更新 PIL 中的像素颜色后图像未保存

Question

I've generated barcode images in PNG format using HuBarcode, and have modified them to add a border of rgb(150, 150, 150) in hopes of accessing the pixels of that color to change them. I can access the pixel and get confirmation via print that the color is being changed, but when I open the image file, nothing has changed. The pixels are still rgb(150, 150, 150) . Here's a code snippet I'm using. I can add more of my code if it will be helpful:

def add_colored_border(self, barcode):
  img = Image.open(barcode)
  img = img.convert('RGB')
  px = img.load()
  for x in range(img.size[0]):
    for y in range(img.size[1]):
      pixel = px[x, y]
      if pixel == (150, 150, 150):  
        pixel = (0, 0, 255)
  img.save('testing.png')

Answer 1

You need to copy the modified pixel value back into the pixel access object. Eg,

if px[x, y] == (150, 150, 150):
    px[x, y] = (0, 0, 255)

There's an example in the old PIL docs for Image.load() .

If you're modifying a lot of pixels, you may wish to use getdata() and putdata() . Those links are to the docs of the new fork of PIL known as Pillow, but those functions are also available in the old PIL.

---

pixel = px[x, y]
if pixel == (150, 150, 150):  
    pixel = (0, 0, 255)

doesn't do what you want because pixel = (0, 0, 255) creates a new tuple and binds that to the name pixel , it doesn't modify the tuple in the PixelAccess object at px[x, y] .

Tuples are immutable, so they can't be modified - if you want to change them you need to replace them with a new tuple. Python does let you modify lists in a way similar to what you were trying, since lists are mutable. To make this work, we need to use a list of lists, we can't use a simple list of integers or strings since Python integers and strings are immutable.

a = [[i] for i in range(5)]
print a
b = a[2]
print b
b[0] = 7
print b
print a

output

[[0], [1], [2], [3], [4]]
[2]
[7]
[[0], [1], [7], [3], [4]]  

This works because I'm modifying the contents of b . But if I do

c = a[3]
c = [11]
print a    

output

[[0], [1], [7], [3], [4]]

Now a is unchanged. The assignment to c binds a new list to c , it doesn't touch the a[3] object that c was previously bound to.

For more on this important topic please see the excellent illustrated article Facts and myths about Python names and values by SO member Ned Batchelder.