### 13.7 Profiler Example

Below, we will give a short example of a profiler session. See also Profiling for the documentation of the profiler functions in detail. Consider the code:

         global N A;

         N = 300;
         A = rand (N, N);

         function xt = timesteps (steps, x0, expM)
           global N;

           if (steps == 0)
             xt = NA (N, 0);
           else
             xt = NA (N, steps);
             x1 = expM * x0;
             xt(:, 1) = x1;
             xt(:, 2 : end) = timesteps (steps - 1, x1, expM);
           endif
         endfunction

         function foo ()
           global N A;

           initial = @(x) sin (x);
           x0 = (initial (linspace (0, 2 * pi, N)))';

           expA = expm (A);
           xt = timesteps (100, x0, expA);
         endfunction

         function fib = bar (N)
           if (N <= 2)
             fib = 1;
           else
             fib = bar (N - 1) + bar (N - 2);
           endif
         endfunction

If we execute the two main functions, we get:

         tic; foo; toc;
         ⇒ Elapsed time is 2.37338 seconds.

         tic; bar (20); toc;
         ⇒ Elapsed time is 2.04952 seconds.

But this does not give much information about where this time is spent; for instance, whether the single call to `expm` is more expensive or the recursive time-stepping itself. To get a more detailed picture, we can use the profiler.

         profile on;
         foo;
         profile off;

         data = profile ('info');
         profshow (data, 10);

This prints a table like:

            #  Function Attr     Time (s)        Calls
         ---------------------------------------------
            7      expm             1.034            1
            3  binary *             0.823          117
           41  binary \             0.188            1
           38  binary ^             0.126            2
           43 timesteps    R        0.111          101
           44        NA             0.029          101
           39  binary +             0.024            8
           34      norm             0.011            1
           40  binary -             0.004          101
           33   balance             0.003            1

The entries are the individual functions which have been executed (only the 10 most important ones), together with some information for each of them. The entries like ‘`binary *`’ denote operators, while other entries are ordinary functions. They include both built-ins like `expm` and our own routines (for instance `timesteps`). From this profile, we can immediately deduce that `expm` uses up the largest proportion of the processing time, even though it is only called once. The second expensive operation is the matrix-vector product in the routine `timesteps`. 1

Timing, however, is not the only information available from the profile. The attribute column shows us that `timesteps` calls itself recursively. This may not be that remarkable in this example (since it's clear anyway), but could be helpful in a more complex setting. As to the question of why is there a ‘`binary \`’ in the output, we can easily shed some light on that too. Note that `data` is a structure array (Structure Arrays) which contains the field `FunctionTable`. This stores the raw data for the profile shown. The number in the first column of the table gives the index under which the shown function can be found there. Looking up `data.FunctionTable(41)` gives:

           scalar structure containing the fields:

             FunctionName = binary \
             TotalTime =  0.18765
             NumCalls =  1
             IsRecursive = 0
             Parents =  7
             Children = [](1x0)

Here we see the information from the table again, but have additional fields `Parents` and `Children`. Those are both arrays, which contain the indices of functions which have directly called the function in question (which is entry 7, `expm`, in this case) or been called by it (no functions). Hence, the backslash operator has been used internally by `expm`.

Now let's take a look at `bar`. For this, we start a fresh profiling session (`profile on` does this; the old data is removed before the profiler is restarted):

         profile on;
         bar (20);
         profile off;

         profshow (profile ('info'));

This gives:

            #            Function Attr     Time (s)        Calls
         -------------------------------------------------------
            1                 bar    R        2.091        13529
            2           binary <=             0.062        13529
            3            binary -             0.042        13528
            4            binary +             0.023         6764
            5             profile             0.000            1
            8               false             0.000            1
            6              nargin             0.000            1
            7           binary !=             0.000            1
            9 __profiler_enable__             0.000            1

Unsurprisingly, `bar` is also recursive. It has been called 13,529 times in the course of recursively calculating the Fibonacci number in a suboptimal way, and most of the time was spent in `bar` itself.

Finally, let's say we want to profile the execution of both `foo` and `bar` together. Since we already have the run-time data collected for `bar`, we can restart the profiler without clearing the existing data and collect the missing statistics about `foo`. This is done by:

         profile resume;
         foo;
         profile off;

         profshow (profile ('info'), 10);

As you can see in the table below, now we have both profiles mixed together.

            #  Function Attr     Time (s)        Calls
         ---------------------------------------------
            1       bar    R        2.091        13529
           16      expm             1.122            1
           12  binary *             0.798          117
           46  binary \             0.185            1
           45  binary ^             0.124            2
           48 timesteps    R        0.115          101
            2 binary <=             0.062        13529
            3  binary -             0.045        13629
            4  binary +             0.041         6772
           49        NA             0.036          101

#### Footnotes

[1] We only know it is the binary multiplication operator, but fortunately this operator appears only at one place in the code and thus we know which occurrence takes so much time. If there were multiple places, we would have to use the hierarchical profile to find out the exact place which uses up the time which is not covered in this example.