Inverse of numpy's bincount function

Question

Given an array of integer counts c, how can I transform that into an array of integers inds such that np.all(np.bincount(inds) == c) i

Answer 1

The following is about twice as fast on my machine than the currently accepted answer; although I must say I am surprised by how well np.repeat does. I would expect it to suffer a lot from temporary object creation, but it does pretty well.

import numpy as np
c = np.array([1,3,2,2])
p = np.cumsum(c)
i = np.zeros(p[-1],np.int)
np.add.at(i, p[:-1], 1)
print np.cumsum(i)

Answer 2

no numpy needed :

c = [1,3,2,2]
reduce(lambda x,y: x + [y] * c[y], range(len(c)), [])

Answer 3

using numpy.repeat :

np.repeat(np.arange(c.size), c)