How to check if an array is in another array in Python

Question

I thought that in would be good for this but it returns true in places where it shouldn't. For example:

import numpy as np

a = np.array([])

for i in range(3):
    for j in range(3):
        a = np.append(a,[i,j])
a = np.reshape(a,(9,2))
print(a)

print([[0,40]] in a)

will print true. I cannot understand why it does this... is it because 0 is in the list? I'd like to have something that only prints true if the entire array is in the list.

I want to have my list

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

and only return true if exactly [0,x] (same shape same order) is in it.

Side note: you can more efficiently construct a by three = np.arange(3.0); np.array([np.repeat(three, 3), np.tile(three, 3)]).T. — John Zwinck
– John Zwinck, Commented Jun 10, 2018 at 10:40

John Zwinck · Accepted Answer · 2018-06-10 10:50:46Z

6

You can do it this way:

([0, 40] == a).all(1).any()

The first step is to compute a 2D boolean array of where the matches are. Then you find the rows where all elements are true. Then you check if any rows are fully matching.

answered Jun 10, 2018 at 10:50

John Zwinck

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Amir Dezhkam · Accepted Answer · 2018-06-10 15:01:58Z

1

This code can help you:

my_list = [0, 40]
print(all(b in a for b in my_list))

answered Jun 10, 2018 at 15:01

Amir Dezhkam

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Comments

Katie · Accepted Answer · 2020-01-28 05:28:13Z

0

You can use np.isin(a_array, b_array) ,it returned a boolean array. for example:

import numpy as np
a_array =np.array([1,2,3,4])
b_array = np.array([3,4,5])
bool_a_b = np.isin(a_array, b_array)
print(bool_a_b)

Output:

[False False True True]

answered Jan 28, 2020 at 5:28

Katie

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