trying to accounting the number of zeros in random script

hello lets say i have this random script w=randint(1000,1) and i want to know how many 'zeros' there are in this script what should i do to account the 'zeros'? thanks!!

Answer 1

[Paulo Silva]

a=randint(1000,1);
numel(a(a==0))

or

sum(a==0)

or

numel(a)-nnz(a)

[Matt Fig]

itsik says,

thank u for your answer but my problem is i want to know how MANY zeros there are all over the matrix lets say i have a=0 1 0 0 0 0 0 0 1 1 1 1 1 0 0

so im trying to make a code that accounting the numbers of zeros and will give me a=8 like there is 8 zeros thanks!

[Matt Fig]

I wonder if you tried Paulo's answer. The solutions he gives are standard solutions to this problem and return the number of zeros in the array... Did you try them???

a = [0 1 0 0 0 0 0 0 1 1 1 1 1 0 0] % 9 zeros, not 8!

sum(a(:)==0) % This DOES count the zeros!

ans =

9

[Matt Tearle]

Minor quibbles.

randint is deprecated. To make a random vector of 0s and 1s you can do

a = randi(2,10000,1)-1;

On my machine, running 11a, the most efficient approach is

nnz(a==0)

followed by

sum(a==0)

and then

numel(a(a==0))

Answer 2

[Yella]

[EDIT: Tue Jun 14 05:17:56 UTC 2011 - Reformat - MKF]

Its simple make a for loop and use and conditional loop like "if else" when ever value is '0' in the matrix increment a variable if not continue the loop.

k=0;
for i= 1:1:n
if (f(i)=0)
    k=k+1;
else continue
end
end

[Matt Fig]

Yella, you have an error. The correct comparison operator is '=='.

if f(i)==0

This could be replaced by:

if ~f(i)

[Yella]

Thanks matt

Answer 3

[itsik]

thanks all of u but i used this code: a=randint(1000,1); numel(a(a==0)) and it helped me!!!