How to crop or remove white background from an image

Question

I am trying to compare images using OpenCV and Python.

Consider these images:

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Both feature an identical pair of shoes, set to a white background. The only difference being that the first has a taller background than the second.

I want to know how to programmatically crop the white backgrounds of both so that I'm left with only the pair of shoes.

I must add that it won't be possible for me to manually crop the backgrounds.

Answer 1

You requirement in the comment: The shoes are on a white background. I would like to completely get rid of the border; as in be left with a rectangular box with either a white or a transparent background, having the length and width of the shoes in the picture.

Then my steps to crop the target regions:

1. Convert to gray, and threshold
2. Morph-op to remove noise
3. Find the max-area contour
4. Crop and save it

#!/usr/bin/python3
# Created by Silencer @ Stackoverflow 
# 2018.01.23 14:41:42 CST
# 2018.01.23 18:17:42 CST
import cv2
import numpy as np

## (1) Convert to gray, and threshold
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
th, threshed = cv2.threshold(gray, 240, 255, cv2.THRESH_BINARY_INV)

## (2) Morph-op to remove noise
kernel = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (11,11))
morphed = cv2.morphologyEx(threshed, cv2.MORPH_CLOSE, kernel)

## (3) Find the max-area contour
cnts = cv2.findContours(morphed, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[-2]
cnt = sorted(cnts, key=cv2.contourArea)[-1]

## (4) Crop and save it
x,y,w,h = cv2.boundingRect(cnt)
dst = img[y:y+h, x:x+w]
cv2.imwrite("001.png", dst)

Result:

Answer 2

Kinght's solution works well. In my case, I also have CMYK images. When I crop them, I get incorrect (vivid colors) output. And it seems OpenCV doesn't support CMYK. So I needed a way to convert CMYK images to RGB, and then open it with OpenCV. This is how I handled it:

import cv2
import numpy

from PIL import Image
from PIL import ImageCms

# force opening truncated/corrupt image files
from PIL import ImageFile
ImageFile.LOAD_TRUNCATED_IMAGES = True

img = "shoes.jpg"

img = Image.open(img)
if img.mode == "CMYK":
    # color profiles can be found at C:\Program Files (x86)\Common Files\Adobe\Color\Profiles\Recommended
    img = ImageCms.profileToProfile(img, "USWebCoatedSWOP.icc", "sRGB_Color_Space_Profile.icm", outputMode="RGB")
# PIL image -> OpenCV image; see https://stackoverflow.com/q/14134892/2202732
img = cv2.cvtColor(numpy.array(img), cv2.COLOR_RGB2BGR)

## (1) Convert to gray, and threshold
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
th, threshed = cv2.threshold(gray, 240, 255, cv2.THRESH_BINARY_INV)

## (2) Morph-op to remove noise
kernel = cv2.getStructuringElement(cv2.MORPH_ELLIPSE, (11,11))
morphed = cv2.morphologyEx(threshed, cv2.MORPH_CLOSE, kernel)

## (3) Find the max-area contour
cnts = cv2.findContours(morphed, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[-2]
cnt = sorted(cnts, key=cv2.contourArea)[-1]

## (4) Crop and save it
x,y,w,h = cv2.boundingRect(cnt)
dst = img[y:y+h, x:x+w]

# add border/padding around the cropped image
# dst = cv2.copyMakeBorder(dst, 10, 10, 10, 10, cv2.BORDER_CONSTANT, value=[255,255,255])

cv2.imshow("image", dst)
cv2.waitKey(0)
cv2.destroyAllWindows()

# create/write to file
# cv2.imwrite("001.png", dst)

Answer 3

by PIL you can convert white background to transparent:

from PIL import Image
  
def convertImage():
    img = Image.open("hi.png")
    img = img.convert("RGBA")
  
    datas = img.getdata()
  
    newData = []
  
    for item in datas:
        if item[0] == 255 and item[1] == 255 and item[2] == 255:
            newData.append((255, 255, 255, 0))
        else:
            newData.append(item)
  
    img.putdata(newData)
    img.save("./New.png", "PNG")
    print("Successful")
  
convertImage()

and here is the output sample:

Answer 4

This link worked perfectly for me for a similar problem, although it uses PIL. Note that it will result in a rectangular image, bounded by the top/right/bottom/left-most pixels that are not white. In your case, it should give identical images with the same size.

I am guessing the code could be easily adapted to work with OpenCV functions only.

Answer 5

I found this on github.

https://imagemagick.org/script/download.php

import pgmagick

def remove_background(image, background=None):
    """Returns a copy of `image` that only contains the parts that is distinct
       from the background. If background is None, returns parts that are
       distinct from white."""
    if background is None:
        background = pgmagick.Image(image.size(), 'white')
    elif isinstance(background, pgmagick.Image):
        blob = pgmagick.Blob()
        background.write(blob)
        background = pgmagick.Image(blob, image.size())
    else:
        background = pgmagick.Image(image.size(), background)
    background.composite(image, 0, 0, pgmagick.CompositeOperator.DifferenceCompositeOp)
    background.threshold(25)
    blob = pgmagick.Blob()
    image.write(blob)
    image = pgmagick.Image(blob, image.size())
    image.composite(background, 0, 0, pgmagick.CompositeOperator.CopyOpacityCompositeOp)
    return image

Answer 6

Although the question has been answered thoroughly already, I would like to share a simple version that relies on numpy only:

import numpy as np

def remove_background(image, bg_color=255):
    # assumes rgb image (w, h, c)
    intensity_img = np.mean(image, axis=2)

    # identify indices of non-background rows and columns, then look for min/max indices
    non_bg_rows = np.nonzero(np.mean(intensity_img, axis=1) != bg_color)
    non_bg_cols = np.nonzero(np.mean(intensity_img, axis=0) != bg_color)
    r1, r2 = np.min(non_bg_rows), np.max(non_bg_rows)
    c1, c2 = np.min(non_bg_cols), np.max(non_bg_cols)

    # return cropped image
    return image[r1:r2+1, c1:c2+1, :]