.8,1,0)
#separate training and test sets
trainset <- iris[iris$train==1,] testset <- iris[iris$train==0,]
#get column index of train flag
trainColNum <- grep("train",names(trainset))
#remove train flag column from train and test sets
trainset <- trainset[,-trainColNum]
testset <- testset[,-trainColNum]
#get column index of predicted variable in dataset
typeColNum <- grep("Species",names(iris))
#build model – linear kernel and C-classification (soft margin) with default cost (C=1)
svm_model <- svm(Species~ ., data=trainset, method="C-classification", kernel="linear")
svm_model
Call:
svm(formula = Species ~ ., data = trainset, method = “C-classification”, kernel = “linear”)
Parameters:
SVM-Type: C-classification
SVM-Kernel: linear
cost: 1
gamma: 0.25
Number of Support Vectors: 24
#training set predictions
pred_train <-predict(svm_model,trainset)
mean(pred_train==trainset$Species) [1] 0.9826087 #test set predictions pred_test <-predict(svm_model,testset) mean(pred_test==testset$Species)
[1] 0.9142857

The output from the SVM model show that there are 24 support vectors. If desired, these can be examined using the SV variable in the model – i.e via svm_model$SV. The test prediction accuracy indicates that the linear performs quite well on this dataset, confirming that it is indeed near linearly separable. To check performance by class, one can create a confusion matrix as described in my post on random forests. I’ll leave this as an exercise for you. Another point is that we have used a soft-margin classification scheme with a cost C=1. You can experiment with this by explicitly changing the value of C. Again, I’ll leave this for you an exercise. Before proceeding to the RBF kernel, I should mention a point that an alert reader may have noticed. The predicted variable, Species, can take on 3 values (setosa, versicolor and virginica). However, our discussion above dealt with a binary (2 valued) classification problem. This brings up the question as to how the algorithm deals multiclass classification problems – i.e those involving datasets with more than two classes. The libsvm algorithm (which svm uses) does this using a one-against-one classification strategy. Here’s how it works: 1. Divide the dataset (assumed to have N classes) into N(N-1)/2 datasets that have two classes each. 2. Solve the binary classification problem for each of these subsets 3. Use a simple voting mechanism to assign a class to each data point. Basically, each data point is assigned the most frequent classification it receives from all the binary classification problems it figures in. With that said for the unrealistic linear classifier, let’s move to the real world. In the code below, I build SVM models using three different kernels 1. Linear kernel (this is for comparison with the following 2 kernels). 2. RBF kernel with default values for the parameters $C$ and $\gamma$. 3. RBF kernel with optimal values for $C$ and $\gamma$. The optimal values are obtained using the tune.svm function (also available in e1071), which essentially builds models for multiple combinations of parameter values and selects the best. OK, lets go: #load required library (assuming e1071 is already loaded) library(mlbench) #load Sonar dataset data(Sonar) #set seed to ensure reproducible results set.seed(42) #split into training and test sets Sonar[,”train”] <- ifelse(runif(nrow(Sonar))<0.8,1,0) #separate training and test sets trainset <- Sonar[Sonar$train==1,]
testset <- Sonar[Sonar$train==0,] #get column index of train flag trainColNum <- grep("train",names(trainset)) #remove train flag column from train and test sets trainset <- trainset[,-trainColNum] testset <- testset[,-trainColNum] #get column index of predicted variable in dataset typeColNum <- grep("Class",names(Sonar)) #build model – linear kernel and C-classification with default cost (C=1) svm_model <- svm(Class~ ., data=trainset, method="C-classification", kernel="linear") #training set predictions pred_train <-predict(svm_model,trainset) mean(pred_train==trainset$Class)
[1] 0.969697
#test set predictions
pred_test <-predict(svm_model,testset)
mean(pred_test==testset$Class) [1] 0.6046512 I’ll leave you to examine the contents of the model. The important point to note here is that the performance of the model with the test set is quite dismal compared to the previous case. This simply indicates that the linear kernel is not appropriate here. Let’s take a look at what happens if we use the RBF kernel with default values for the parameters: #build model: radial kernel, default params svm_model <- svm(Class~ ., data=trainset, method="C-classification", kernel="radial") #print params svm_model$cost
[1] 1
svm_model$gamma [1] 0.01666667 #training set predictions pred_train <-predict(svm_model,trainset) mean(pred_train==trainset$Class)
[1] 0.9878788
#test set predictions
pred_test <-predict(svm_model,testset)
mean(pred_test==testset$Class) [1] 0.7674419 That’s a pretty decent improvement from the linear kernel. Let’s see if we can do better by doing some parameter tuning. To do this we first invoke tune.svm and use the parameters it gives us in the call to svm: #find optimal parameters in a specified range tune_out <- tune.svm(x=trainset[,-typeColNum],y=trainset[,typeColNum],gamma=10^(-3:3),cost=c(0.01,0.1,1,10,100,1000),kernel="radial") #print best values of cost and gamma tune_out$best.parameters$cost [1] 10 tune_out$best.parameters$gamma [1] 0.01 #build model svm_model <- svm(Class~ ., data=trainset, method="C-classification", kernel="radial",cost=tune_out$best.parameters$cost,gamma=tune_out$best.parameters$gamma) #training set predictions pred_train <-predict(svm_model,trainset) mean(pred_train==trainset$Class)
[1] 1
#test set predictions
pred_test <-predict(svm_model,testset)
mean(pred_test==testset\$Class)
[1] 0.8139535

Which is fairly decent improvement on the un-optimised case.

### Wrapping up

This bring us to the end of this introductory exploration of SVMs in R. To recap, the distinguishing feature of SVMs in contrast to most other techniques is that they attempt to construct optimal separation boundaries between different categories.

SVMs  are quite versatile and have been applied to a wide variety of domains ranging from chemistry to pattern recognition. They are best used in binary classification scenarios. This brings up a question as to where SVMs are to be preferred to other binary classification techniques such as logistic regression. The honest response is, “it depends” – but here are some points to keep in mind when choosing between the two. A general point to keep in mind is that SVM  algorithms tend to be expensive both in terms of memory and computation, issues that can start to hurt as the size of the dataset increases.

Given all the above caveats and considerations, the best way  to figure out whether an SVM approach will work for your problem may be to do what most machine learning practitioners do: try it out!

Written by K

February 7, 2017 at 8:27 pm