NumPy: Make filter array from a (N,2) shaped array

Question

I have the following array:

[[4 9]
 [5 4]
 ...
 [2 9]]

I want to filter this arr array, in a way, that I only have the elements of it, where both elements are between 0 and 7, and discard the rest. My solution, so far, has been to create a filter array, to index it with:

filter_array = ((arr >= 0) & (arr <= 7))

My problem is, this returns an array of the same shape as arr:

[[ True False]
 [ True True]
 ...
 [ True False]]

Which I can't use to index the original array in the way that I want. I want to discard the entire line, if any of the elements are not between the values I want:

#desired output:
[ False
  True
  ...
  False ]

I want to solve this in a "numpy-ish" manner, since the array is quite large, so performance is important. (I don't want to just iterate over it with some for loops)

Bruno Mello · Accepted Answer · 2020-04-02 11:06:37Z

1

You can sum in the axis=1 and see if it sums to 2:

filtered_array = (filtered_array.sum(1)==2)

Another way, where you can use and operator:

filtered_array = filter_array[:,0] & filter_array[:,1]

Lastly:

filter_array = filter_array.all(1)

This latter for me is the best way, but you can choose what works for you.

answered Apr 2, 2020 at 11:06

Bruno Mello

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2 Comments

FBruzzesi Over a year ago

In the first solution you proposed, you can just multiply each row to avoid checking the sum, namely filtered_array.prod(1) will give you 1 as result if and only if both elements are True.

Bruno Mello Over a year ago

@FBruzzesi, That's true, I guess it is kind of the same thing .all(1) do under the hoods.

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