Find the column with the least number of 1s (ones) in it | Find column with at least one non zero element

Consider I have a m by n matrix with binary data

Is there a easy way to write a code that can identify the column with the least number of ones in them.

My current code identifies the column with zero 1s, which is not what I seek

H = [0 0 1 1 0 1 0 1 0 1 1 0;
     0 0 0 1 0 1 1 0 1 0 1 0;
     0 1 1 0 0 1 1 1 0 0 0 1;
     0 1 1 0 1 0 1 0 1 0 0 0;
     0 0 0 1 1 0 0 1 1 0 0 0;
     0 1 0 0 1 0 0 0 0 0 1 1];
%For ex, here the column with least number of ones is column 10.
%whereas column 1 has all zero elements which is not the result I want.

%This is my code which gives wrong results
 a = sum(H);

amax = max(a);
amin = min(a);

Answer 1

[Walter Roberson]

count = sum(H);
idx = find(count);
[~, relidx] = min(count(idx)) ;
mincol = idx(relidx);

[Walter Roberson]

count = sum(H);
count(count==0) = nan;
[~, mincol] = min(count) ;

[Rishi Balasubramanian]

This works. Thank You.

Answer 2

[Image Analyst]

If you have multiple columns where the count is the same minimum count, then you can't use min() because it will find only the FIRST occurrence. You need to use find():

H = [0 0 1 1 0 1 0 1 0 1 1 0;
     0 0 0 1 0 1 1 0 1 0 1 0;
     0 1 0 0 0 1 1 1 0 0 0 1;
     0 1 0 0 1 0 1 0 1 0 0 0;
     0 0 0 1 1 0 0 1 1 0 0 0;
     0 1 0 0 1 0 0 0 0 0 1 1];

count = sum(H == 1, 1) % Can handle non-1 values also, in case they occur.
count(count==0) = nan; % Tell it to ignore a count of zero.
minCount = min(count) % Find the min count other than 0.
% Find ALL the columns where the min count can occur.
% Unfortunately, min() only finds the FIRST column.
columns = find(count == minCount) % Correctly returns both column 3 and column 10 which both have a count of 1

count =

0 3 1 3 3 3 3 3 3 1 3 2

minCount =

1

columns =

3 10