# Calculate median values of slices of array associated with histogram bins

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dormant on 15 Nov 2023
Commented: Star Strider on 15 Nov 2023
I have an array (array1) which I want to put into histogram bins. I have a second array (array2) which is the same length.
I can sum the values of array2 in each bin with this:
[count, edges, idx] = histcounts(array1,edges);
binsums = accumarray(idx,array2);
But how can I calculate the median values of array2 in each bin?

Star Strider on 15 Nov 2023
I would use another accumarray call:
binmedians = accumarray(idx+1, (1:numel(idx)).', [], @(x)median(array2(x)))
Creating data and correcting for ‘idx’ having 0 for some elements (making them inappropriate index values) —
array1 = randn(1E+3,1);
array2 = randn(1E+3,1);
edges = 1:9;
[count, edges, idx] = histcounts(array1,edges);
binsums = accumarray(idx+1,array2)
binsums = 4×1
-13.0331 20.8357 -1.5279 -2.0722
binmedians = accumarray(idx+1, (1:numel(idx)).', [], @(x)median(array2(x)))
binmedians = 4×1
-0.0237 0.2213 0.1113 -1.0361
.
dormant on 15 Nov 2023
Thank you very much. Some of the values going into the median ar NaNs, but I can handle them.
Star Strider on 15 Nov 2023
As always, my pleasure!
Use the missingflag option that works best for you.
One such:
binmedians = accumarray(idx+1, (1:numel(idx)).', [], @(x)median(array2(x),'omitmissing'))
There are several others.
.

Steven Lord on 15 Nov 2023
I would use the groupsummary function with the grouping information from the idx variable.
If all you want is the bin numbers, you could also use the discretize function instead of histcounts.
dormant on 15 Nov 2023
Thanks you very much. I'll try this as well as the solution suggested by Star Strider.

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