CODEWITHSUNDEEP
python-2.7opencv
Richie Close
Richie Close

Reputation: 77

How do I crop an image in opencv and python based on percentage

I want to take an image and crop it by 75% from the center of the original image. To be honest I'm at a bit of a loss on it. I had thought about getting the original image size

height, width = image.shape

getting a percentage value and cropping:

cropped_y_start = int((height * 0.75))
cropped_y_end = int((height * 0.25))
cropped_x_start = int((width * 0.75))
cropped_x_end = int((width * 0.25))

print cropped_x_start
print cropped_x_end

crop_img = image[cropped_y_start:cropped_y_end, cropped_x_start:cropped_x_end]

There are multiple problems with this but the main one being its not based off the center of the image. Any advice on this would be much appreciated.

Upvotes: 3

Views: 6953

Answers (1)

Kyle Hounslow
Kyle Hounslow

Reputation: 104

A simple way is to first get your scaled width and height, and then crop from the image center to plus/minus the scaled width and height divided by 2.
Here is an example:

import cv2

def crop_img(img, scale=1.0):
    center_x, center_y = img.shape[1] / 2, img.shape[0] / 2
    width_scaled, height_scaled = img.shape[1] * scale, img.shape[0] * scale
    left_x, right_x = center_x - width_scaled / 2, center_x + width_scaled / 2
    top_y, bottom_y = center_y - height_scaled / 2, center_y + height_scaled / 2
    img_cropped = img[int(top_y):int(bottom_y), int(left_x):int(right_x)]
    return img_cropped

img = cv2.imread('lena.jpg')
img_cropped = crop_img(img, 0.75)

Output:

Original

Original

Cropped by 75%

Cropped by 75%

Upvotes: 9

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