# Cumulative Summation down a matrix in loop

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IDN 2022 年 2 月 9 日
コメント済み: Adam Danz 2022 年 2 月 10 日
Hello!
I have 2 matrix, I would like to sum Matrix B values cummulative given condition. The condition is that it starts to sum once Matrix A = -1 and stops when Matrix A = 1 and it goes on and on all the way down to the end of the data set. Thanks for the help!
Matrix A Matrix B CumSum
0
0
0
-1 0
-1 0.02
-1 -12.09
-1 6.61
1 1.1 -4.36 CumSum
0 0
-1 0
-1 -6.8
-1 -26.87
-1 2.67
-1 -9.99
-1 9.28
-1 -3.17
1 8.6 7.39 CumSum
0
0
0
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Turlough Hughes 2022 年 2 月 9 日
Do you mean -4.36 and -26.28?
IDN 2022 年 2 月 9 日

those are the totals when cumum runs by summing values in Matrix B. So -4.36 and -26.28 is the expected answer when all set an done. Thanks for helping!

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### 採用された回答

Adam Danz 2022 年 2 月 9 日

If and only if the groups marked by A=-1 to A=1 are not interruped by any other values in A and a 1 does not appear before the first -1, the solution is,
Warning: Column headers from the file were modified to make them valid MATLAB identifiers before creating variable names for the table. The original column headers are saved in the VariableDescriptions property.
Set 'VariableNamingRule' to 'preserve' to use the original column headers as table variable names.
T = 20×3 table
MatrixA MatrixB CumSum _______ _______ ______ 0 NaN NaN 0 NaN NaN 0 NaN NaN -1 0 NaN -1 0.02 NaN -1 -12.09 NaN -1 6.61 NaN 1 1.1 -4.36 0 0 NaN -1 0 NaN -1 -6.8 NaN -1 -26.87 NaN -1 2.67 NaN -1 -9.99 NaN -1 9.28 NaN -1 -3.17 NaN
startIdx = find([false;diff(T.MatrixA==-1)==1]);
stopIdx = find([false;diff(T.MatrixA==1)==1]);
groupSums = arrayfun(@(start,stop)sum(T.MatrixB(start:stop)),startIdx,stopIdx)
groupSums = 2×1
-4.3600 -26.2800
Or perhaps you want,
T.CumSum(stopIdx) = groupSums
T = 20×3 table
MatrixA MatrixB CumSum _______ _______ ______ 0 NaN NaN 0 NaN NaN 0 NaN NaN -1 0 NaN -1 0.02 NaN -1 -12.09 NaN -1 6.61 NaN 1 1.1 -4.36 0 0 NaN -1 0 NaN -1 -6.8 NaN -1 -26.87 NaN -1 2.67 NaN -1 -9.99 NaN -1 9.28 NaN -1 -3.17 NaN
##### 4 件のコメント2 件の古いコメントを表示2 件の古いコメントを非表示
IDN 2022 年 2 月 9 日
Got it!
Adam Danz 2022 年 2 月 10 日
Just saw your message now. Glad you worked it out. That line merely places data within the table T so nothing should be coming out horizontally.

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### その他の回答 (1 件)

Highphi 2022 年 2 月 9 日
matrixA = [0 0 0 -1 -1 -1 -1 1 0 -1 -1 -1 -1 -1 -1 -1 1 0 0 0]'; % for ref
matrixB = [0 0 0 0 0.02 -12.09 6.61 1.1 0 0 -6.8 -26.87 2.67 -9-99 9.28 -3.17 8.6 0 0 0]'; % for ref
state = 0;
matrixSums = zeros(size(matrixA,1), 1);
for i = 1:size(matrixA, 1)
temp = matrixA(i);
if (temp == -1) && (state == 0)
state = 1;
cumSum = matrixB(i);
elseif (state == 1) && (temp ~= 1)
cumSum = cumSum + matrixB(i);
elseif (state == 1) && (temp == 1)
cumSum = cumSum + matrixB(i);
matrixSums(i) = cumSum;
state = 0;
end
end
outMat = [matrixA, matrixB, matrixSums]
outMat = 20×3
0 0 0 0 0 0 0 0 0 -1.0000 0 0 -1.0000 0.0200 0 -1.0000 -12.0900 0 -1.0000 6.6100 0 1.0000 1.1000 -4.3600 0 0 0 -1.0000 0 0
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IDN 2022 年 2 月 9 日
Thanks, this looks good as well...i just dont have the toolbox required for the "state" functionality
Highphi 2022 年 2 月 9 日
you shouldn't need a toolbox, it's just a variable to indicate your status

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