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"_blank" href="https://github.com/hldh214/validCode/" rel="external nofollow" >https://github.com/hldh214/validCode/
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希望本文所述对大家Python程序设计有所帮助。