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

Question

I have an array of numbers and I'd like to create another array that represents the rank of each item in the first array. I'm using Python and NumPy.

For example:

array = [4,2,7,1]
ranks = [2,1,3,0]

Here's the best method I've come up with:

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

Are there any better/faster methods that avoid sorting the array twice?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Use argsort() twice will do it:

>>> array = [4,2,7,1]
>>> ranks = numpy.array(array).argsort().argsort()
>>> ranks
array([2, 1, 3, 0])

Answer 5

For a vectorized version of an averaged rank, see below. I love np.unique, it really widens the scope of what code can and cannot be efficiently vectorized. Aside from avoiding python for-loops, this approach also avoids the implicit double loop over 'a'.

import numpy as np

a = np.array( [4,1,6,8,4,1,6])

a = np.array([4,2,7,2,1])
rank = a.argsort().argsort()

unique, inverse = np.unique(a, return_inverse = True)

unique_rank_sum = np.zeros_like(unique)
np.add.at(unique_rank_sum, inverse, rank)
unique_count = np.zeros_like(unique)
np.add.at(unique_count, inverse, 1)

unique_rank_mean = unique_rank_sum.astype(np.float) / unique_count

rank_mean = unique_rank_mean[inverse]

print rank_mean

Answer 6

I tried to extend both solution for arrays A of more than one dimension, supposing you process your array row-by-row (axis=1).

I extended the first code with a loop on rows; probably it can be improved

temp = A.argsort(axis=1)
rank = np.empty_like(temp)
rangeA = np.arange(temp.shape[1])
for iRow in xrange(temp.shape[0]): 
    rank[iRow, temp[iRow,:]] = rangeA

And the second one, following k.rooijers suggestion, becomes:

temp = A.argsort(axis=1)
rank = temp.argsort(axis=1)

I randomly generated 400 arrays with shape (1000,100); the first code took about 7.5, the second one 3.8.

Answer 7

Apart from the elegance and shortness of solutions, there is also the question of performance. Here is a little benchmark:

import numpy as np
from scipy.stats import rankdata
l = list(reversed(range(1000)))

%%timeit -n10000 -r5
x = (rankdata(l) - 1).astype(int)
>>> 128 µs ± 2.72 µs per loop (mean ± std. dev. of 5 runs, 10000 loops each)

%%timeit -n10000 -r5
a = np.array(l)
r = a.argsort().argsort()
>>> 69.1 µs ± 464 ns per loop (mean ± std. dev. of 5 runs, 10000 loops each)

%%timeit -n10000 -r5
a = np.array(l)
temp = a.argsort()
r = np.empty_like(temp)
r[temp] = np.arange(len(a))
>>> 63.7 µs ± 1.27 µs per loop (mean ± std. dev. of 5 runs, 10000 loops each)

Answer 8

I tried the above methods, but failed because I had many zeores. Yes, even with floats duplicate items may be important.

So I wrote a modified 1D solution by adding a tie-checking step:

def ranks (v):
    import numpy as np
    t = np.argsort(v)
    r = np.empty(len(v),int)
    r[t] = np.arange(len(v))
    for i in xrange(1, len(r)):
        if v[t[i]] <= v[t[i-1]]: r[t[i]] = r[t[i-1]]
    return r

# test it
print sorted(zip(ranks(v), v))

I believe it's as efficient as it can be.

Answer 9

argsort and slice are symmetry operations.

try slice twice instead of argsort twice. since slice is faster than argsort

array = numpy.array([4,2,7,1])
order = array.argsort()
ranks = np.arange(array.shape[0])[order][order]

Answer 10

I liked the method by k.rooijers, but as rcoup wrote, repeated numbers are ranked according to array position. This was no good for me, so I modified the version to postprocess the ranks and merge any repeated numbers into a combined average rank:

import numpy as np
a = np.array([4,2,7,2,1])
r = np.array(a.argsort().argsort(), dtype=float)
f = a==a
for i in xrange(len(a)):
   if not f[i]: continue
   s = a == a[i]
   ls = np.sum(s)
   if ls > 1:
      tr = np.sum(r[s])
      r[s] = float(tr)/ls
   f[s] = False

print r  # array([ 3. ,  1.5,  4. ,  1.5,  0. ])

I hope this might help others too, I tried to find anothers solution to this, but couldn't find any...

Answer 11

More general version of one of the answers:

In [140]: x = np.random.randn(10, 3)

In [141]: i = np.argsort(x, axis=0)

In [142]: ranks = np.empty_like(i)

In [143]: np.put_along_axis(ranks, i, np.repeat(np.arange(x.shape[0])[:,None], x.shape[1], axis=1), axis=0)

See How to use numpy.argsort() as indices in more than 2 dimensions? to generalize to more dims.