Comparing elements of numpy arrays in python

Question

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

unutbu · Answer 1 · 2011-07-08T19:06:05.563

8

The numpy way is to use np.allclose:

np.allclose(a,b)

Though for integers,

not (a-b).any()

is quicker.

edited Jul 08 '11 at 19:06

answered Jul 08 '11 at 19:01

unutbu

score 5 · Accepted Answer · answered Jul 08 '11 at 18:39

5

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)

answered Jul 08 '11 at 18:39

SiggyF

1

I think .any makes more sense for `!=` and .all for `==`. – SiggyF Jul 08 '11 at 19:44

score 3 · Answer 3 · answered Jul 08 '11 at 18:35

3

convert to a list:

if list(output[x][y]) != [150,25,75]

answered Jul 08 '11 at 18:35

jcomeau_ictx

You can also compare two arrays of the same shape, which gives you an array of True/False values. – Thomas K Jul 08 '11 at 18:39

score 0 · Answer 4 · answered Jul 08 '11 at 18:40

0

you could try:

a = output[x][y]
b = [150,25,75]

if not all([i == j for i,j in zip(a, b)]):

answered Jul 08 '11 at 18:40

TorelTwiddler

4 Answers4