"In" operator for numpy arrays?

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

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.

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