“In” operator for numpy arrays?

How can I do the “in” operation on a numpy array?
(Return True if an element is present in the given numpy array)

For strings, lists and dictionaries, the functionality is intuitive to understand.

Here’s what I got when I applied that on a numpy array

a
array([[[2, 3, 0],
    [1, 0, 1]],

   [[3, 2, 0],
    [0, 1, 1]],

   [[2, 2, 0],
    [1, 1, 1]],

   [[1, 3, 0],
    [2, 0, 1]],

   [[3, 1, 0],
    [0, 2, 1]]])

b = [[3, 2, 0],
    [0, 1, 1]]

b in a
True
#Aligned with the expectation

c = [[300, 200, 0],
    [0, 100, 100]]

c in a
True
#Not quite what I expected

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

You could compare the input arrays for equality, which will perform broadcasted comparisons across all elements in a at each position in the last two axes against elements at corresponding positions in the second array. This will result in a boolean array of matches, in which we check for ALL matches across the last two axes and finally check for ANY match, like so –

((a==b).all(axis=(1,2))).any()

Sample run

1) Inputs :

In [68]: a
Out[68]: 
array([[[2, 3, 0],
        [1, 0, 1]],

       [[3, 2, 0],
        [0, 1, 1]],

       [[2, 2, 0],
        [1, 1, 1]],

       [[1, 3, 0],
        [2, 0, 1]],

       [[3, 1, 0],
        [0, 2, 1]]])

In [69]: b
Out[69]: 
array([[3, 2, 0],
       [0, 1, 1]])

2) Broadcasted elementwise comparisons :

In [70]: a==b
Out[70]: 
array([[[False, False,  True],
        [False, False,  True]],

       [[ True,  True,  True],
        [ True,  True,  True]],

       [[False,  True,  True],
        [False,  True,  True]],

       [[False, False,  True],
        [False, False,  True]],

       [[ True, False,  True],
        [ True, False,  True]]], dtype=bool)

3) ALL match across last two axes and finally ANY match :

In [71]: (a==b).all(axis=(1,2))
Out[71]: array([False,  True, False, False, False], dtype=bool)

In [72]: ((a==b).all(axis=(1,2))).any()
Out[72]: True

Following similar steps for c in a –

In [73]: c
Out[73]: 
array([[300, 200,   0],
       [  0, 100, 100]])

In [74]: ((a==c).all(axis=(1,2))).any()
Out[74]: False

Method 2

This question is fairly old, but if you’re like me, you might have thought there was no numpy equivalent for in by reading it.

Numpy 1.13 was released in 2017, and with it came the function isin(), which now nicely solves the problem:

import numpy as np

a = np.array([[[2, 3, 0],
               [1, 0, 1]],
              [[3, 2, 0],
               [0, 1, 1]],
              [[2, 2, 0],
               [1, 1, 1]],
              [[1, 3, 0],
               [2, 0, 1]],
              [[3, 1, 0],
               [0, 2, 1]]])

b = [[3, 2, 0],
     [0, 1, 1]]

print np.isin(b,a)

# [[ True  True  True]
#  [ True  True  True]]

c = [[300, 200, 0],
    [0, 100, 100]]

print np.isin(c,a)

# [[False False  True]
#  [ True False False]]

You’ll probably want to use np.all() at the end if you’re looking for the entire element to be in the test array.

Method 3

Since I do not have enough reputation to comment but the answer suggested by @rp0 using np.isin and then np.all will not work with numpy arrays.
Counter Example:

a = np.array([[[2, 3, 0],
               [1, 0, 1]],
              [[3, 2, 0],
               [0, 1, 1]],
              [[2, 2, 0],
               [1, 1, 1]],
              [[1, 3, 0],
               [2, 0, 1]],
              [[3, 1, 0],
               [0, 2, 1]]])
b = [[3, 2, 0],
     [0, 1, 1]]

c = [[3, 3, 3],
     [3, 3, 3]]

print(np.isin(b,a))
[[ True  True  True]
 [ True  True  True]]

print(np.isin(c,a))
[[ True  True  True]
 [ True  True  True]]

For completeness, I converted the array to python list using tolist and then used normal ‘in’ operator and its worked just as expected.

print(b in a.lolist())
True

print(c in a.tolist())
False

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0