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.
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Matt J
2016년 10월 25일
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).
You mean you want
sum A(i,1:A(i,1))
for each i?
채택된 답변
Chris Dischner
2016년 10월 25일
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,:));
end
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
댓글 수: 2
Matt J
2016년 10월 27일
편집: Matt J
2016년 10월 27일
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.
추가 답변 (2개)
Teja Muppirala
2016년 10월 28일
result = [unique(A(:,1)) grpstats(A(:,2:end),A(:,1),@any)] % Keep it binary
or
result = [unique(A(:,1)) grpstats(A(:,2:end),A(:,1),@sum)] % Take the sum
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