Cumulative Summation down a matrix in loop

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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
  2 Commenti
Turlough Hughes
Turlough Hughes il 9 Feb 2022
Do you mean -4.36 and -26.28?
IDN
IDN il 9 Feb 2022
Modificato: IDN il 9 Feb 2022
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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Risposta accettata

Adam Danz
Adam Danz il 9 Feb 2022
Modificato: Adam Danz il 9 Feb 2022
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,
T = readtable('Sum Sample.xlsx')
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 Commenti
Adam Danz
Adam Danz il 10 Feb 2022
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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Più risposte (1)

Highphi
Highphi il 9 Feb 2022
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
  2 Commenti
IDN
IDN il 9 Feb 2022
Thanks, this looks good as well...i just dont have the toolbox required for the "state" functionality
Highphi
Highphi il 9 Feb 2022
you shouldn't need a toolbox, it's just a variable to indicate your status

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