Rank items in an array using Python/NumPy, without sorting array twice

Use argsort twice, first to obtain the order of the array, then to obtain ranking:

array = numpy.array([4,2,7,1])order = array.argsort()ranks = order.argsort()

When dealing with 2D (or higher dimensional) arrays, be sure to pass an axis argument to argsort to order over the correct axis.

This question is a few years old, and the accepted answer is great, but I think the following is still worth mentioning. If you don't mind the dependency on scipy, you can use scipy.stats.rankdata:

In [22]: from scipy.stats import rankdataIn [23]: a = [4, 2, 7, 1]In [24]: rankdata(a)Out[24]: array([ 3.,  2.,  4.,  1.])In [25]: (rankdata(a) - 1).astype(int)Out[25]: array([2, 1, 3, 0])

A nice feature of rankdata is that the method argument provides several options for handling ties. For example, there are three occurrences of 20 and two occurrences of 40 in b:

In [26]: b = [40, 20, 70, 10, 20, 50, 30, 40, 20]

The default assigns the average rank to the tied values:

In [27]: rankdata(b)Out[27]: array([ 6.5,  3. ,  9. ,  1. ,  3. ,  8. ,  5. ,  6.5,  3. ])

method='ordinal' assigns consecutive ranks:

In [28]: rankdata(b, method='ordinal')Out[28]: array([6, 2, 9, 1, 3, 8, 5, 7, 4])

method='min' assigns the minimum rank of the tied values to all the tied values:

In [29]: rankdata(b, method='min')Out[29]: array([6, 2, 9, 1, 2, 8, 5, 6, 2])

See the docstring for more options.

Use advanced indexing on the left-hand side in the last step:

array = numpy.array([4,2,7,1])temp = array.argsort()ranks = numpy.empty_like(temp)ranks[temp] = numpy.arange(len(array))

This avoids sorting twice by inverting the permutation in the last step.