Amazon验证码机器学习破解

### Amazon验证码识别 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 [Amazon验证码识别](https://github.com/TTyb/AmazonCaptcha) > 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 Image for 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 = False foundletter = False start = 0 end = 0 letters = [] 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, e