Deleting row in numpy array based on condition

Question

I have a 2D numpy array of shape [6,2] and I want to remove the subarrays with the third element containing 0.

array([[0, 2, 1], #Input
       [0, 1, 1],
       [1, 1, 0],
       [1, 0, 2],
       [0, 2, 0],
       [2, 1, 2]])

array([[0, 2, 1], #Output
       [0, 1, 1],
       [1, 0, 2],
       [2, 1, 2]])

My code is positives = gt_boxes[np.where(gt_boxes[range(gt_boxes.shape[0]),2] != 0)]

It works but is there a simplified method to this?

Chris · Accepted Answer · 2017-08-01 02:09:37Z

14

You can use boolean indexing.

In [413]: x[x[:, -1] != 0]
Out[413]: 
array([[0, 2, 1],
       [0, 1, 1],
       [1, 0, 2],
       [2, 1, 2]])

x[:, -1] will retrieve the last column
x[:, -1] != 0 returns a boolean mask
Use the mask to index into the original array

edited Aug 1, 2017 at 2:09

Chris

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answered Aug 1, 2017 at 2:06

cs95

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2 Comments

Chris Over a year ago

Dang it. I didn't get to finish writing my solution, and it was still longer than this anyways. +1

cs95 Over a year ago

@ChristianDean Yo that's how I feel 99% of the time in this tag :p

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