Check value of numpy array at index

Question

I have recently started working in numpy. I am trying to test if a 2d array contains a specific subarray. The code below returns an error. How can I fix this?

import numpy as np

testArray = np.array([[None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0]])

for i in range(len(testArray)):
    if (testArray[i] == [None, 0]):
        print(i)

score 1 · Answer 1 · answered Dec 10 '18 at 20:04

Without iterating, you can use all:

>>> testArray[(testArray == [None,0]).all(1)]
array([[None, 0],
       [None, 0],
       [None, 0],
       [None, 0],
       [None, 0],
       [None, 0],
       [None, 0],
       [None, 0]], dtype=object)

Or if you just want to see whether that subarray exists, use any in addition:

>>> (testArray == [None,0]).all(1).any()
True

genhernandez · Answer 2 · 2018-12-10T20:04:28.327

0

The error you are seeing is a Value Error. When you compare the numpy arrays, you get an array consisting of boolean values. The problem was that you were using that array to as your condition, which resulted in the value error since the truth-value of arrays is ambiguous. You can resolve by using .any or .all, depending on whether you care if all the elements are present in your array.

Try this:

testArray = np.array([[None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0]]) 

for i in range(len(testArray)): 
    containsValue = (testArray[i] == [None, 0]).all()
    if (containsValue): 
        print(i)

edited Dec 10 '18 at 20:04

answered Dec 10 '18 at 20:01

genhernandez

453
1
5
19

You may as well be using lists, this doesn't use any numpy features. It will actually be slower than using lists. – roganjosh Dec 10 '18 at 20:03
They didn't ask to optimize for speed, they just said they had an error in their code. – genhernandez Dec 10 '18 at 20:05
Presumably they would not be layering in an external dependency to make their code _slower_ than native Python. – roganjosh Dec 10 '18 at 20:06
I can't pretend to know what the asker is thinking. They had an error in their code and my answer resolves said error. – genhernandez Dec 10 '18 at 20:08

not_a_bot_no_really_82353 · Answer 3 · 2018-12-10T20:13:07.493

0

Well, one of your '[None, 0]' is a list which you are comparing to an array which doesn't make much sense. So if you want to fix the code then you can access the values by:

import numpy as np

testArray = np.array([[None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0], [None, 0]])

for i in range(len(testArray)):
    if (testArray[i][0] == None and testArray[i][1] == 0):
        print(i)

edited Dec 10 '18 at 20:13

answered Dec 10 '18 at 20:03

not_a_bot_no_really_82353

589
5
18

As the other answer, there's absolutely no use for numpy in this approach – roganjosh Dec 10 '18 at 20:05
Where did they ask to use numpy in the approach? – genhernandez Dec 10 '18 at 20:06
It's a numpy array and they state that they're learning numpy? – roganjosh Dec 10 '18 at 20:07
2

one is a list the other one a numpy array, I will adjust my answer – not_a_bot_no_really_82353 Dec 10 '18 at 20:11
He specifically stated that he is using numpy. I just fixed the code, I don't know why he wants to do it that way. But he never told us what he actually wants to achieve so, I do not know. – not_a_bot_no_really_82353 Dec 10 '18 at 20:14
I am connecting dots and suggesting that the appropriate answer is one that makes use of vectorization. Treating numpy arrays as lists is almost always terrible performance. It's not unreasonable to assume that they are looking for improved efficiency by the fact they're using the library. – roganjosh Dec 10 '18 at 20:16
Might be true, but I believe he is just messing around with numpy to learn the stuff. – not_a_bot_no_really_82353 Dec 10 '18 at 20:20

Check value of numpy array at index

3 Answers3