How do I count the number of zeros in a matrix?

I have data imported from excel into a matrix and want to count the number of zeros in the file to work out the time.

Answer 1

[Azzi Abdelmalek]

A=[0 0 1;0 2 4;7 8 1]
idx=A==0
out=sum(idx(:))

or

out=nnz(~A)

[AMAL ALGEDIR]

Thank you very much. your answer always useful and simple.

[rehan anwar]

it will not works when matrice is [0 90 180]

[Walter Roberson]

Azzi's code suggestions both output 1 for [0 90 180], which would appear to be the correct answer; what result were you expecting?

Answer 2

[Mischa Kim]

Caroline, use

 numberOfZeros = sum(data(:)==0)

[Elyasaf Elbaz]

This is perfect!

[Gabriela Garcia]

Can this be modified so that it searches for the number of zeros in a specific column of the matrix? I want to see the number of zeros each column has.

[Walter Roberson]

sum(data(:,COLUMN)==0)

If you want to know column-by-column, then

sum(data == 0, 1)

or most compactly (but less clear)

sum(~data,1)

If you are sure that there are more than one rows, then

sum(~data)

is even more compact.

Answer 3

[Image Analyst]

The way I thought of is different still:

numberOfZeros = numel(data) - nnz(data);

Now how that allows you to "to work out the time", I have no idea. What do you mean by that? Is one of the rows or columns a time or date number?

[Franchesca]

I will end up calculating jump height however to do this I need to calculate the time each participant was off the force plate (so when the reading is 0) in one of the columns. How would I specify a column for matlab to count the number of zeros. Would it simply be:

numberOfZeros = numel(data(7,1)) - nnz(data);

[Image Analyst]

Nope. That's not a good idea. You need to count the number of zero stretches, not the number of zeros. Good thing you explained the larger context so we can see that what you asked for is probably not what you need. What if a person jumped 3 times: she jumped high the first time so that there were 3 zeros the first jump, 2 zeros at the second jump, and just a single zero for the third jump? Well that's 3 jumps, not 6 like you'd get if you counted zeros.

That's why it's important to give the context at the start so people don't give you wrong answers, like we all did up til now. Often a person asks very specifically "How do I do X?" and then later it becomes known that the person wanted outcome Z and thought X was the best approach. Then people realize that approach X is not the correct or best approach, and the poster should use approach Y. I think that's the situation here.

If you have the Image Processing Toolbox, you can simply count the number of jumps this way:

[labeledRegions, numberOfJumps] = bwlabel(data == 0);

If you don't have that toolbox you can count the number of "jump starts" by taking the difference and counting how many are more than 0.

diffs = diff(data); % data(k+1)-data(k);
jumpStarts = (diffs > 0) & data(1:end-1) == 0;
numberOfJumps = sum(jumpStarts);

or something like that.

Answer 4

[palaniraj p]

x=[ 1 4 0; 0 3 3; 5 0 0]
 nr=size(x,1);
 nc=size(x,1);
 xz=0; 
for ir=1:nr
 for ic=1:nc 
if x(ir,ic)==0 
xz=xz+1; 
end 
end 
end 
xz

Answer 5

[Salam Ismaeel]

% for a vector 
sum(A == 0)
% for a 2D matrix
sum(sum(A == 0)

[Walter Roberson]

Would there be a reason you would use this code instead of what Mischa suggested in

https://www.mathworks.com/matlabcentral/answers/126267-how-do-i-count-the-number-of-zeros-in-a-matrix#answer_133813

[Salam Ismaeel]

Not the same, they are slightly different!

[Walter Roberson]

Under what circumstances would they produce different results?

[Salam Ismaeel]

Both will produce the same results, but in different way? Please try them! It is not a big deal!

[Walter Roberson]

In order to use your code, you need

if isvector(A)
    out = sum(A == 0);
elseif ismatrix(A)
    out = sum(sum(A == 0));
elseif ndims(A) == 3
    out = sum(sum(sum(A == 0)));
elseif ndims(A) == 4
    out = sum(sum(sum(sum(A == 0))));
else
    error('sorry, code only handles up to 4 dimensions')
end

Where-as Mischa's code would be simply

out = sum(A(:) == 0);

which would handle all number of dimensions including more than 4 dimensions.

Under what circumstances do you see your code as being advantageous compared to Mischa's ?

[Salam Ismaeel]

If you read my comments correctly, I said for a vector and for a 2D matrix only!