Sort numpy array by row and order matching values based on original array

Question

I have a 2D numpy array and I would like to sort the rows based on first column values. The trouble is the way it is formatted:

Column I am sorting by: 0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,etc ->you can see it repeats itself

Basically I want to group the 1s, then 2s, then 3s, then 4s. The ordering of the matching values matter: I want the 1st '1' row to be the first one that appeared in the unsorted array, followed by the one that shows up next, etc. I use this command: sortedData= myData[myData[:,0].argsort()]

Unfortunately, it doesn't not appear to order matching columns based on the original ordering of the array. Are there certain options I can turn on to enable this?

Thanks!

Warren Weckesser · Accepted Answer · 2014-10-10 19:09:15Z

3

You can change the sorting algorithm used by argsort with the kind argument.

Use

sortedData= myData[myData[:,0].argsort(kind='mergesort')]

to preserve the order of the equal items. (Merge sort is a stable sorting algorithm.)

edited Oct 10, 2014 at 19:09

answered Oct 10, 2014 at 19:00

Warren Weckesser

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1 Comment

Andi Over a year ago

As per numpy v.1.15.0, select kind="stable"`. mergesort`` is just for backwards compatibility. numpy.org/doc/stable/reference/generated/numpy.argsort.html

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