numpy efficient array multiplication

Question

I have a three dimensional array img of shape [1200,1600,3] and a two dimensional array labels of shape [1200,1600]. The first array is from an image, the second one is from labels in the image. Location [i,j] in the img array corresponds to an image pixel. I want to create a new array of the same dimension as the img array, such that for the pixels with label 0, the original array is unchanged, but all other pixels are whitened (255,255,255).

The code I am using is:

 import numpy as np

 newimg=np.zeros((img.shape[0],img.shape[1],img.shape[2]))
 for i in range(0,img.shape[0]):
     for j in range(0,img.shape[1]):
          if labels[i][j]==0:
             newimg[i][j]=img[i][j]
     else:
         newimg[i][j]=np.array([255,255,255])

Is there a faster way of doing this?

Joe Kington · Accepted Answer · 2014-10-02 21:03:13Z

3

Generally speaking, you'd do something similar to:

newimg = img.copy()
newimg[labels != 0, :] = 255

or alternatively:

newimg = np.where(labels[..., None] != 0, img, 255)

edited Oct 2, 2014 at 21:03

answered Oct 2, 2014 at 20:53

Joe Kington

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2 Comments

Joe Kington Over a year ago

@DSM - Your answer is actually quite a bit more detailed than mine. Feel free to undelete it!

DSM Over a year ago

I think we're whitening the non-zero labels.

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