Numpy.Array in Python list?

15
votes

I've got a list (used as a stack) of numpy arrays. Now I want to check if an array is already in the list. Had it been tuples for instance, I would simply have written something equivalent to (1,1) in [(1,1),(2,2)]. However, this does not work for numpy arrays; np.array([1,1]) in [np.array([1,1]), np.array([2,2])] is an error (ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()). The error message does not help here AFAIK, as it is referring to comparing arrays directly.

I have a hard time beliving it wouldn't be possible, but I suppose there's something I'm missing.

5
pythonnumpy
I have a hard time believing the simplest method requires 2 function calls and 1 list comprehension... This functionality seems common enough to warrant it's own built-in function - Rufus

5 Answers

25
votes

To test if an array equal to a is contained in the list my_list, use

any((a == x).all() for x in my_list)
2
votes

If you are looking for the exact same instance of an array in the stack regardless of whether the data is the same, then you need to this:

id(a) in map(id, my_list)
2
votes

Sven's answer is the right choice if you want to compare the actual content of the arrays. If you only want to check if the same instance is contained in the list you can use

any(a is x for x in mylist)

One benefit is that this will work for all kinds of objects.

0
votes

What about this: 

a = array([1, 1])

l = [np.array([1,1]), np.array([2,2])]
list(map(lambda x: np.array_equal(x, a), l)

[True, False]
-1
votes

You can convert the array into a list by using tolist() and then do the check:

my_list = [[1,1], [2,2]]

print(np.array([1,1]).tolist() in my_list)
print(np.array([1,2]).tolist() in my_list)