Finding the index of a numpy array in a list

Question

import numpy as np
foo = [1, "hello", np.array([[1,2,3]]) ]

I would expect

foo.index( np.array([[1,2,3]]) ) 

to return

2

but instead I get

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

anything better than my current solution? It seems inefficient.

def find_index_of_array(list, array):
    for i in range(len(list)):
        if np.all(list[i]==array):
            return i

find_index_of_array(foo, np.array([[1,2,3]]) )
# 2

Answer 1

The reason for the error here is obviously because numpy's ndarray overrides == to return an array rather than a boolean.

AFAIK, there is no simple solution here. The following will work so long as the
np.all(val == array) bit works.

next((i for i, val in enumerate(lst) if np.all(val == array)), -1)

Whether that bit works or not depends critically on what the other elements in the array are and if they can be compared with numpy arrays.

Answer 2

How about this one?

arr = np.array([[1,2,3]])
foo = np.array([1, 'hello', arr], dtype=np.object)

# if foo array is of heterogeneous elements (str, int, array)
[idx for idx, el in enumerate(foo) if type(el) == type(arr)]

# if foo array has only numpy arrays in it
[idx for idx, el in enumerate(foo) if np.array_equal(el, arr)]

Output:

[2]

Note: This will also work even if foo is a list. I just put it as a numpy array here.