Transform a set of numbers in numpy so that each number gets converted into a number of other numbers which are less than it

Solution 1:

What you actually need to do is get the inverse of the sorting order of your array:

import numpy as np
x = np.random.rand(10)
y = np.empty(x.size,dtype=np.int64)
y[x.argsort()] = np.arange(x.size)

Example run (in ipython):

In [367]: x
Out[367]: 
array([ 0.09139335,  0.29084225,  0.43560987,  0.92334644,  0.09868977,
        0.90202354,  0.80905083,  0.4801967 ,  0.99086213,  0.00933582])

In [368]: y
Out[368]: array([1, 3, 4, 8, 2, 7, 6, 5, 9, 0])

Alternatively, if you want to get the number of elements greater than each corresponding element in x, you have to reverse the sorting from ascending to descending. One possible option to do this is to simply swap the construction of the indexing:

y_rev = np.empty(x.size,dtype=np.int64)
y_rev[x.argsort()] = np.arange(x.size)[::-1]

another, as @unutbu suggested in a comment, is to map the original array to the new one:

y_rev = x.size - y - 1

Solution 2:

Here's one approach using np.searchsorted -

np.searchsorted(np.sort(x),x)

Another one mostly based on @Andras Deak's solution using argsort() -

x.argsort().argsort()

Sample run -

In [359]: x
Out[359]: 
array([ 0.62594394,  0.03255799,  0.7768568 ,  0.03050498,  0.01951657,
        0.04767246,  0.68038553,  0.60036203,  0.3617409 ,  0.80294355])

In [360]: np.searchsorted(np.sort(x),x)
Out[360]: array([6, 2, 8, 1, 0, 3, 7, 5, 4, 9])

In [361]: x.argsort().argsort()
Out[361]: array([6, 2, 8, 1, 0, 3, 7, 5, 4, 9])

Solution 3:

In addition to the other answers another solution using boolean indexing could be:

sum(x > i for i in x)

For your example:

In [10]: x
Out[10]: 
array([ 0.62594394,  0.03255799,  0.7768568 ,  0.03050498,  0.01951657,
        0.04767246,  0.68038553,  0.60036203,  0.3617409 ,  0.80294355])

In [10]: y = sum(x > i for i in x)
In [11]: y
Out[10]: array([6, 2, 8, 1, 0, 3, 7, 5, 4, 9])