Extend / replicate a value by column when found in array

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cedric W
cedric W 2018-9-6
编辑： Guillaume 2018-9-6
Say I have a matrix 5x5 with binary entries 0 and 1
A=[1 1 1 1 1;1 0 1 1 1;1 0 1 0 0;0 1 1 0 1;1 1 1 0 0]
I'd like to have as a result, a matrix where when the array is parsed by column, if 0 is found, then all rows indexed after are 0:
B=[1 1 1 1 1;1 0 1 1 1;1 0 1 0 0;0 0 1 0 0;0 0 1 0 0]
Any clue for a fast computing solution ? FYI every column is independent.
  2 个评论
cedric W
cedric W 2018-9-6
That's why I tried to give an example: A is the input matrix, B is the output. Say columns are #of simulations, rows are days. If on a column,on one day the value is 0, then for all the next days it will remain at 0. And you loop for the number so simulations you have. Simulations (columns therefore) are independent. Hope this is clearer.

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Guillaume
Guillaume 2018-9-6
编辑：Guillaume 2018-9-6
This is trivially achieved with cumprod since as soon as a 0 is encountered in a column the cumulative product is 0 from then on:
A = [1 1 1 1 1;1 0 1 1 1;1 0 1 0 0;0 1 1 0 1;1 1 1 0 0]
B = cumprod(A)

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