Python PIL - 像素 rgb 颜色到十六进制

Question

我正在从这里寻找示例和问题。我创造了东西。我希望 rgb 在像素上十六进制。我的代码，但我没有解决，

#-*- coding: utf-8 -*-
import Image
def read(ch):
    return list(ch.getdata())
def hex2rgb(v):
    v = v.lstrip('#')
    lv = len(v)
    return tuple(int(v[i:i+lv/3], 16) for i in range(0, lv, lv/3))
def rgb2hex(rgb):
    return '#%02x%02x%02x' % rgb
imj = Image.open('sample.png','r')
x,y = imj.size
pix = list(imj.getdata())
if imj.mode in ('RGBA','LA') or (imj.mode == 'P' and 'transparency' in imj.info):   
    red,green,blue,alfa = imj.convert('RGBA').split()
    #rgb = imj.convert('RGBA').split()[:-1]
    r,g,b,a = read(red),read(green),read(blue),read(alfa)
for r_,g_,b_ in r,g,b:
    print rgb2hex((r_,g_,b_))

错误在这里为 r，g，b 中的 r_，g_，b_：但是我怎么能解决不知道？

感谢您的关注？善行..

Answer 1

问题：

for r_,g_,b_ in r,g,b:

修复：

for r_, g_, b_ in zip(r, g, b):

但说实话，你不必要地使事情复杂化：

#!/usr/bin/python
from PIL import Image
def rgb2hex(r, g, b):
    return '#{:02x}{:02x}{:02x}'.format(r, g, b)
img = Image.open('sample.png')
if img.mode in ('RGBA', 'LA') or (img.mode == 'P' and 'transparency' in img.info):   
    pixels = list(img.convert('RGBA').getdata())
    for r, g, b, a in pixels: # just ignore the alpha channel
       print rgb2hex(r, g, b)

更新：获取像素的 x 和 y 位置以及其十六进制颜色

#!/usr/bin/python
from PIL import Image
def rgb2hex(r, g, b):
    return '#{:02x}{:02x}{:02x}'.format(r, g, b)
img = Image.open('sample.png')
if img.mode in ('RGBA', 'LA') or (img.mode == 'P' and 'transparency' in img.info):   
    pixels = img.convert('RGBA').load()
    width, height = img.size
    for x in range(width):
        for y in range(height):
            r, g, b, a = pixels[x, y]
            print 'x = %s, y = %s, hex = %s' % (x, y, rgb2hex(r, g, b))