在破解Amazon的验证码的时候，利用机器学习得到验证码破解精度超过70%，主要是训练样本不够，如果在足够的样本下达到90%是非常有可能的。

update后，样本数为2800多，破解精度达到90%以上，perfect!

文档结构为

-- iconset1   -- ...-- jpg   -- img   -- jpg -- ... -- error.txt -- py -- crack.py

需要的库

pip3 install pillow or easy_install Pillow

必须文件下载地址

1.读取图片，打印图片的结构直方图

遍历出所有的jpg文件，

import os# 找出文件夹下所有xml后缀的文件def listfiles(rootdir, prefix='.xml'): file = [] for parent, dirnames, filenames in os.walk(rootdir): if parent == rootdir: for filename in filenames: if filename.endswith(prefix): file.append(rootdir + filename) return file else: pass if __name__ == '__main__': path = "../jpg/img/" jpgname = listfiles(path, "jpg")

jpgname为一个数组，将文件夹中的jpg文件全部遍历出来

['../jpg/img/056567f5e15f8d5f46bc5e07905009fd.jpg', '../jpg/img/05796993cf0a3c779b6fe83db2a27ac3.jpg', '../jpg/img/073847b62252c63829850cb1bd49601e.jpg', '../jpg/img/07aafc4694264509135490b85630aaf5.jpg', '../jpg/img/07d126e49e42143e0d21a0dafd522ac8.jpg', '../jpg/img/07dbfd0bd41d11e9475a96bc724e9f56.jpg', '../jpg/img/07fb8e7163e2ebd36e90c209502051ed.jpg', '../jpg/img/08ff7dc78f348ad7e4309eda9588a5f5.jpg', '../jpg/img/09dc3340f3c4a77c61cd18da7b3eca82.jpg', '../jpg/img/0b354ba9e9a132075fcc3dff6f517106.jpg', '../jpg/img/0bdca69fec2089cfaa46b458f5e483c3.jpg', '../jpg/img/0d0b1d778e00a1c84001d5838b9f5ef1.jpg', '../jpg/img/0d14f8838c30f6b54f266d9eb02e1b93.jpg', '../jpg/img/0e8d3e12d36d39314acfcd3bb8c3970a.jpg',...]

读取图片，得到图片的结构直方图

from PIL import Imagefor item in jpgname:    newjpgname = []    im = Image.open(item)    print(item)    # jpg不是最低像素，gif才是，所以要转换像素 im = im.convert("P") # 打印像素直方图 his = im.histogram()

像素直方图打印结果为

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 2, 0, 1, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 3, 1, 3, 3, 0, 0, 0, 0, 0, 0, 1, 0, 3, 2, 132, 1, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 15, 0, 1, 0, 1, 0, 0, 8, 1, 0, 0, 0, 0, 1, 6, 0, 2, 0, 0, 0, 0, 18, 1, 1, 1, 1, 1, 2, 365, 115, 0, 1, 0, 0, 0, 135, 186, 0, 0, 1, 0, 0, 0, 116, 3, 0, 0, 0, 0, 0, 21, 1, 1, 0, 0, 0, 2, 10, 2, 0, 0, 0, 0, 2, 10, 0, 0, 0, 0, 1, 0, 625]

该数组长度为255，每一个元素代表（0-255）颜色的多少，例如最后一个元素为625，即255（代表的是白色）最多，组合在一起

values = {}for i in range(0, 256):    values[i] = his[i] # 排序，x:x[1]是按照括号内第二个字段进行排序,x:x[0]是按照第一个字段 temp = sorted(values.items(), key=lambda x: x[1], reverse=True) # print(temp)

打印结果为

[(255, 625), (212, 365), (220, 186), (219, 135), (169, 132), (227, 116), (213, 115), (234, 21), (205, 18), (184, 15), (241, 10), (248, 10), (191, 8), (198, 6), (155, 3), (157, 3), (158, 3), (167, 3), (228, 3), (56, 2), (67, 2), (91, 2), (96, 2), (109, 2), (122, 2), (127, 2), (134, 2), (140, 2), (168, 2), (176, 2), (200, 2), (211, 2), (240, 2), (242, 2), (247, 2), (43, 1), (44, 1), (53, 1), (61, 1), (68, 1), (79, 1), (84, 1), (92, 1), (101, 1), (103, 1), (104, 1), (107, 1), (121, 1), (126, 1), (129, 1), (132, 1), (137, 1), (149, 1), (151, 1), (153, 1), (156, 1), (165, 1), (170, 1), (171, 1), (175, 1), (186, 1), (188, 1), (192, 1), (197, 1), (206, 1), (207, 1), (208, 1), (209, 1), (210, 1), (215, 1), (223, 1), (235, 1), (236, 1), (253, 1), (0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0), (11, 0), (12, 0), (13, 0), (14, 0), (15, 0), (16, 0), (17, 0), (18, 0), (19, 0), (20, 0), (21, 0), (22, 0), (23, 0), (24, 0), (25, 0), (26, 0), (27, 0), (28, 0), (29, 0), (30, 0), (31, 0), (32, 0), (33, 0), (34, 0), (35, 0), (36, 0), (37, 0), (38, 0), (39, 0), (40, 0), (41, 0), (42, 0), (45, 0), (46, 0), (47, 0), (48, 0), (49, 0), (50, 0), (51, 0), (52, 0), (54, 0), (55, 0), (57, 0), (58, 0), (59, 0), (60, 0), (62, 0), (63, 0), (64, 0), (65, 0), (66, 0), (69, 0), (70, 0), (71, 0), (72, 0), (73, 0), (74, 0), (75, 0), (76, 0), (77, 0), (78, 0), (80, 0), (81, 0), (82, 0), (83, 0), (85, 0), (86, 0), (87, 0), (88, 0), (89, 0), (90, 0), (93, 0), (94, 0), (95, 0), (97, 0), (98, 0), (99, 0), (100, 0), (102, 0), (105, 0), (106, 0), (108, 0), (110, 0), (111, 0), (112, 0), (113, 0), (114, 0), (115, 0), (116, 0), (117, 0), (118, 0), (119, 0), (120, 0), (123, 0), (124, 0), (125, 0), (128, 0), (130, 0), (131, 0), (133, 0), (135, 0), (136, 0), (138, 0), (139, 0), (141, 0), (142, 0), (143, 0), (144, 0), (145, 0), (146, 0), (147, 0), (148, 0), (150, 0), (152, 0), (154, 0), (159, 0), (160, 0), (161, 0), (162, 0), (163, 0), (164, 0), (166, 0), (172, 0), (173, 0), (174, 0), (177, 0), (178, 0), (179, 0), (180, 0), (181, 0), (182, 0), (183, 0), (185, 0), (187, 0), (189, 0), (190, 0), (193, 0), (194, 0), (195, 0), (196, 0), (199, 0), (201, 0), (202, 0), (203, 0), (204, 0), (214, 0), (216, 0), (217, 0), (218, 0), (221, 0), (222, 0), (224, 0), (225, 0), (226, 0), (229, 0), (230, 0), (231, 0), (232, 0), (233, 0), (237, 0), (238, 0), (239, 0), (243, 0), (244, 0), (245, 0), (246, 0), (249, 0), (250, 0), (251, 0), (252, 0), (254, 0)]

将占比最多的10个颜色筛选出来

# 占比最多的10种颜色for j, k in temp[:10]:    print(j, k)    # 255 12177 # 0 772 # 254 94 # 1 40 # 245 10 # 12 9 # 236 9 # 243 9 # 2 8 # 6 8 # 255是白底，0是黑色，可以打印来看看0和254

2.构造新的无杂质图片

生成一张白底啥都没有的图片

# 获取图片大小，生成一张白底255的图片im2 = Image.new("P", im.size, 255)

利用上一步占比最多的颜色可以看出，255是白底，0是黑色，可以打印来看看0和254

最后证明0是黑色字母，254是斑点，可以舍弃！

将这些颜色根据宽和高的坐标以此写入新生成的白底照片中

# 获取图片大小，生成一张白底255的图片im2 = Image.new("P", im.size, 255)for y in range(im.size[1]): # 获得y坐标 for x in range(im.size[0]): # 获得坐标(x,y)的RGB值 pix = im.getpixel((x, y)) # 这些是要得到的数字 # 事实证明只要0就行，254是斑点 if pix == 0: # 将黑色0填充到im2中 im2.putpixel((x, y), 0) # 生成了一张黑白二值照片 # im2.show()

黑白二值照片

3.切割图片

x代表图片的宽，y代表图片的高

对图片进行纵向切割

# 纵向切割# 找到切割的起始和结束的横坐标inletter = Falsefoundletter = Falsestart = 0end = 0letters = []for x in range(im2.size[0]): for y in range(im2.size[1]): pix = im2.getpixel((x, y)) if pix != 255: inletter = True if foundletter == False and inletter == True: foundletter = True start = x if foundletter == True and inletter == False: foundletter = False end = x letters.append((start, end)) inletter = False

打印结果为

# [(27, 47), (48, 71), (73, 101), (102, 120), (122, 147), (148, 166)]

(27, 47)代表从x=27到x=47纵向切割成一条状

保存字段到本地，这里就是training_samples.py文件里面的内容，为的就是生成训练样本，这里生成的样本有2800多！

# 保存切割下来的字段count = 0for letter in letters:    # (切割的起始横坐标，起始纵坐标，切割的宽度，切割的高度)    im3 = im2.crop((letter[0], 0, letter[1], im2.size[1])) # 随机生成0-10000的数字 a = random.randint(0, 10000) # 更改成用时间命名 im3.save("../jpg/letter/%s.gif" % (time.strftime('%Y%m%d%H%M%S', time.localtime()) + str(a))) count += 1

字段样式

4.训练识别

使用的是 AI与向量空间图像识别

将标准图片转换成向量坐标a，需要识别的图片字段为向量坐标b，cos(a,b)值越大说明夹角越小，越接近重合

空间两向量计算公式

编写的夹角公式为

# 夹角公式import mathclass VectorCompare: # 计算矢量大小 # 计算平方和 def magnitude(self, concordance): total = 0 # concordance.iteritems:报错'dict' object has no attribute 'iteritems' # concordance.items() for word, count in concordance.items(): total += count ** 2 return math.sqrt(total) # 计算矢量之间的 cos 值 def relation(self, concordance1, concordance2): topvalue = 0 # concordance1.iteritems:报错'dict' object has no attribute 'iteritems' # concordance1.items() for word, count in concordance1.items(): # if concordance2.has_key(word):报错'dict' object has no attribute 'has_key' # 改成word in concordance2 if word in concordance2: # 计算相乘的和 topvalue += count * concordance2[word] return topvalue / (self.magnitude(concordance1) * self.magnitude(concordance2))

转换验证码图片为向量：

# 将图片转换为矢量def buildvector(im): d1 = {} count = 0 for i in im.getdata(): d1[count] = i count += 1 return d1

打印结果

{0: 255, 1: 255, 2: 255, 3: 255, 4: 255, 5: 255, 6: 255, 7: 255, 8: 255, 9: 255, 10: 255, 11: 255, 12: 255, 13: 255, 14: 255, 15: 255, 16: 255, 17: 255, 18: 255, 19: 255, 20: 255, 21: 255, 22: 255, 23: 255, 24: 255, 25: 255, 26: 255, 27: 255, 28: 255, 29: 255, 30: 255, 31: 255, 32: 255, 33: 255, 34: 255, 35: 255, 36: 255, 37: 255, 38: 255, 39: 255, 40: 255, 41: 255, 42: 255, 43: 255, 44: 255, 45: 255, 46: 255, 47: 255, 48: 255, 49: 255, 50: 255, 51: 255, 52: 255, 53: 255, 54: 255, 55: 255, 56: 255, 57: 255, 58: 255, 59: 255, 60: 255, 61: 255, 62: 255, 63: 255, 64: 255, 65: 255, 66: 255, 67: 0, 68: 0, 69: 0, 70: 255, 71: 255, 72: 255, 73: 255, 74: 0, 75: 0, 76: 0, 77: 255, 78: 0, 79: 255, 80: 255, 81: 0, 82: 0, 83: 0, 84: 0, 85: 0, 86: 0, 87: 255, 88: 255, 89: 0, 90: 255, 91: 255, 92: 255, 93: 0, 94: 0, 95: 255, 96: 0, 97: 255, 98: 0, 99: 255, 100: 255, 101: 0, 102: 0, 103: 0, 104: 0, 105: 0, 106: 0, 107: 255, 108: 255, 109: 0, 110: 0, 111: 0, 112: 0, 113: 0, 114: 255, 115: 255, 116: 255, 117: 0, 118: 0, 119: 0, 120: 255, 121: 0, 122: 255, 123: 255, 124: 255, 125: 0, 126: 0, 127: 0, 128: 255, 129: 0, 130: 0, 131: 255, 132: 255, 133: 0, 134: 0, 135: 0, 136: 255, 137: 0, 138: 0, 139: 0, 140: 0, 141: 0, 142: 0, 143: 255, 144: 255, 145: 0, 146: 0, 147: 0, 148: 0, 149: 0, 150: 0, 151: 255, 152: 255, 153: 255, 154: 255, 155: 0, 156: 0, 157: 0, 158: 255, 159: 255, 160: 255, 161: 255, 162: 255, 163: 255, 164: 255, 165: 255, 166: 255, 167: 255, 168: 255, 169: 255, 170: 255, 171: 255, 172: 255, 173: 255, 174: 255, 175: 255}

加载训练集，且把训练集也变成向量

v = VectorCompare()iconset = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'] import os imageset = [] for letter in iconset: for img in os.listdir('../iconset1/%s/' % (letter)): temp = [] if img != "Thumbs.db" and img != ".DS_Store": temp.append(buildvector(Image.open("../iconset1/%s/%s" % (letter, img)))) imageset.append({letter: temp})

开始识别验证码

# 开始破解训练count = 0for letter in letters:    # (切割的起始横坐标，起始纵坐标，切割的宽度，切割的高度) im3 = im2.crop((letter[0], 0, letter[1], im2.size[1])) guess = [] # 将切割得到的验证码小片段与每个训练片段进行比较 for image in imageset: for x, y in image.items(): if len(y) != 0: guess.append((v.relation(y[0], buildvector(im3)), x)) # 排序选出夹角最小的（即cos值最大）的向量，夹角越小则越接近重合，匹配越接近 guess.sort(reverse=True) print("", guess[0])

排序选出夹角最小的（即cos值最大）的向量，夹角越小则越接近重合，匹配越接近

guess.sort(reverse=True)print("", guess[0]) count += 1

运行结果

结果显示前面是匹配度，后面是匹配的字母

将图片的名字改成识别后的名字

# 得到拼接后的验证码识别图像newname = str("".join(newjpgname))os.rename(item, path + newname + ".jpg")

效果为

完整源码在

更多学习请看