How to crop an image in OpenCV using Python

It's very simple. Use numpy slicing.

import cv2
img = cv2.imread("lenna.png")
crop_img = img[y:y+h, x:x+w]
cv2.imshow("cropped", crop_img)
cv2.waitKey(0)

i had this question and found another answer here: copy region of interest

If we consider (0,0) as top left corner of image called im with left-to-right as x direction and top-to-bottom as y direction. and we have (x1,y1) as the top-left vertex and (x2,y2) as the bottom-right vertex of a rectangle region within that image, then:

roi = im[y1:y2, x1:x2]

here is a comprehensive resource on numpy array indexing and slicing which can tell you more about things like cropping a part of an image. images would be stored as a numpy array in opencv2.

:)

This code crops an image from x=0,y=0 to h=100,w=200.

import numpy as np
import cv2

image = cv2.imread('download.jpg')
y=0
x=0
h=100
w=200
crop = image[y:y+h, x:x+w]
cv2.imshow('Image', crop)
cv2.waitKey(0) 

Note that, image slicing is not creating a copy of the cropped image but creating a pointer to the roi. If you are loading so many images, cropping the relevant parts of the images with slicing, then appending into a list, this might be a huge memory waste.

Suppose you load N images each is >1MP and you need only 100x100 region from the upper left corner.

Slicing:

X = []
for i in range(N):
    im = imread('image_i')
    X.append(im[0:100,0:100]) # This will keep all N images in the memory. 
                              # Because they are still used.

Alternatively, you can copy the relevant part by .copy(), so garbage collector will remove im.

X = []
for i in range(N):
    im = imread('image_i')
    X.append(im[0:100,0:100].copy()) # This will keep only the crops in the memory. 
                                     # im's will be deleted by gc.

After finding out this, I realized one of the comments by user1270710 mentioned that but it took me quite some time to find out (i.e., debugging etc). So, I think it worths mentioning.