Python3实现的简单验证码识别功能示例
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2022-03-30 20:00:55
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这篇文章主要介绍了Python3实现的简单验证码识别功能,涉及Python针对验证码图片识别处理相关操作技巧,需要的朋友可以参考下
本文实例讲述了Python3实现的简单验证码识别功能。分享给大家供大家参考,具体如下:
这次的需求是自动登录某机构网站, 其验证码很具特色, 很适合做验证码识别入门demo, 先贴主要代码, 其中图片对比使用了编辑距离算法, 脚本使用了pillow库
from PIL import Image import requests import re splitter = re.compile(r'\d{30}') # 分割二值化后的图片 # distance('11110000', '00000000') # 比较两个字符串有多少位不同, 返回不同的位数 def distance(string1, string2): d_str1 = len(string1) d_str2 = len(string2) d_arr = [[0] * d_str2 for i in range(d_str1)] for i in range(d_str1): for j in range(d_str2): if string1[i] == string2[j]: if i == 0 and j == 0: d_arr[i][j] = 0 elif i != 0 and j == 0: d_arr[i][j] = d_arr[i - 1][j] elif i == 0 and j != 0: d_arr[i][j] = d_arr[i][j - 1] else: d_arr[i][j] = d_arr[i - 1][j - 1] else: if i == 0 and j == 0: d_arr[i][j] = 1 elif i != 0 and j == 0: d_arr[i][j] = d_arr[i - 1][j] + 1 elif i == 0 and j != 0: d_arr[i][j] = d_arr[i][j - 1] + 1 else: d_arr[i][j] = min(d_arr[i][j - 1], d_arr[i - 1][j], d_arr[i - 1][j - 1]) + 1 current = max(d_arr[d_str1 - 1][d_str2 - 1], abs(d_str2 - d_str1)) # print("Levenshtein Distance is",current) # print(current) return current # 去除字符串里面连续的1 def no_one(string): n_arr = splitter.findall(string) n_arr = filter(lambda each_str: each_str != '111111111111111111111111111111', n_arr) n_result = '' for n_each in n_arr: n_result += str(n_each) return n_result opener = requests.session() res = opener.get('http://60.211.254.236:8402/Ajax/ValidCodeImg.ashx').content with open('verify.gif', 'wb') as v: v.write(res) img = Image.open('verify.gif') img = img.convert('L') size = img.size # img = img.point(table, '1') img_arr = img.load() # for x in range(size[0]): # for y in range(size[1]): # if img_arr[x, y] > 210: # img_arr[x, y] = 1 # else: # img_arr[x, y] = 0 # img.save('after.gif') inc = 0 str1 = '' str2 = '' str3 = '' cur_str = '' for x in range(size[0]): for y in range(size[1]): if img_arr[x, y] > 210: cur_str += '1' else: cur_str += '0' # print(img_arr[i, j], end='') # cur_str += str(img_arr[x, y]) inc += 1 # if inc % 18 == 0: # print('\n----') # else: # print('') if inc == 18: str1 = cur_str cur_str = '' elif inc == 36: str2 = cur_str cur_str = '' elif inc == 54: str3 = cur_str cur_str = '' str1 = str1[:-60] str2 = str2[:-60] str3 = str3[:-60] str1 = no_one(str1) str2 = no_one(str2) str3 = no_one(str3) str1 = str1.strip('1') str2 = str2.strip('1') str3 = str3.strip('1') # print(str1) # print(str3) with open('./dict/plus') as plus: with open('./dict/minus') as minus: p = plus.read() m = minus.read() is_add = 1 if distance(p, str2) < distance(m, str2) else 0 arr1 = [] arr3 = [] for each in range(1, 10): with open('./dict/{}'.format(each)) as f: ff = f.read() arr1.append([each, distance(ff, str1)]) arr3.append([each, distance(ff, str3)]) arr1 = sorted(arr1, key=lambda item: item[1]) arr3 = sorted(arr3, key=lambda item: item[1]) result = arr1[0][0] + arr3[0][0] if is_add else arr1[0][0] - arr3[0][0] print(result) # login_url = 'http://60.211.254.236:8402/Ajax/Login.ashx?Method=G3_Login' # login_data = { # 'loginname': usn, # 'password': pwd, # 'validcode': result, # # } # opener.get(login_url, login_data)
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