I have a very big Matlab simulation project in my hands, which I wanted to optimize, since I'm running it many times to tune parameters and the like.

Using Matlab's profile I identified one function that is eating up most of my time, specifically the output(i,1) assignment line.

This function is called a LOT, where input is a 10x1 double passed as an argument, and output is also a 10x1 vector.

function output = my_function(input)

a = size(input,1);
output = input*0;
dens = density(input);

% for each i, output(i) is the maximum between output(i+1) and mean(output(i+1:end))
for i = 1:a-1
    output(i,1)= max(mean(dens(i+1:a,1)),dens(i+1,1));
output(a,1) = dens(a,1);


density() function:

function dens = density(input)

dens = 1000*(1 - (input+288.9414)./(508929.2.*(input+68.12963)).*(input-3.9863).^2);


My ideas:

  • I think vectorization would maybe help to get rid of the loop, but I'm not familiar at all with the technique.
  • Is there a faster/alternative way to calculate the mean (maybe without Matlab's built-in function call?)
  • \$\begingroup\$ I'm assuming you have your own density function defined? It isn't defined in the standard MatLab functions \$\endgroup\$
    – syb0rg
    Commented Dec 5, 2014 at 18:15
  • \$\begingroup\$ Yes, density is an external function I declared. It is not relevant for my question, as dens is also a 10x1 double vector \$\endgroup\$ Commented Dec 5, 2014 at 18:16
  • \$\begingroup\$ The line that eats up 90% of the resources of the script is output(i,1)= max(mean(dens(i+1:a,1)),dens(i+1,1));, thus the vectorization I mentioned \$\endgroup\$ Commented Dec 5, 2014 at 18:31
  • 1
    \$\begingroup\$ What density does matters because you might be able to calculate output(i,1) from values in output rather than taking a mean of values in dens. My offhand thought is that you are iterating in the wrong direction. Given what you are doing, you should iterate down not up and calculate the mean as you go. Someone like syb0rg with more knowledge of Matlab could probably give you better help if you offered a working example with all code. \$\endgroup\$
    – Brythan
    Commented Dec 5, 2014 at 21:17
  • \$\begingroup\$ @Brythan, @sybOrg I'm so sorry, I really thought it wasn't necessary. I added the density function in the last edit \$\endgroup\$ Commented Dec 5, 2014 at 22:14

3 Answers 3



If you look at all the calls to mean, you see that what you are calculating in each loop is a "Cumulative moving average", but in the oposite direction of what one usually would do.

A cumulative moving average can be calculated quite simply using the cumulative sum, cumsum, and dividing by the number of elements. Since you're starting with all elements and up with only one, we need to flip it.

The running mean, (or cumulative moving average) can be calculated like this:

  1. First flip dens upside down
  2. Take the cumulative sum of it
  3. Divide each element by the number of elements counted
  4. Flip it back up


running_mean = flipud(cumsum(flipud(dens))./(1:numel(dens)).');

Now, we need to find which element is larger, the cumulative sum, or the original density variable. You are not including the first element of the cumulative sum, and you are appending dens(end) at the end of the vector.

The maximum value of each row of a matrix can be found using max(A,[],2), where the number two represent the second dimension (columns). So, the values we want to use are:

max([running_mean(2:end), dens(2:end)],[],2)

In addition, we need to append the last element, so:

output = [max([running_mean(2:end), dens(2:end)],[],2); dens(end)];

To summarize, the entire operation requires no loops, one call to cumsum, and one call to max. In addition we need to flip it twice, but that takes virtually no time.

In total, your function can be written like this:

function output = my_function(vector)

n = numel(vector);    % numel, not length
output = zeros(n,1)   % prior to R2015b, use: output(n,1) = 0

% Create an anonymous function called density:
density = @(vector) 1000*(1 - (vector+288.9414)./(508929.2.*(vector+68.12963)).*(vector-3.9863).^2);

% Calculate the density:
dens = density(vector);

% Calculate the running mean, and create the final output vector:
running_mean = flipud(cumsum(flipud(dens))./(1:numel(dens)).');
output = [max([running_mean(2:end), dens(2:end)],[],2); dens(end)];


Time in Octave shows the improvement, as a function of vector size. For vectors with only ten elements, the execution time is 25% of the original code.

enter image description here

Review of your code:

  1. Don't use input as a variable name! input is used to ask users for a command line input. Using it as a variable name makes the function useless, but will also potentially create strange errors or warnings. (I called it vector in this review. You should pick a more descriptive name)

    The same goes with max, mean, length etc. Don't use function names as variable names.

  2. Your input is a vector, so no need to specify that you always want the first column. Use linear indexing instead of subscripts. It's much faster, and you won't mess up the dimensions (oops, it was a column vector, not a row vector).

  3. Don't use size(vector,1), and don't use length(vector), use numel(vector). It's faster and more robust.

  4. You are pre-allocating output. Very good! That's a common bottleneck.

  5. output = vector*0; is clever. Prior to Matlab R2015b zeros(n,1) was actually far from the fastest way to create a vector of zeros! In fact, output(n,1) = 0 was up to 500 times faster! This is because zeros creates a matrix, then fills the elements with zeros, where as the latter just creates a matrix, where the values are zero by default.

  6. In your loop, don't use i as the iterator, it is used for the imaginary unit (sqrt(-1)). It's common to use for instance ii, jj etc.

  7. If your density function is only what you have posted there, then you can simply create an anonymous function inside the main function:

    density = @(vector) 1000*(1 - (vector+288.9414)./(508929.2.*(vector+68.12963)).*(vector-3.9863).^2);

    Call it like:

    dens = density(vector);

Review of reviews:

There are a few things I don't like about the other reviews that I want to point out:

  1. "You don't need to specify that 1 when indexing a vector, MATLAB knows". Many Matlab users use max(A), when they want to find the maximum value of all columns in a matrix. It's much simpler than max(A, [], 1). You might however come into trouble if A has varying number of rows (and matrices often do). Because the moment that matrix has only one row, max(A) will give the maximum value of the entire vector, instead of each (one row) column. This can give you a major headache!

  2. "Why not use length?" Well, in this case, length is much better than size, but numel should still be the preferred choice. length is better than size for two reasons. Using size limits the function to only work on column vectors only, and will fail on row vectors. Second, numel is much faster than length for large vectors, and both are much faster than size for all vectors.

  3. "Don't use the profiler, use tic/toc". True, the profiler includes the time it takes to run itself, but it's still the easiest way to identify bottlenecks. For exact timing however, don't use tic/toc. timeit was introduced in Matlab R2013b, and is much better than tic/toc. If you are using an older version, the function is available at the File Exchange. tic/toc is also dangerous, because it will reset the timer if tic is called inside another function. Have a look at this question on SO from three years ago.

  • 1
    \$\begingroup\$ That's a major improvement, nice answer! \$\endgroup\$
    – Mast
    Commented Jul 12, 2016 at 14:06
  • 1
    \$\begingroup\$ Although it has been a long time since I touched this Matlab code, it is a very instructional and detailed answer. Very much appreciated! \$\endgroup\$ Commented Jul 12, 2016 at 14:08

A few notes:

  • I'm confused about the design of your function. If you are only processing column vectors, why not just process a row vector instead? That's what I would expect when calling this function.

     my_function([1:2:20]') % how I have to call your function; convoluted and unexpected
     my_function(1:2:20) % how I expect to call your function; cleaner
  • Vectorization is definitely the way to go for efficiency in MatLab. One way to vectorize code is to use the : operator, which it seems you know how to use.

    Unfortunately, we can not nest colon operators in Matlab, so I do not know of a way to vectorize your code. Break apart how the colon operator works and how your code works to see why

    % (i + 1):a where i is 1:(a - 1) [a is the length, 10 in this case]
    2 3 4 5 6 7 8 9 10 % 2:10
    3 4 5 6 7 8 9 10 % 3:10
    4 5 6 7 8 9 10 % 4:10
    5 6 7 8 9 10 % 5:10
    % and so on
    % (2:10):10
    2 3 4 5 6 7 8 9 10 % doesn't quite work... have to use a loop

    There may be some other way to accomplish what you are trying to do, but I can't think of a different and more efficient way to approach it

  • You should be putting brackets around the return value(s) in your function declaration. They can be omitted, but it's better practice to keep them there.

  • Your current way of getting the size of your vector isn't how I would go about it.

    a = size(input,1);

    You are always going to have one of the dimensions be of size 1, correct? So why not use the length() function instead, which always returns the greatest of the number of dimensions passed to it.

    a = length(input);
  • You have a clever way of getting a zeroed out vector of the same length.

    output = input*0;

    I'm not sure how the performance would compare, but using the function zeros() may be faster.

  • You don't need to specify that 1 when indexing a vector, MatLab knows.

    a = size(input,1);
    a = size(input); % almost the same, is the same for your purposes
  • Don't use MatLab's profiler to measure the performance of your code. Why? Because in measuring the code, it also happens to measure itself running the profiling tests and compiling the results, which can take up quite some time on some machines (including mine).

    The more accepted way to measure how long code takes to run is to use tic and toc, as such:

    >> tic; my_function([1:2:20]'); toc;
    Elapsed time is 0.000976 seconds.

    Make sure you put them on the same line though like I did above, or it will also count the time it takes you to type out the function and execute it, as well as typing out toc and executing that.

    >> tic;
    >> my_function([1:2:20]');
    >> toc;
    Elapsed time is 13.184153 seconds.
  • \$\begingroup\$ I've edited the question with a solution proposal. Not sure why the residual happens... \$\endgroup\$ Commented Dec 8, 2014 at 11:48

Warning: I don't have Matlab installed and haven't done much with it, so my syntax may be off.

% for each i, output(i) is the maximum between output(i+1) and mean(output(i+1:end))
for i = 1:a-1
    output(i,1)= max(mean(dens(i+1:a,1)),dens(i+1,1));
output(a,1) = dens(a,1);

If I understand that correctly, the % line is a comment. Then you have a for loop from 1 to a-1 where you assign all but the last entry of output. After the loop finishes, you assign the last entry of output.

The downside of this approach is that you have to sum up and divide the array for every entry. I think it would be better to do the end first and keep a running sum. I'm not sure of syntax, but I would expect that to look something like

count = 1;
sum = dens(a,1);
output(a,1) = dens(a,1);

for i = a-1:-1:2
    output(i,1) = max((sum / count), dens(i+1,1));
    sum = sum + dens(i,1);
    count = count + 1;

output(1,1) = max(sum / count), dens(2,1));

Again, apologies if my syntax doesn't work. Hopefully this illustrates what I'm trying to do.

I stuck with the syntax that you were using in case I didn't understand syb0rg's comments properly.

Note: if you do this, don't forget to re-profile. It may be that there is an optimization on calculating the mean that outstrips the theoretical advantage of this approach.

  • \$\begingroup\$ The main point that bugged me with the question was that the mean function is called all the time, and it could probably be computed beforehand, to make the main loop faster. Your approach is similar to what I had in mind. \$\endgroup\$ Commented Jul 15, 2015 at 20:34

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