Get elements of a numpy array that satisfy some condition

Question

I have a numpy array full of integers, let's say

[[1,2],[3,4]]

I want to get a binary array containing 1 if the element satisfies belongs to a list, and 0 otherwise.

If I write

condition = arr == 2

I get

[[false, true], [false, false]]

which is what I want.

But what if I want to keep the elements 2 and 3 ? I tried

condition = arr in [2,3]

but it doesn't work, I get a

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I would like to do that for any possible list.

Is there any efficient way to do that? I know how to do it brutally, but I'm need to be efficient.

Thank you very much!

`condition = arr == 2` returns that: `array([[False, True], [False, False]])` — Mihai Alexandru-Ionut, Oct 03 '18 at 14:25
Check [this](https://stackoverflow.com/questions/10062954/valueerror-the-truth-value-of-an-array-with-more-than-one-element-is-ambiguous) — Irfanuddin, Oct 03 '18 at 14:25

score 3 · Accepted Answer · answered Oct 03 '18 at 14:33

3

how about:

np.isin(arr,[2,3])

output:

array([[False,  True],
       [ True, False]])

answered Oct 03 '18 at 14:33

Binyamin Even

That's perfect, thank you! – Adrien Nivaggioli Oct 03 '18 at 14:37
@AdrienNivaggioli, have you tried my answer in order to see if it's working ? – Mihai Alexandru-Ionut Oct 03 '18 at 14:47
@MihaiAlexandru-Ionut of course your solution is working. It is also mentioned in the comments. But I think (with all modesty:) ) that my solution is more elegant, and requires less code, especially if the second list is large. – Binyamin Even Oct 03 '18 at 14:49

score 0 · Answer 2 · answered Oct 03 '18 at 14:48

0

My 2 cents. :-)

arr = np.array([[1,2],[3,4]])
np.logical_or(arr==2,arr==3)

output:

array([[False,  True],
      [ True, False]])

answered Oct 03 '18 at 14:48

Sidon

2 Answers2