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]])

• timgeb: Your problem description is too vague. Give us an example with a small quadratic matrix and the expected result.
• Đorđe Ivanović: I am sorry for that. Is it clearer now?
• timgeb: Yes, much better!
• Divakar: Did either of the posted solutions work for you?
• Đorđe Ivanović: Sorry for late comment, your answer is great. Thank you once again

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

Answer 3

Another solution using numpy broadcasting:

np.min(np.where(npArr[None,:,:] == arr[:,None,None], 0, a),0)
Out[730]: 
array([[1, 4, 0, 0, 0],
 [2, 0, 6, 6, 1],
 [0, 0, 1, 0, 0],
 [0, 0, 2, 4, 0]])

• MSeifert: You actually shouldn't try this with huge arrays because you create huge intermediate arrays. For an npArr of size (4096,4096) and arr of size 3 this creates a 3 * 4096 * 4096 array (50 million elements). And because you use where and the condition you actually have several of those huge arrays (I would guess 3 but I'm not sure). That can quickly exhaust the RAM.