Faster way to loop through each pixel in an image in python?-CodePudding

Need to run though an image in python and essentially calculate the average location of all acceptable pixels within a certain boundary. The image is black and white. The acceptable pixels have a value of 255 and the unacceptable pixels have a value of zero. The image is something like 2592x1944 and takes maybe 15 seconds to run. This will need to be looped several times. Is there a faster way to do this?

goodcount = 0
sumx=0
sumy=0
xindex=0
yindex=0

for row in mask:
    yindex =1
    xindex=0
    for n in row:
        xindex =1
        if n == 255:
            goodcount  = 1
            sumx  = xindex
            sumy  = yindex

if goodcount != 0:

    y = int(sumy / goodcount)
    x = int(sumx / goodcount)

CodePudding user response：

np.where() will return arrays of the indices where a condition is true, which we can average (adding 1 to match your indexing) and cast to an integer:

if np.any(mask == 255):
    y, x = [int(np.mean(indices   1)) for indices in np.where(mask == 255)]

CodePudding user response：

I think you are looking for the centroid of the white pixels, which OpenCV will find for you very fast:

#!/usr/bin/env python3

import cv2
import numpy as np

# Load image as greyscale
im = cv2.imread('star.png')

# Make greyscale version
grey = cv2.cvtColor(im, cv2.COLOR_BGR2GRAY)

# Ensure binary
_, grey = cv2.threshold(grey,127,255,0)

# Calculate moments
M = cv2.moments(grey)

# Calculate x,y coordinates of centre
cX = int(M["m10"] / M["m00"])
cY = int(M["m01"] / M["m00"])

# Mark centroid with circle, and tell user
cv2.circle(im, (cX, cY), 10, (0, 0, 255), -1)
print(f'Centroid at location: {cX},{cY}')

# Save output file
cv2.imwrite('result.png', im)

Sample Output

Centroid at location: 1224,344

That takes 381 microseconds on the 1692x816 image above, and rises to 1.38ms if I resize the image to the dimensions of your image... I'm calling that a 10,800x speedup