Increment Numpy array with repeated indices

I have a Numpy array and a list of indices whose values I would like to increment by one. This list may contain repeated indices, and I would like the increment to scale with the number of repeats of each index. Without repeats, the command is simple:

a=np.zeros(6).astype('int')
b=[3,2,5]
a[b]+=1

With repeats, I've come up with the following method.

b=[3,2,5,2]                     # indices to increment by one each replicate
bbins=np.bincount(b)
b.sort()                        # sort b because bincount is sorted
incr=bbins[np.nonzero(bbins)]   # create increment array
bu=np.unique(b)                 # sorted, unique indices (len(bu)=len(incr))
a[bu]+=incr

Is this the best way? Is there are risk involved with assuming that the np.bincount and np.unique operations would result in the same sorted order? Am I missing some simple Numpy operation to solve this?

Solution 1:

In numpy >= 1.8, you can also use the at method of the addition 'universal function' ('ufunc'). As the docs note:

For addition ufunc, this method is equivalent to a[indices] += b, except that results are accumulated for elements that are indexed more than once.

So taking your example:

a = np.zeros(6).astype('int')
b = [3, 2, 5, 2]

…to then…

np.add.at(a, b, 1)

…will leave a as…

array([0, 0, 2, 1, 0, 1])

Solution 2:

After you do

bbins=np.bincount(b)

why not do:

a[:len(bbins)] += bbins

(Edited for further simplification.)

Solution 3:

If b is a small subrange of a, one can refine Alok's answer like this:

import numpy as np
a = np.zeros( 100000, int )
b = np.array( [99999, 99997, 99999] )

blo, bhi = b.min(), b.max()
bbins = np.bincount( b - blo )
a[blo:bhi+1] += bbins

print a[blo:bhi+1]  # 1 0 2