Vectorized assignment in Numpy

Question

Let's assume I have a large 2D numpy array, e.g. 1000x1000 elements. I also have two 1D integer arrays of length L, and a float 1D arrray of the same length. If I want to simply assign floats to different positions in the original array according to integer array, I could write:

mat = np.zeros((1000,1000))
int1 = np.random.randint(0,999,size=(50000,))
int2 = np.random.randint(0,999,size=(50000,))
f = np.random.rand(50000)
mat[int1,int2] = f

But if there were collisions i.e. multiple floats corresponding to single location, all but the last would be overwritten. Is there a way to somehow aggregate all the collisions, e.g. mean or median of all the floats falling at the same location? I would like to take advantage of vectorization and hopefully avoid interpreter loops.

Thanks!

Answer 1

Building on hpaulj's suggestion, here's how to get the mean value in case of collisions:

import numpy as np

mat = np.zeros((2,2))
int1 = np.zeros(2, dtype=int)
int2 = np.zeros(2, dtype=int)
f = np.array([0,1])

np.add.at(mat, [int1, int2], f)
n = np.zeros((2,2))
np.add.at(n, [int1, int2], 1)
mat[int1, int2] /= n[int1, int2]
print(mat)

array([[0.5, 0. ],
       [0. , 0. ]])

Answer 2

You can manipulate your data in pandas and then assign.

Starting from

mat = np.zeros((1000,1000))
a = np.random.randint(0,999,size=(50000,))
b = np.random.randint(0,999,size=(50000,))
c = np.random.rand(50000)

You can define a function

def get_aggregated_collisions(a,b,c):
    df = pd.DataFrame({'x':a, 'y':b, 'v':c})
    df['coord'] = df[['x','y']].apply(tuple,1)
    d = df.groupby('coord').agg({"v":'mean','x':'first', 'y':'first'}).to_dict('list')
    return d

and then

d = get_aggregated_collisions(a,b,c)
mat[d['x'], d['y']] = d['v']

The whole operation (including generating the matrixes, np.random etc) ran quite ok

1.05 s ± 30.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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The idea behind making a tuple of coordinates was to have a hashable option to group values by their coordinates. Maybe there is even a smarter way to do this :) always open to suggestions.