'Automatically cropping an image with python/PIL

Can anyone help me figure out what's happening in my image auto-cropping script? I have a png image with a large transparent area/space. I would like to be able to automatically crop that space out and leave the essentials. Original image has a squared canvas, optimally it would be rectangular, encapsulating just the molecule.

here's the original image: Original Image

Doing some googling i came across PIL/python code that was reported to work, however in my hands, running the code below over-crops the image.

import Image
import sys

image=Image.open('L_2d.png')
image.load()

imageSize = image.size
imageBox = image.getbbox()

imageComponents = image.split()

rgbImage = Image.new("RGB", imageSize, (0,0,0))
rgbImage.paste(image, mask=imageComponents[3])
croppedBox = rgbImage.getbbox()
print imageBox
print croppedBox
if imageBox != croppedBox:
    cropped=image.crop(croppedBox)
    print 'L_2d.png:', "Size:", imageSize, "New Size:",croppedBox
    cropped.save('L_2d_cropped.png')

the output is this:script's output

Can anyone more familiar with image-processing/PLI can help me figure out the issue?



Solution 1:[1]

You can use numpy, convert the image to array, find all non-empty columns and rows and then create an image from these:

import Image
import numpy as np

image=Image.open('L_2d.png')
image.load()

image_data = np.asarray(image)
image_data_bw = image_data.max(axis=2)
non_empty_columns = np.where(image_data_bw.max(axis=0)>0)[0]
non_empty_rows = np.where(image_data_bw.max(axis=1)>0)[0]
cropBox = (min(non_empty_rows), max(non_empty_rows), min(non_empty_columns), max(non_empty_columns))

image_data_new = image_data[cropBox[0]:cropBox[1]+1, cropBox[2]:cropBox[3]+1 , :]

new_image = Image.fromarray(image_data_new)
new_image.save('L_2d_cropped.png')

The result looks like cropped image

If anything is unclear, just ask.

Solution 2:[2]

Install Pillow

pip install Pillow

and use as

from PIL import Image
    
image=Image.open('L_2d.png')

imageBox = image.getbbox()
cropped = image.crop(imageBox)
cropped.save('L_2d_cropped.png')

When you search for boundaries by mask=imageComponents[3], you search only by blue channel.

Solution 3:[3]

I tested most of the answers replied in this post, however, I was ended up my own answer. I used anaconda python3.

from PIL import Image, ImageChops

def trim(im):
    bg = Image.new(im.mode, im.size, im.getpixel((0,0)))
    diff = ImageChops.difference(im, bg)
    diff = ImageChops.add(diff, diff, 2.0, -100)
    #Bounding box given as a 4-tuple defining the left, upper, right, and lower pixel coordinates.
    #If the image is completely empty, this method returns None.
    bbox = diff.getbbox()
    if bbox:
        return im.crop(bbox)

if __name__ == "__main__":
    bg = Image.open("test.jpg") # The image to be cropped
    new_im = trim(bg)
    new_im.show()

Solution 4:[4]

Here's another version using pyvips.

import sys
import pyvips

image = pyvips.Image.new_from_file(sys.argv[1])
left, top, width, height = image.find_trim(threshold=2, background=[255, 255, 255])
image = image.crop(left, top, width, height)
image.write_to_file(sys.argv[2])

The pyvips trimmer is useful for photographic images. It does a median filter, subtracts the background, finds pixels over the threshold, and removes up to the first and last row and column outside this set. The median and threshold mean it is not thrown off by things like JPEG compression, where noise or invisible compression artefacts can confuse other trimmers.

If you don't supply the background argument, it uses the pixel at (0, 0). threshold defaults to 10, which is about right for JPEG.

Here it is running on an 8k x 8k pixel NASA earth image:

$ time ./trim.py /data/john/pics/city_lights_asia_night_8k.jpg x.jpg
real    0m1.868s
user    0m13.204s
sys     0m0.280s
peak memory: 100mb

Before:

Earth at night before crop

After:

Earth after crop

There's a blog post with some more discussion here.

Solution 5:[5]

This is an improvement over snew's reply, which works for transparent background. With mathematical morphology we can make it work on white background (instead of transparent), with the following code:

from PIL import Image
from skimage.io import imread
from skimage.morphology import convex_hull_image
from skimage.color import rgb2gray
im = imread('L_2d.jpg')
plt.imshow(im)
plt.title('input image')
plt.show()
# create a binary image
im1 = 1 - rgb2gray(im)
threshold = 0.5
im1[im1 <= threshold] = 0
im1[im1 > threshold] = 1
chull = convex_hull_image(im1)
plt.imshow(chull)
plt.title('convex hull in the binary image')
plt.show()
imageBox = Image.fromarray((chull*255).astype(np.uint8)).getbbox()
cropped = Image.fromarray(im).crop(imageBox)
cropped.save('L_2d_cropped.jpg')
plt.imshow(cropped)
plt.show()

enter image description here enter image description here enter image description here

Solution 6:[6]

pilkit already contains processor for automatic cropping TrimBorderColor. SOmething like this should work:

from pilkit.lib import Image
from pilkit.processors import TrimBorderColor

img = Image.open('/path/to/my/image.png')
processor = TrimBorderColor()
new_img = processor.process(img)

https://github.com/matthewwithanm/pilkit/blob/b24990167aacbaab3db6d8ec9a02f9ad42856898/pilkit/processors/crop.py#L33

Solution 7:[7]

Came across this post recently and noticed the PIL library has changed. I re-implemented this with openCV:

import cv2

def crop_im(im, padding=0.1):
    """
    Takes cv2 image, im, and padding % as a float, padding,
    and returns cropped image.
    """
    bw = cv2.cvtColor(im, cv2.COLOR_BGR2GRAY)
    rows, cols = bw.shape
    non_empty_columns = np.where(bw.min(axis=0)<255)[0]
    non_empty_rows = np.where(bw.min(axis=1)<255)[0]
    cropBox = (int(min(non_empty_rows) * (1 - padding)),
                int(min(max(non_empty_rows) * (1 + padding), rows)),
                int(min(non_empty_columns) * (1 - padding)),
                int(min(max(non_empty_columns) * (1 + padding), cols)))
    cropped = im[cropBox[0]:cropBox[1]+1, cropBox[2]:cropBox[3]+1 , :]

    return cropped

im = cv2.imread('testimage.png')
cropped = crop_im(im)
cv2.imshow('', cropped)
cv2.waitKey(0)

Solution 8:[8]

I know that this post is old but, for some reason, none of the suggested answers worked for me. So I hacked my own version from existing answers:

import Image
import numpy as np
import glob
import shutil
import os

grey_tolerance = 0.7 # (0,1) = crop (more,less)

f = 'test_image.png'
file,ext = os.path.splitext(f)

def get_cropped_line(non_empty_elms,tolerance,S):
    if (sum(non_empty_elms) == 0):
        cropBox = ()
    else:
        non_empty_min = non_empty_elms.argmax()
        non_empty_max = S - non_empty_elms[::-1].argmax()+1
        cropBox = (non_empty_min,non_empty_max)
    return cropBox

def get_cropped_area(image_bw,tol):
    max_val = image_bw.max()
    tolerance = max_val*tol
    non_empty_elms = (image_bw<=tolerance).astype(int)
    S = non_empty_elms.shape
    # Traverse rows
    cropBox = [get_cropped_line(non_empty_elms[k,:],tolerance,S[1]) for k in range(0,S[0])]
    cropBox = filter(None, cropBox)
    xmin = [k[0] for k in cropBox]
    xmax = [k[1] for k in cropBox]
    # Traverse cols
    cropBox = [get_cropped_line(non_empty_elms[:,k],tolerance,S[0]) for k in range(0,S[1])]
    cropBox = filter(None, cropBox)
    ymin = [k[0] for k in cropBox]
    ymax = [k[1] for k in cropBox]
    xmin = min(xmin)
    xmax = max(xmax)
    ymin = min(ymin)
    ymax = max(ymax)
    ymax = ymax-1 # Not sure why this is necessary, but it seems to be.
    cropBox = (ymin, ymax-ymin, xmin, xmax-xmin)
    return cropBox

def auto_crop(f,ext):
    image=Image.open(f)
    image.load()
    image_data = np.asarray(image)
    image_data_bw = image_data[:,:,0]+image_data[:,:,1]+image_data[:,:,2]
    cropBox = get_cropped_area(image_data_bw,grey_tolerance)
    image_data_new = image_data[cropBox[0]:cropBox[1]+1, cropBox[2]:cropBox[3]+1 , :]
    new_image = Image.fromarray(image_data_new)
    f_new = f.replace(ext,'')+'_cropped'+ext
    new_image.save(f_new)

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 ryanjdillon
Solution 2 Nico Schlömer
Solution 3 Richard
Solution 4
Solution 5 mustafa candan
Solution 6 igo
Solution 7
Solution 8 Charles