Sum matrix elements group by indices in Python

Question

I have two matrix (same row and column): one with float values, which are grouped by indices in the other matrix. As a result, I want a dictionary or a list with the sums of the elements for each index. Indices always start at 0.

A = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]])
B = np.array([[1,3,1,2],[3,0,2,2]])

RESULT = {0: -0.41, 1: 0.07, 2: 0.03, 3: 0.11}

I found this solution, but I'm searching for a faster one. I'm working with matrix with 784 x 300 cells and this algorithm takes ~28ms to complete.

import numpy as np

def matrix_sum_by_indices(indices,matrix):
    a = np.hstack(indices)
    b = np.hstack(matrix)
    sidx = a.argsort()
    split_idx = np.flatnonzero(np.diff(a[sidx])>0)+1
    out = np.split(b[sidx], split_idx)
    return [sum(x) for x in out]

If you can help me find a better and plain solution to this problem, I'll be grateful!

EDIT: I made a mistake, time to complete is ~8ms in a 300*10 matrix, but ~28ms in a 784x300.

EDIT2: My A elements are float64, so bincount give me ValueError.

Molecule

8ms? I would say that is pretty fast. How fast are you looking for? – Foreclosure

@Foreclosure I'm looking for something like under the ms. Because this procedure is in a ~6000*100 loops. – Molecule

You're probably looking for bincount in some form. – Dampier

Answer 1

You can make use of bincount here:

a = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]])
b = np.array([[1,3,1,2],[3,0,2,2]])

N = b.max() + 1
id = b + (N*np.arange(b.shape[0]))[:, None] # since you can't apply bincount to a 2D array
np.sum(np.bincount(id.ravel(), a.ravel()).reshape(a.shape[0], -1), axis=0)

Output:

array([-0.41,  0.07,  0.03,  0.11])

As a function:

def using_bincount(indices, matrx):
    N = indices.max() + 1
    id = indices + (N*np.arange(indices.shape[0]))[:, None] # since you can't apply bincount to a 2D array
    return np.sum(np.bincount(id.ravel(), matrx.ravel()).reshape(matrx.shape[0], -1), axis=0)

Timings on this sample:

In [5]: %timeit using_bincount(b, a)
31.1 µs ± 1.74 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [6]: %timeit matrix_sum_by_indices(b, a)
61.3 µs ± 2.62 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [88]: %timeit scipy.ndimage.sum(a, b, index=[0,1,2,3])
54 µs ± 218 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

(scipy.ndimage.sum should be faster on much larger samples)

Dampier

This is working fine, but if my a elements are float64? – Molecule

@Molecule can you post an example of how you're creating your matrix? – Dampier

@user348302 I'm working with a K-Means Neural Network, so i don't have a generator for the indices matrix. – Molecule

And are you positive you are passing the arguments in the correct order? indices are b, values are a? I'll post a function that matches your input names – Dampier

Yeah for sure, I'm passing the args in the correct order.. the problem (known in the Python Doc, under bincount) are the float64 – Molecule

But you are passing an integer array, while using floats as the weights – Dampier

Integer array and float64 weights – Molecule

Answer 2

The numpy_indexed package has efficient and simple solutions to this problem (disclaimer: I am its author):

import numpy_indexed as npi
keys, values = npi.group_by(B.flatten()).sum(A.flatten())

Crocodile

How can i use this in Jupyter Notebook? – Molecule

There is a conda-forge package and a pip installer available; notebook or not should make no difference. – Crocodile

Answer 3

The following solution, relying on scipy.ndimage.sum is highly optimized for speed:

import numpy as np
A = np.array([[0.52,0.25,-0.45,0.13], [-0.14,-0.41,0.31,-0.41]])
B = np.array([[1,3,1,2], [3,0,2,2]])
import scipy.ndimage
print(scipy.ndimage.sum(A, B, index=[0,1,2,3]))

You may have to work a little for having the index parameter be exactly what you want. It is the list of the indices you want to get in the result. Maybe the following will be a good starting point:

print(scipy.ndimage.sum(A,B, index=np.unique(B)))

but if you know by advance the list of all indices, it will be more efficient to hard-code it here.

Shirker

How can I use this with float64? – Molecule

You can have whatever you want in the A array; is it what you are asking for? – Shirker

Nevermind, It works with float64. It's faster than mine (~10ms), but I'm looking for a faster one, if it's possible – Molecule