From numpy array, get indices of values list

Question

like this:

>> arr = np.array([[0, 50], [100, 150], [200, 250]]) 
>>> values = [100, 200, 300] 

>>> arr in values

expect:

array([[False, False],
       [ True, False],
       [ True, False]])

result:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I wrote following code and it works, but this code cannot accept changing length of list

(arr==values[0]) | (arr==values[1]) | (arr==values[2])

Answer 1

Use np.isin:

import numpy as np

arr = np.array([[0, 50], [100, 150], [200, 250]])
values = [100, 200, 300]

np.isin(arr, values)

result:

array([[False, False],
       [ True, False],
       [ True, False]])

Answer 2

This should work but only for 2 level depth:

import numpy as np


def is_in(arr, values):
    agg = []
    for a in arr:
        if isinstance(a, list):
            result = is_in(a, values)
            agg.append(result)
        else:
            result = np.isin(arr, values)
            return result
    return agg

arr = np.array([[0, 50], [100, 150], [200, 250]])
values = [100, 200, 300]

print(is_in(arr, values))

Answer 3

also change name of that variable from values to something different to valuess for example.

if valuess in arr.values :
print(valuess)

OR

Use a lambda function.

Let's say you have an array:

nums = [0,1,5]
Check whether 5 is in nums:

(len(filter (lambda x : x == 5, nums)) > 0)