如何在python中对图像进行操作

Question

I am trying to complete a challenge where i use an equation to construct a new image (d) from other images. Then i must get the flag in the image (d). The given images are a.png, b.png c.png and y.png and they can be found here: https://drive.google.com/drive/folders/1bZOm_0apr5ZmaRNf9R5UVIEmtMuYSphn?usp=sharing

The equation: d = y - 21a - 3b + 41c

My current code

from PIL import Image

imagey = Image.open('y.png')
imagea = Image.open('a.png')
imageb = Image.open('b.png')
imagec = Image.open('c.png')
size = width, height = imagey.size

new = Image.new('RGB', size)

imgy = imagey.load()
imga = imagea.load()
imgb = imageb.load()
imgc = imagec.load()
data = new.load()

for x in range(width):
    for y in range(height):
        they = imgy[x, y]
        thea = imga[x, y]
        theb = imgb[x, y]
        thec = imgc[x, y]

        new_color = ((int(they[0])) & ~(int((21 * thea[0])) ^ int((3 * theb[0])) ^ int(~(41 * thec[0]))),
                     (int(they[1])) & ~(int((21 * thea[1])) ^ int((3 * theb[1])) ^ int(~(41 * thec[1]))),
                     (int(they[2])) & ~(int((21 * thea[2])) ^ int((3 * theb[2])) ^ int(~(41 * thec[2]))))
        data[x, y] = new_color

new.save('final.png')
new.show()

Answer 1

If you would convert Pillow image to numpy array or you would use OpenCV or imageio to load image (and get directly numpy array ) then you could do directly

new = imagey - 21*imagea - 3*imageb + 41*imagec

Result:

Not ideal but much better than with your code.

It can be problem with overflow . It may create array with 8bits values and calculations can gives 16bits or 32bits values which can be reduced to 8bits in every calculation.

Full working code:

import imageio

imagey = imageio.imread('y.png')
imagea = imageio.imread('a.png')
imageb = imageio.imread('b.png')
imagec = imageio.imread('c.png')

new = imagey - 21*imagea - 3*imageb + 41*imagec

imageio.imwrite('final.png', new)

# --- imageio doesn't have function to display it ---

import matplotlib.pyplot as plt

plt.imshow(new)
plt.show()

---

EDIT:

If I use OpenCV then I get ideal result

Full working code:

import cv2

imagey = cv2.imread('y.png')
imagea = cv2.imread('a.png')
imageb = cv2.imread('b.png')
imagec = cv2.imread('c.png')

new = imagey - 21*imagea - 3*imageb + 41*imagec

cv2.imwrite('final.png', new)

# --- show window with image and wait for press any key ---

cv2.imshow('Image', new)
cv2.waitKey(0)
cv2.destroyAllWindows()

---

EDIT:

By the way: version which converts PIL Image to numpy array and later it converts back to PIL Image - but it gives the same result as imageio .

from PIL import Image
import numpy as np

imagey = Image.open('y.png')
imagea = Image.open('a.png')
imageb = Image.open('b.png')
imagec = Image.open('c.png')

arr_y = np.array(imagey)
arr_a = np.array(imagea)
arr_b = np.array(imageb)
arr_c = np.array(imagec)

arr_new = arr_y - 21*arr_a - 3*arr_b + 41*arr_c

new = Image.fromarray(arr_new)

new.save('final.png')
new.show()

---

BTW:

If I check images on Linux using program file then it shows that b.png and c.png are JPEG , not PNG .

$ file b.png

b.png: JPEG image data, JFIF standard 1.01, resolution (DPI), 
density 300x300, segment length 16, 
Exif Standard: [TIFF image data, big-endian, direntries=0], baseline, 
precision 8, 960x640, components 3

---

I found that cv2.imread() gives little different values for c.png (which is JPG file) then other modules - and I don't mean that cv2 gives colors in BGR instead of RGB - and later this gives correct result. Probably cv2 uses different C library to read JPG .