使用python(OpenCV)在图像中进行像素迭代非常慢 [英] Iterations through pixels in an image are terribly slow with python (OpenCV)

问题描述

我知道使用OpenCV和C ++遍历像素并访问它们的值.现在，我正在尝试自己学习python ，我试图在python中做同样的事情.但是，当我运行以下代码时，需要花费很多时间(〜7-10秒)来显示图像.即使显示图像后，脚本也可以继续运行几秒钟.

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.

我在SO的> ，但我无法理解在我的情况下如何使用numpy(因为我是python的初学者)以及是否真的需要它?

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?

代码说明:我只是想在图像的左侧和右侧放置黑色像素.

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)

推荐答案

(注意:我对opencv不熟悉，但这似乎是一个numpy问题)

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

非常慢"的部分是您要在python字节码中循环，而不是让numpy以C速度循环.

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

尝试直接分配给遮盖要归零区域的3维切片.

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

上面显示的是对左侧进行归零；在右侧进行相同的操作对于读者来说是一种练习.

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

如果numpy切片语法使您不满意，建议您通读

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

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