Set specific values to zero in numpy array

Question

I have numpy array/matrix of shape (4096,4096) and an array of elements that should be set to zero. I have found function numpy.in1d that works fine but is very slow for my computations. I was wondering if exists some faster way for this execution because I need to repeat this on very large number of matrices so every optimization is helpful.

Here is example:

The numpy array looks like this:

npArr = np.array([
    [1, 4, 5, 5, 3],
    [2, 5, 6, 6, 1],
    [0, 0, 1, 0, 0],
    [3, 3, 2, 4, 3]])

and another array is :

arr = np.array([3,5,8])

The numpy array npArr should look like after the replacement:

array([[ 1,  4,  0,  0,  0],
       [ 2,  0,  6,  6,  1],
       [ 0,  0,  1,  0,  0],
       [ 0,  0,  2,  4,  0]])

Answer 1

If you have numba you can solve this with a custom function that doesn't need an intermediate mask:

import numpy as np
import numba as nb

@nb.njit
def replace_where(arr, needle, replace):
    arr = arr.ravel()
    needles = set(needle)
    for idx in range(arr.size):
        if arr[idx] in needles:
            arr[idx] = replace

This gives the correct result for your example:

npArr = np.array([[1, 4, 5, 5, 3],
                  [2, 5, 6, 6, 1],
                  [0, 0, 1, 0, 0],
                  [3, 3, 2, 4, 3]])

arr = np.array([3,5,8])

replace_where(npArr, arr, 0)
print(npArr)
# array([[1, 4, 0, 0, 0],
#        [2, 0, 6, 6, 1],
#        [0, 0, 1, 0, 0],
#        [0, 0, 2, 4, 0]])

And it should be really, really fast. I timed it for several array sizes and it was 5-20 times faster (depending on the sizes, especially the arr size) than np.in1d.

Answer 2

Here's an alternative using np.searchsorted -

def in1d_alternative_2D(npArr, arr):
    idx = np.searchsorted(arr, npArr.ravel())
    idx[idx==len(arr)] = 0
    return arr[idx].reshape(npArr.shape) == npArr

It assumes arr to be sorted. If it's not, we need to sort and then use the posted method.

Sample run -

In [90]: npArr = np.array([[1, 4, 5, 5, 3],
    ...:     [2, 5, 6, 6, 1],
    ...:     [0, 0, 1, 0, 0],
    ...:     [3, 3, 2, 14, 3]])
    ...: 
    ...: arr = np.array([3,5,8])
    ...: 

In [91]: in1d_alternative_2D(npArr, arr)
Out[91]: 
array([[False, False,  True,  True,  True],
       [False,  True, False, False, False],
       [False, False, False, False, False],
       [ True,  True, False, False,  True]], dtype=bool)

In [92]: npArr[in1d_alternative_2D(npArr, arr)] = 0

In [93]: npArr
Out[93]: 
array([[ 1,  4,  0,  0,  0],
       [ 2,  0,  6,  6,  1],
       [ 0,  0,  1,  0,  0],
       [ 0,  0,  2, 14,  0]])

Benchmarking against numpy.in1d

Equivalent solution using np.in1d would be :

np.in1d(npArr, arr).reshape(npArr.shape)

Let's time our proposed one against it and also verify results for the sizes mentioned in the question.

In [85]: # (4096, 4096) shaped 'npArr' and search array 'arr' of 1000 elems
    ...: npArr = np.random.randint(0,10000,(4096,4096))
    ...: arr = np.sort(np.random.choice(10000, 1000, replace=0 ))
    ...: 

In [86]: out1 = np.in1d(npArr, arr).reshape(npArr.shape)
    ...: out2 = in1d_alternative_2D(npArr, arr)
    ...: 

In [87]: np.allclose(out1, out2)
Out[87]: True

In [88]: %timeit np.in1d(npArr, arr).reshape(npArr.shape)
1 loops, best of 3: 3.04 s per loop

In [89]: %timeit in1d_alternative_2D(npArr, arr)
1 loops, best of 3: 1 s per loop