I want to compare two 1x3 arrays such as:
if output[x][y] != [150,25,75]
(output here is a 3x3x3 so output[x][y] is only a 1x3).
I'm getting an error that says:
ValueError: The truth value of an array with more than one element is ambiguous.
Does that mean I need to do it like:
if output[y][x][0] == 150 and output[y][x][1] == 25 and output[y][x][2] == 75:
or is there a cleaner way to do this?
I'm using Python v2.6
The numpy way is to use np.allclose:
np.allclose(a,b)
Though for integers,
not (a-b).any()
is quicker.
You should also get the message:
Use a.any() or a.all()
This means that you can do the following:
if (output[x][y] != [150,25,75]).all():
That is because the comparison of 2 arrays or an array with a list results in a boolean array. Something like:
array([ True, True, True], dtype=bool)
!= and .all for ==.convert to a list:
if list(output[x][y]) != [150,25,75]
you could try:
a = output[x][y]
b = [150,25,75]
if not all([i == j for i,j in zip(a, b)]):
