Iterations through pixels in an image are terribly slow with python (OpenCV)

Question

I am aware of iterating through pixels and accessing their values using OpenCV with C++. Now, i am trying to learn python myself and i tried to do the same thing in python. But when i am running the following code, it takes a lot of time (~7-10 seconds) to display the image. And the script keeps running on for few more seconds even after displaying the image.

I found a similar question here at SO but i am not able to understand how do i use numpy in my case (because i am a beginner in python) and whether or not it is really required?

Code Explanation: I am just trying to put the black pixels on the left and right side of the image.

import numpy as np
import cv2 as cv

#reading an image
img = cv.imread('image.jpg')
height, width, depth = img.shape

for i in range(0, height):
    for j in range(0, (width/4)):
        img[i,j] = [0,0,0]  

for i in range(0, height):
    for j in range(3*(width/4), width):
        img[i,j] = [0,0,0]        

cv.imshow('image',img)

cv.waitKey(0)

roippi · Accepted Answer · 2014-10-19 21:22:50Z

(note: I'm not familiar with opencv, but this appears to be a numpy issue)

The "terribly slow" part is that you're looping in python bytecode, rather than letting numpy loop at C speed.

Try directly assigning to a (3-dimensional) slice that masks the region you want to zero out.

import numpy as np

example = np.ones([500,500,500], dtype=np.uint8)

def slow():
     img = example.copy()
     height, width, depth = img.shape
     for i in range(0, height):             #looping at python speed...
         for j in range(0, (width//4)):     #...
             for k in range(0,depth):       #...
                 img[i,j,k] = 0
     return img


def fast():
     img = example.copy()
     height, width, depth = img.shape
     img[0:height, 0:width//4, 0:depth] = 0 # DO THIS INSTEAD
     return img 

np.alltrue(slow() == fast())
Out[22]: True

%timeit slow()
1 loops, best of 3: 6.13 s per loop

%timeit fast()
10 loops, best of 3: 40 ms per loop

The above shows zeroing out the left side; doing the same for the right side is an exercise for the reader.

If the numpy slicing syntax trips you up, I suggest reading through the indexing docs.

thanks for youe reply. As i have mentioned that i am completely new to python so, it would be very helpful for me if you could explain the basic idea of your code.
I'm not sure what you're asking about; I explained that you need to assign to a 3D slice, and commented the line where I do that. Be more specific about what you don't understand.
Or more simply, img[:,:width/4,:] = 0, and for the other edge, img[:,-width/4:,:] = 0
And if you're feeling like doing it in one line, img[:,np.r_[:width/4,-width/4:],:] = 0
In this line: img[0:height, 0:width//4, 0:depth] = 0 , can 0 be a custom function to be applied for every pixel?

Collectives™ on Stack Overflow

Iterations through pixels in an image are terribly slow with python (OpenCV)

1 Answer 1

5 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

5 Comments

Linked

Related