Find the cell where the sum equals a threshold

Hello,

I have a 32 column by one row array. I want to sum across the values in each column until I match a threshold value. The result should provide the column # (which will be between 1:32)

I've tried a few things and havent been able to iron this down.

Here's the data:

A = [0 0 4520 51418 101386 90907 78735 65863 45766 26831 30586 14864 7905 4741 2840 2475 872 290 99 44 15 2 0 0 0 1 0 2 0 0 0 0];

Threshold value is = 0.5*sum(A).

So I want to find the cell where A = 265081.

The result should provide the cell number (or bin if you will) where this value occurs while summing from the first cell of A to the last.

[Adam Danz]

"The result should provide the cell number (or bin if you will) where this value occurs while summing from the first cell of A to the last.... So I want to find the cell where A = 265081."

This will never happen with those data. The cumulative sum of A is below and the value 265081 is not a member of that vector.

The best you can do is return the first element that is above or below that value.

 cumsum(A)
ans =
  Columns 1 through 11
           0           0        4520       55938      157324      248231      326966      392829      438595      465426      496012
  Columns 12 through 22
      510876      518781      523522      526362      528837      529709      529999      530098      530142      530157      530159
  Columns 23 through 32
      530159      530159      530159      530160      530160      530162      530162      530162      530162      530162

Answer 1

[Adam Danz]

Compare the cumulative sum against your threshold. The line below returns the first index of A that is greater than or equal to your threshold.

A = [0	0	4520	51418	101386	90907	78735	65863	45766	26831	30586	14864	7905	4741	2840	2475	872	290	99	44	15	2	0	0	0	1	0	2	0	0	0	0];
threshold = 5000; 

idx = find(cumsum(A) >= threshold,1); 

If you'd rather have the index of the last element that is less than or equal to your threshold,

idx = find(cumsum(A) <= threshold,1,'last'); 

[Eric Escoto]

Thanks, Adam

The explanation really helps. Yes, in this case I think the first index greater than or equal to would work best. This should trap the value in the index of interest, and return that index.

Much appreciated!

[Adam Danz]

Thanks for the feedback. Be aware if A contains negative values. In that case, the cumulative sum will not be monotonically increasing which may lead to the possibility that you'll have >1 element that passes through your threshold.