Increment Numpy array with repeated indices

Question

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?

Note that numpy.zeros(6).astype('int') is better written as numpy.zeros(6, int). — Eric O. Lebigot
– Eric O. Lebigot, Commented Jan 5, 2010 at 8:39

gojomo · Accepted Answer · 2015-03-06 08:01:36Z

43

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])

answered Mar 6, 2015 at 8:01

gojomo

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3 Comments

Sibbs Gambling Over a year ago

This solution is the most elegant one AFAIK!

John Deighan Over a year ago

I'm trying to do this with a matrix, but I'm getting an error: arr = np.array([[1,3,5],[7,9,11]]); lIndexes = [[0,1],[1,0],[1,2]] np.add.at(arr, lIndexes, 1) Any ideas?

gojomo Over a year ago

I'd suggest posting this new question about multi-dimensional indexing as a separate question.

Alok Singhal · Accepted Answer · 2010-01-05 06:28:54Z

6

After you do

bbins=np.bincount(b)

why not do:

a[:len(bbins)] += bbins

(Edited for further simplification.)

edited Jan 5, 2010 at 6:28

answered Jan 5, 2010 at 6:08

Alok Singhal

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4 Comments

Eric O. Lebigot Over a year ago

Would this not be slower, when b contains just a few large bin numbers?

Alok Singhal Over a year ago

Yes, it will be slower than a simple Python loop in that case, but still faster than OP's code. I did a quick timing test with b = [99999, 99997, 99999], and a = np.zeros(1000, 'int'). Timings are: OP: 2.5 ms, mine: 495 us, simple loop: 84 us.

fideli Over a year ago

This works well. A simple loop has generally been slower in my program. Thanks.

ajwood Over a year ago

Is there a similar way to accomplish this in a multi-dimensional case?

denis · Accepted Answer · 2010-01-09 17:52:41Z

1

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

answered Jan 9, 2010 at 17:52

denis

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