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 dependence on scipy
, you can use scipy.stats.rankdata
:
In [22]: from scipy.stats import rankdata
In [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.