Cumulative Summation down a matrix in loop

Question

IDN 2022-2-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1646415-cumulative-summation-down-a-matrix-in-loop

评论： Adam Danz 2022-2-10

Sum Sample.xlsx

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

-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

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Turlough Hughes 2022-2-9

Do you mean -4.36 and -26.28?

IDN 2022-2-9

编辑：IDN 2022-2-9

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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Answer 1

Adam Danz 2022-2-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1646415-cumulative-summation-down-a-matrix-in-loop#answer_892345

编辑：Adam Danz 2022-2-9

在 MATLAB Online 中打开

Sum Sample.xlsx

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
MatrixAMatrixBCumSum____________________

       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
MatrixAMatrixBCumSum____________________

       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 

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IDN 2022-2-9

Got it!

Adam Danz 2022-2-10

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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Answer 2

Highphi 2022-2-9

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1646415-cumulative-summation-down-a-matrix-in-loop#answer_892350

在 MATLAB Online 中打开

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