Crop an image in the centre using PIL

问题:

How can I crop an image in the center? Because I know that the box is a 4-tuple defining the left, upper, right, and lower pixel coordinate but I don't know how to get these coordinates so it crops in the center.

回答1:

Assuming you know the size you would like to crop to (new_width X new_height):

import Image
im = Image.open(<your image>)
width, height = im.size   # Get dimensions

left = (width - new_width)/2
top = (height - new_height)/2
right = (width + new_width)/2
bottom = (height + new_height)/2

# Crop the center of the image
im = im.crop((left, top, right, bottom))

This will break if you attempt to crop a small image larger, but I'm going to assume you won't be trying that (Or that you can catch that case and not crop the image).

回答2:

One potential problem with the proposed solution is in the case there is an odd difference between the desired size, and old size. You can't have a half pixel on each side. One has to choose a side to put an extra pixel on.

If there is an odd difference for the horizontal the code below will put the extra pixel to the right, and if there is and odd difference on the vertical the extra pixel goes to the bottom.

import numpy as np

def center_crop(img, new_width=None, new_height=None):        

    width = img.shape[1]
    height = img.shape[0]

    if new_width is None:
        new_width = min(width, height)

    if new_height is None:
        new_height = min(width, height)

    left = int(np.ceil((width - new_width) / 2))
    right = width - int(np.floor((width - new_width) / 2))

    top = int(np.ceil((height - new_height) / 2))
    bottom = height - int(np.floor((height - new_height) / 2))

    if len(img.shape) == 2:
        center_cropped_img = img[top:bottom, left:right]
    else:
        center_cropped_img = img[top:bottom, left:right, ...]

    return center_cropped_img

回答3:

This is the function I was looking for:

from PIL import Image
im = Image.open("test.jpg")

crop_rectangle = (50, 50, 200, 200)
cropped_im = im.crop(crop_rectangle)

cropped_im.show()

Taken from another answer

回答4:

Crop center and around:

def im_crop_around(img, xc, yc, w, h):
    img_width, img_height = img.size  # Get dimensions
    left, right = xc - w / 2, xc + w / 2
    top, bottom = yc - h / 2, yc + h / 2
    left, top = round(max(0, left)), round(max(0, top))
    right, bottom = round(min(img_width - 0, right)), round(min(img_height - 0, bottom))
    return img.crop((left, top, right, bottom))


def im_crop_center(img, w, h):
    img_width, img_height = img.size
    left, right = (img_width - w) / 2, (img_width + w) / 2
    top, bottom = (img_height - h) / 2, (img_height + h) / 2
    left, top = round(max(0, left)), round(max(0, top))
    right, bottom = round(min(img_width - 0, right)), round(min(img_height - 0, bottom))
    return img.crop((left, top, right, bottom))

回答5:

I feel like the simplest solution that is most suitable for most applications is still missing. The accepted answer has an issue with uneven pixels and especially for ML algorithms, the pixel count of the cropped image is paramount.

In the following example, I would like to crop an image to 224/100, from the center. I do not care if the pixels are shifted to the left or right by 0.5, as long as the output picture will always be of the defined dimensions. It avoids the reliance on math.*.

from PIL import Image
import matplotlib.pyplot as plt


im = Image.open("test.jpg")
left = int(im.size[0]/2-224/2)
upper = int(im.size[1]/2-100/2)
right = left +224
lower = upper + 100

im_cropped = im.crop((left, upper,right,lower))
print(im_cropped.size)
plt.imshow(np.asarray(im_cropped))

The output is before cropping (not shown in code):

after:

The touples show the dimensions.