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Counting occurrences of each column in a matrix

Asked by Daniel Ring on 8 Oct 2018
Latest activity Commented on by Daniel Ring on 8 Oct 2018
I have a matrix with 16 columns and a very large number of rows. Each column has two possible outcomes in it (for example 1 or 0 in one column, .6 and .4 in another). I would like to count each of these occurrences in the columns. Preferably, I'd like to have a vector that counts each outcome of its respective column. Thank you!
[EDITED, Jan, moved from section for answers]
Example: A = [1 .6 .7 .8; 0 .4 .3 .2; 1 .6 .7 .8; 1 .6 .3 .8]
ans = [3 3 2 2 ; 1 1 2 2]

  3 Comments

an example?
Jan
on 8 Oct 2018
The example is not clear: 0.8 occurs 3 times, so shouldn't the last value of the output be 3 or perhaps 1? Why is the first column of the output [3;1] and not [1;3]?
I'm sorry, it should be ans = [3 3 2 3 ; 1 1 2 1] The first column in [3;1] because 1 occurs 3 times and 0 occurs 1 time.

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3 Answers

Answer by ANKUR KUMAR on 8 Oct 2018
 Accepted Answer

Since you have not given any sample data, I am taking some random data.
A=randi(50,50,50);
A(A<=25)=0.4;
A(A>25)=0.6;
A contains only 0.4 and 0.6 only, spreaded out completely. The below program gives you the occurrence of 0.4 and 0.6 in every coloumn.
nums1=arrayfun(@(x) length(find(A(:,x)==0.4)),1:size(A,2));
nums2=arrayfun(@(x) length(find(A(:,x)==0.6)),1:size(A,2));
occur=[nums1' nums2'];
nums2 can also be calculated as
nums2=size(A,1)-nums1;

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Answer by Jan
on 8 Oct 2018
Edited by Jan
on 8 Oct 2018

Maybe:
A = [1 .6 .7 .8; ...
0 .4 .3 .2; ...
1 .6 .7 .8; ...
1 .6 .3 .8];
Count = sum(A == A(1, :), 1); % Auto-expand: >= R2016b
Result = [Count; size(A, 1) - Count];
For older Matlab versions:
Count = sum(bsxfun(@eq, A, A(1, :)), 1);

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Answer by dpb
on 8 Oct 2018

Several ways to do this; probably easiest is via converting to categorical with the unique values in each column transformed to a single pair of categories--[0|1], [Y|N], [HI|LO], ..., whatever makes sense for the meanings.
Then histogram on those variables to return counts.
Or, use unique and the returned (optional) second index array to bin over.

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