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## Answers

There is a function to find the number of nonzero matrix elements nnz. You can use this function on a logical matrix, which will return the number of true.

In this case, we apply nnz on the matrix A==0, hence the elements of the logical matrix are true, if the original element was 0, false for any other element than 0.

A = [1, 3, 1;
0, 0, 2;
0, 2, 1];
nnz(A==0) %// returns 3, i.e. the number of zeros of A (the amount of true in A==0)

The credits for the benchmarking belong to Divarkar.

# Benchmarking

Using the following paramters and inputs, one can benchmark the solutions presented here with timeit.

Input sizes

- Small sized datasize - 1:10:100
- Medium sized datasize - 50:50:1000
- Large sized datasize - 500:500:4000

Varying % of zeros

- ~10% of zeros case - A = round(rand(N)*5);
- ~50% of zeros case - A = rand(N);A(A<=0.5)=0;
- ~90% of zeros case - A = rand(N);A(A<=0.9)=0;

The results are shown next -

**1) Small Datasizes**

**2. Medium Datasizes**

**3. Large Datasizes**

**Observations**

If you look closely into the NNZ and SUM performance plots for medium and large datasizes, you would notice that their performances get closer to each other for 10% and 90% zeros cases. For 50% zeros case, the performance gap between SUM and NNZ methods is comparatively wider.

As a general observation across all datasizes and all three fraction cases of zeros,
SUM method seems to be the undisputed winner. Again, an interesting thing was observed here that the general case solution sum(A(:)==0) seems to be better in performance than sum(~A(:)).

some basic matlab to know: the (:) operator will flatten any matrix into a column vector , ~ is the NOT operator flipping zeros to ones and non zero values to zero, then we just use sum:

sum(~A(:))

This should be also about **10** times faster than the length(find... scheme, in case efficiency is important.

**Edit:** in the case of NaN values you can resort to the solution:

sum(A(:)==0)

I'll add something to the mix as well. You can use histc and compute the histogram of the entire matrix. You specify the second parameter to be which bins the numbers should be collected at. If we just want to count the number of zeroes, we can simply specify 0 as the second parameter. However, if you specify a matrix into histc, it will operate along the columns but we want to operate on the entire matrix. As such, simply transform the matrix into a column vector A(:) and use histc. In other words, do this:

histc(A(:), 0)

This should be equivalent to counting the number of zeroes in the entire matrix A.

Well I don't know if I'm answering well the question but you could code it as follows :

% Random Matrix
M = [1 0 4 8 0 6;
0 0 7 4 8 0;
8 7 4 0 6 0];
n = size(M,1); % Number of lines of M
p = size(M,2); % Number of columns of M
nbrOfZeros = 0; % counter
for i=1:n
for j=1:p
if M(i,j) == 0
nbrOfZeros = nbrOfZeros + 1;
end
end
end
nbrOfZeros

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