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
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from PIL import Imageimport requestsimport resplitter = 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1def 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_resultopener = requests.session()res = opener.get('http://60.211.254.236:8402/Ajax/ValidCodeImg.ashx').contentwith 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 = 0str1 = ''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 0arr1 = []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) |
字库已经部署到GitHub地址：https://github.com/hldh214/validCode/
希望本文所述对大家Python程序设计有所帮助。
原文链接：https://blog.csdn.net/hldh214/article/details/51364933
