Cumulative Summation down a matrix in loop

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IDN
IDN el 9 de Feb. de 2022
Comentada: Adam Danz el 10 de Feb. de 2022
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 comentarios
Turlough Hughes
Turlough Hughes el 9 de Feb. de 2022
Do you mean -4.36 and -26.28?
IDN
IDN el 9 de Feb. de 2022
Editada: IDN el 9 de Feb. de 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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Respuesta aceptada

Adam Danz
Adam Danz el 9 de Feb. de 2022
Editada: Adam Danz el 9 de Feb. de 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 comentarios
IDN
IDN el 9 de Feb. de 2022
Got it!
Adam Danz
Adam Danz el 10 de Feb. de 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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Más respuestas (1)

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

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