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

4 vues (au cours des 30 derniers jours)
dormant le 15 Nov 2023
Commenté : Star Strider le 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?
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### Réponse acceptée

Star Strider le 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
.
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dormant le 15 Nov 2023
Thank you very much. Some of the values going into the median ar NaNs, but I can handle them.
Star Strider le 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.
.

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### Plus de réponses (1)

Steven Lord le 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.
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dormant le 15 Nov 2023
Thanks you very much. I'll try this as well as the solution suggested by Star Strider.

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