# Comparing two NumPy arrays for equality, element-wise

`(A==B).all()`

test if all values of array (A==B) are True.

Note: maybe you also want to test A and B shape, such as `A.shape == B.shape`

**Special cases and alternatives** (from dbaupp's answer and yoavram's comment)

It should be noted that:

- this solution can have a strange behavior in a particular case: if either
`A`

or`B`

is empty and the other one contains a single element, then it return`True`

. For some reason, the comparison`A==B`

returns an empty array, for which the`all`

operator returns`True`

. - Another risk is if
`A`

and`B`

don't have the same shape and aren't broadcastable, then this approach will raise an error.

In conclusion, if you have a doubt about `A`

and `B`

shape or simply want to be safe: use one of the specialized functions:

`np.array_equal(A,B) # test if same shape, same elements valuesnp.array_equiv(A,B) # test if broadcastable shape, same elements valuesnp.allclose(A,B,...) # test if same shape, elements have close enough values`

The `(A==B).all()`

solution is very neat, but there are some built-in functions for this task. Namely `array_equal`

, `allclose`

and `array_equiv`

.

(Although, some quick testing with `timeit`

seems to indicate that the `(A==B).all()`

method is the fastest, which is a little peculiar, given it has to allocate a whole new array.)

If you want to check if two arrays have the same `shape`

AND `elements`

you should use `np.array_equal`

as it is the method recommended in the documentation.

Performance-wise don't expect that any equality check will beat another, as there is not much room to optimize

`comparing two elements`

. Just for the sake, i still did some tests.

`import numpy as npimport timeitA = np.zeros((300, 300, 3))B = np.zeros((300, 300, 3))C = np.ones((300, 300, 3))timeit.timeit(stmt='(A==B).all()', setup='from __main__ import A, B', number=10**5)timeit.timeit(stmt='np.array_equal(A, B)', setup='from __main__ import A, B, np', number=10**5)timeit.timeit(stmt='np.array_equiv(A, B)', setup='from __main__ import A, B, np', number=10**5)> 51.5094> 52.555> 52.761`

So pretty much equal, no need to talk about the speed.

The `(A==B).all()`

behaves pretty much as the following code snippet:

`x = [1,2,3]y = [1,2,3]print all([x[i]==y[i] for i in range(len(x))])> True`