How to create a loop to sum across columns conditional on index matching?

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Hello, I have a binary matrix, with a corresponding index in the first column. For example:
A = [1 1 0 1 0 0 1 ..0]
1 0 0 0 1 1 0 ..1
2 ...
2 ...
2 ...
3 ...
[... ... ]
I want to sum all binary elements corresponding to 1 in the index, 2 in the index, etc. (i.e. index is =1 in the first and second row,so sum across each column, and so on). For the index, the number of corresponding rows with same index number varies across the dataset. I would appreciate all help and advice on how to do this.
Matlabio on 26 Oct 2016
I actually want to have a conditional sum - if index values match in any 2 or more rows, then sum each column in these rows to create a new summation vector. For instance: assume I have the following data matrix (with index in the 1st column): A = [1 1 0 0 0; 2 0 1 0 0; 2 0 0 0 1] In this case, for i =1, my result would be: 1 0 0 0; for i=2 => 0 1 0 1.

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Accepted Answer

Chris Dischner
Chris Dischner on 25 Oct 2016
What about something like this?
Assuming your index values are sequential:
aMax = max(A(:,1));
for i = 1:aMax
aSum(i,:) = sum(A(A(:,1) == i,:));
Will get you the sums. You can use modular arithmetic to get you to the bit representation
(eg, 1+1+1 = 3 ==> 3 mod 2 = 1
Matt J
Matt J on 27 Oct 2016
Edited: Matt J on 27 Oct 2016
I get the summation vector for each i with some values >1, which should not be the case since it's strictly binary and cannot exceed 1 in any column sum, for any i
The chances are excellent that it is the input data's fault and not the code's, i.e, that you have not successfully fulfilled the condition "cannot exceed 1 in any column sum, for any i". There is no reason this should be the case just because A(:,2:end) are all zeros and ones.

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More Answers (2)

Matt J
Matt J on 27 Oct 2016
Edited: Matt J on 27 Oct 2016
The following produces a result B, such that B(1,:) is the consolidation of all rows of A with index 1, B(2,:) all rows with index 2, etc...
  1 Comment
Matlabio on 3 Nov 2016
Thank you so much! You were actually right - there was an issue with the data which I have corrected, otherwise it worked perfectly.

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Teja Muppirala
Teja Muppirala on 28 Oct 2016
result = [unique(A(:,1)) grpstats(A(:,2:end),A(:,1),@any)] % Keep it binary
result = [unique(A(:,1)) grpstats(A(:,2:end),A(:,1),@sum)] % Take the sum


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