fastest way to iterate over all pixels of an image in python

问题

i have already read an image as an array :

import numpy as np
from scipy import misc
face1=misc.imread('face1.jpg')

face1 dimensions are (288, 352, 3)

i need to iterate over every single pixel and populate a y column in a training set i took the following approach :

Y_training = np.zeros([1,1],dtype=np.uint8)

for i in range(0, face1.shape[0]): # We go over rows number 
    for j in range(0, face1.shape[1]): # we go over columns number
        if np.array_equiv(face1[i,j],[255,255,255]):
           Y_training=np.vstack(([0], Y_training))#0 if blank
        else:
           Y_training=np.vstack(([1], Y_training))

b = len(Y_training)-1
Y_training = Y_training[:b]
np.shape(Y_training)`

Wall time: 2.57 s

As i need to do above process for about 2000 images is there any faster approach where we could decrease running time to milliseconds or naonseconds

回答1:

You can use broadcasting to perform broadcasted comparison against the white pixel : [255, 255, 255] and ALL reduce each row with .all(axis=-1) and finally convert to int dtype. This would give us the output you would have right after exiting the loop.

Thus, one implementation would be -

(~((face1 == [255,255,255]).all(-1).ravel())).astype(int)

Alternatively, a bit more compact version -

1-(face1 == [255,255,255]).all(-1).ravel()

来源：https://stackoverflow.com/questions/45353997/fastest-way-to-iterate-over-all-pixels-of-an-image-in-python