新博客地址：http://gorthon.sinaapp.com/
经过一段时间的思考，觉得在某网站验证码的识别笔记【一】里面写的不太好，后面旋转也不太好弄
，聚类又不想碰也很难。
后来决定改了个方法，就是基于形状来识别，就是根据人的眼睛怎么识别(区分)字符来识别的。
代码写的一团糟……
还是写出来吧，好像我玷污了python的整洁与美观，就这样了，以后再改进吧。
#! /usr/bin/env python
# coding: utf-8
import Image, ImageEnhance, ImageFilter
import math, cPickle
# 随便找个图片先获得以下参数:
GRID_COLOR = (128, 191, 255) #背景表格颜色
BG_COLOR = (227, 218, 237) # 背景颜色
ARC_COLOR = (128, 128, 255) # 干扰线(弧线)颜色
BLACK = (0,0,0)
WHITE = (255, 255, 255)
### 观察到这类验证码没有大写字母和数字
##OMEGA_1 = 'abd'.split(' ') # 圈在下
### a：8个数小于55
### d：8个数大于55，上7后有0
### b：8个数大于55，上7后无0
### 注意a字体不同圈位置不同，另外g可能是上圈，也可能有两个圈
### 这里a是下圈，g是上圈
##OMEGA_2 = 'o'.split(' ') # 只有一个圈
### o：无7(有时候有7)
##OMEGA_3 = 'egpq'.split(' ') # 圈在上
### e：下面有8在右边，右有7
### g：下面有8在左边，右无7
### p：0在左下角，下无8
### q：0在右下角，下无8
##
##OMEGA_4 = 'cfhijklmnrstuvwxyz'.split(' ') # 无圈
def printf(s='', debug=True):
if debug:
print s
def eraseGridAndArc(im):
w, h = im.size
pixels = im.load()
for y in range(h):
for x in range(w):
pixel = pixels[x, y]
if pixel == BG_COLOR or pixel == GRID_COLOR:
im.putpixel((x, y), WHITE)
elif pixel == ARC_COLOR:
up = y > 0 and pixels[x, y-1] or None
down = y < h-1 and pixels[x, y+1] or None
up_down = up or down
im.putpixel((x, y), up_down and up_down or WHITE)
enhancer = ImageEnhance.Contrast(im)
im = enhancer.enhance(255) # 提高对比度,这个参数大点就好
im = im.convert('1') # 二值化
#im.show()
return im
def getPoint(im, end=False):
pixels = im.load()
w, h = im.size
range_w = end and range(w-1, 0, -1) or range(w)
for x in range_w:
for y in range(h):
if pixels[x, y] == 0:
return end and x + 1 or x
def getVerticalProjection(im):
# 得到垂直投影图, 返回投影图数据
pixels = im.load()
w, h = im.size
start_x = getPoint(im)
end_x = getPoint(im, end=True)
graph = [0] * (end_x - start_x)
for x in range(start_x, end_x):
for y in range(h):
pixel = pixels[x, y]
if pixel == 0: # 此列有字符
graph[x - start_x] += 1
return start_x, end_x, graph
def showVerticalProjection(graph):
# 显示垂直投影图
w = len(graph)
h = max(graph)
img = Image.new('1', (w, h))
for x in range(w):
for y in range(h):
if y <= graph[x]:
img.putpixel((x, y), 255)
else:
break
img = img.transpose(Image.FLIP_TOP_BOTTOM)
img.show()
def cut(start_x, end_x, graph, im):
chars = [start_x]
for i, v in enumerate(graph):
if v !=0:
if graph[i - 1] == 0:
chars.append(i + start_x)
elif graph[i - 1] !=0:
chars.append(i + start_x)
chars.append(end_x)
result = list()
for i in range(len(chars) - 1):
result.append((chars[i], chars[i + 1] - 1))
result = [v for (i, v) in enumerate(result) if not i%2] # 去除波谷的0
w, h = im.size
chars = list()
for char_start_x, char_end_x in result:
chars.append(im.crop((char_start_x, 0, char_end_x, h)))
# 纵向切割之后再横向切割:
result = list()
for char in chars:
w, h = char.size
pixels = char.load()
try:
for y in range(h):
for x in range(w):
if pixels[x, y] == 0:
start_y = y
raise
except:
pass
try:
for y in range(h-1, -1, -1):
for x in range(w):
if pixels[x, y] == 0:
end_y = y
raise
except:
pass
result.append(char.crop((0, start_y, w, end_y)))
return result
def unique(s):
r = s[0]
for i in list(s):
if i != r[-1]:
r += i
return r
def otsu(chars, debug):
vcode = '' # 返回值
# 凹凸区域和圈的提取
for im in chars:
global result, w, h
pixels = im.load()
result = im.copy().load()
w, h = im.size
for y in range(h):
for x in range(w):
count = 0
# 八个方向扫描:
lst = (range(x + 1, w), range(x)) # 东、西
for rng in lst:
for i in rng:
if pixels[i, y] == 0: # 黑色
count += 1
break
lst = (range(y + 1, h), range(y)) # 南、北
for rng in lst:
for j in rng:
if pixels[x, j] == 0:
count += 1
break
p = ((1, -1), (1, 1), (-1, 1), (-1, -1)) # 东北、东南、西南、西北四个方向
for m1, m2 in p:
try:
for i in range(999):
if pixels[x + i * m1, y + i * m2] == 0:
count += 1
break
except:
pass
count = count == 8 and 8 or (7 if 4 < count < 8 else 4)
# count为8: 圈区域
# count为5、6或7:凹区域
if result[x, y] != 0:
result[x, y] = count
else: # 对于孤立的0将其置为其左边或右边的数
if 0<x<w-1:
if result[x - 1, y] != 0 and result[x+1, y] != 0:
result[x, y] = result[x-1, y]
#end for x
# end for y
for y in range(h):
for x in range(w):
if result[x, y] == 8 and 0<x<w-1:
if result[x - 1, y] != 8 and result[x+1, y] != 8: # 对于孤立的8将其置为其左边的数
result[x, y] = result[x-1, y]
if result[x, y] == 7 and 0<x<w-1:
if result[x - 1, y] != 7 and result[x+1, y] != 7: # 对于孤立的7将其置为其左边的数
result[x, y] = result[x-1, y]
## # 8的8个方向不应该有7，否则应该将8置为7
## # 对于倾斜m很有效
## for y in range(h-1, -1, -1):
## for x in range(w-1, -1, -1):
## if result[x, y] == 8 and 0<x<w-1 and 0<y<h-1:
## if result[x-1, y-1] == 7 or result[x-1, y] == 7 or result[x, y-1] == 7 or result[x+1, y] == 7 or \
## result[x, y+1] == 7 or result[x+1, y+1] == 7 or result[x+1, y-1] == 7 or result[x-1, y+1] == 7:
## result[x, y] = 7
# 对于干扰线遮挡b的底部的时候会将其判断为h，故去除干扰线时有待改进
# 此类的还有o会识别为n或者u等等
num_8 = 0
for y in range(h):
for x in range(w):
if result[x, y] == 8:
num_8 += 1
# 对于倾斜的n或者u等字符内部会出现8的区域
# 如果8的个数小于一定个数(比如说15个)就将其看成是5、6或7即凹区域
# 通过观察可看出此圈区域的8个数很多(在50个以上), 这里我取的30...
has_circle = True # 有圈
if num_8 < 30:
has_circle = False # 无圈
for y in range(h):
for x in range(w):
if result[x, y] == 8:
result[x, y] = 7
if has_circle:
def has_7():
# 没有7就是0
num_7 = 0
for y in range(h):
for x in range(w):
if result[x, y] == 7:
num_7 += 1
return num_7
if has_7() < 6: # o
# 有时候有7, 这里取小于6个7也为o
# 去除干扰线引起的e
if result[w/2, h/2] == 0:
vcode += 'e'
printf('e', debug)
else:
# 最后3排o的0多于p、q
s = ''
for y in range(h - 3, h):
for x in range(w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > s.count('7'): # o
vcode += 'o'
printf('o', debug)
else: # pq
# 右下角5*4区域0多于7为q
s = ''
for y in range(h - 5, h):
for x in range(w - 4, w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > s.count('7'): # qp
vcode += 'q'
printf('q', debug)
else:
vcode += 'p'
printf('p', debug)
else:
num_8_up = 0 # 记录8在上的个数
num_8_down = 0# 记录8在下的个数
for y in range(h/2):
for x in range(w):
if result[x, y] == 8:
num_8_up += 1
for y in range(h/2, h):
for x in range(w):
if result[x, y] == 8:
num_8_down += 1
if num_8_up > num_8_down: # 圈在上:'egpq'
num_8 = 0 # 整个字符中8的个数,小于50个为e，多于50个为gpq
for y in range(h):
for x in range(w):
if result[x, y] == 8:
num_8 += 1
if num_8 < 50: # e
# 用一条直线从2/3的地方穿过，如果遇到07070这样的情况那么是m
try:
find = False
for y in range(h):
s = ''
for x in range(w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('07070') != -1:
vcode += 'm'
printf('m', debug)
find = True
raise
except:
pass
if not find:
vcode += 'e'
printf('e', debug)
else: # pgq
# 倒数第二排的0个数多于一半为g
zero = 0
y = h - 3
for x in range(w):
if result[x, y] == 0:
zero += 1
if zero >= w / 2: # g
vcode += 'g'
printf('g', debug)
else: # pq
# 右下角5*4区域0多于7为q
s = ''
for y in range(h - 5, h):
for x in range(w - 4, w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > s.count('7'): # qp
vcode += 'q'
printf('q', debug)
else:
vcode += 'p'
printf('p', debug)
else: # 圈在下'abd'
num_8 = 0 # 整个字符中8的个数,小于55个为a，多于55个为bd
for y in range(h):
for x in range(w):
if result[x, y] == 8:
num_8 += 1
if num_8 < 55:
vcode += 'a'
printf('a', debug)
else:
# 7与0的位置关系
try:
for y in range(h/2):
for x in range(w):
if result[x, y] == 7:
if result[x - 1, y] in (0, 8):
vcode += 'b'
printf('b', debug)
else:
vcode += 'd'
printf('d', debug)
raise
except:
pass
else: # 没有圈,'cfhijklmnrstuvwxyz'
# i和j是由两部分组成的，其他字符都是一个整体
def has_2_part():
for y in range(h/3):
for x in range(w):
if result[x, y] == 0:
for y in range(y+1, h/3):
for x in range(w):
if result[x, y] != 0:
yes = True
else:
yes = False
break
if yes:
return True
return False
if has_2_part(): # ij
# 最近几列如果0上面有7则为i否则为j
# 取右下角4X4方格
s = ''
for y in range(h/2, h-3):
for x in range(w-4, w):
t = str(result[x, y])
if t != '0':
t = '7'
s += t
if s.count('7') > s.count('0'):
vcode += 'i'
printf('i', debug)
else:
vcode += 'j'
printf('j', debug)
else: #'cfhklmnrstuvwxyz'
# c的中间向右均无0
def has_zreo():
s = ''
for y in range(h/2 - 3, h/2 + 4):
for x in range(w/2 - 2, w):
s += str(result[x, y])
if s.count('0') < 2:
return False
return True
if not has_zreo(): # c
# 排除r
# r的倒数第二排0个数很少，而c很多(多于宽度的一半)
s = ''
for x in range(w):
s += str(result[x, h - 2])
if s.count('0') > w / 2:
vcode += 'c'
printf('c', debug)
else:
vcode += 'r'
printf('r', debug)
else: # 'fhklmnrstuvwxyz'
# 用一条直线穿过，如果遇到07070比较多的情况那么是m或者是w
find = False
num_07070 = 0
for y in range(h):
s = ''
for x in range(w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('07070') != -1:
num_07070 += 1
if num_07070 >= 3: # m/w否则有可能为右倾的f
# 字符上面中间位置看7和0的个数
# 7多0少：w
# 7少0多：m
s = ''
for y in range(5):
for x in range(w/2 - 2, w/2 + 3):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('7') > s.count('0'):
vcode += 'w'
printf('w', debug)
else:
vcode += 'm'
printf('m', debug)
find = True
else:
find = False
if not find: # fhklnrstuvxyz
# 用一条直线横穿字符，如果得到的unique为070的个数多于6那么为h,k,n, u,v,x,y
# 否则为f,l,r,s,t,z
num_070 = 0
for y in range(h):
s = ''
for x in range(w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
#print s
if s.find('070') != -1:
num_070 += 1
if num_070 < 6: # 'flrstz'
# s倒数几列下半部分很多0
s = ''
for y in range(h/2, h):
for x in range(w- 3, w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > s.count('7'): # s
vcode += 's'
printf('s', debug)
else: # 'flrtz'
# 将字符为分2*3的块，那么东北那块出现070个数>=3的话就是f
num_070 = 0
for x in range(w/2, w):
s = ''
for y in range(2*h/3):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('070') != -1:
num_070 += 1
if num_070 >= 3: # f
vcode += 'f'
printf('f', debug)
else: # lrtz
# 右下角6*3区域内0少的为r
# !!!!!!!有可能是右倾的t!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!末完成!!!!!!!!!!!!!!!!!!!!!!!!!!
s = ''
for y in range(h - 3, h):
for x in range(w - 6, w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('7') > s.count('0'): # r
vcode += 'r'
printf('r', debug)
else: # ltz
# 宽度小于10为右倾的l
if w < 10:
vcode += 'l'
printf('l', debug)
else:
# 左上角4*8区域内0多于7为l，否则为t
# 这里不用考虑z，因为z基本上不出现
s = ''
for y in range(4):
for x in range(8):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > s.count('7'): # l
vcode += 'l'
printf('l', debug)
else:
vcode += 't'
printf('t', debug)
else: # 'hknuvxy'
# n的顶部中间有点其他都没有如中间的3*3内多于3个点
s = ''
for y in range(1, 4):
for x in range(w/2 - 1, w/2 + 2):
s += str(result[x, y])
if s.count('0') > 3: # n
# 排除h
# h右上角5*5内几乎没有0
num_0 = 0
for y in range(5):
for x in range(w-5, w):
if result[x, y] == 0:
num_0 += 1
if num_0 > 2: # l
vcode += 'n'
printf('n', debug)
else:
vcode += 'h'
printf('h', debug)
else:
# k与x的后半部分(纵向切分时),会出现070在5个左右，这里取>=3
num_070 = 0
for x in range(w/2, w):
s = ''
for y in range(h):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('070') != -1:
num_070 += 1
if num_070 >= 3: # kx
# 也有可能是y，判断倒数第5排有没有070即可，没有即为y
s = ''
for x in range(w):
t = str(result[x, h - 5])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('070') != -1:
# x的前半部分(纵向切分时),会出现070在5个左右，这里取>=3
# 对于孤立的7(纵向看)将其置为其下边的数
for x in range(w):
for y in range(h):
if result[x, y] == 7 and 0<y<h-1:
if result[x, y-1] != 7 and result[x, y+1] != 7:
result[x, y] = result[x, y+1]
num_070 = 0
for x in range(w/2):
s = ''
for y in range(h):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
s = unique(s)
if s.find('070') != -1:
num_070 += 1
if num_070 >= 3: # x
vcode += 'x'
printf('x', debug)
else:
vcode += 'k'
printf('k', debug)
else:
vcode += 'y'
printf('y', debug)
else: # 'huvy'
# h的上半部分有几行0很多, 并且中间有0
# h的下半部分中间一列没有0
num_0 = 0
line = -1
for y in range(h/2):
s = ''
for x in range(w):
t = str(result[x, y])
t = t!='0' and '7' or '0'
s += t
if s.count('0') > w/2:
num_0 += 1
line = y
#(0 in (result[w/2, line - 1], result[w/2, line], result[w/2, line - 2])) and
if line > 0 and num_0 > 2 and (
0 not in (result[w/2, h-1], result[w/2, h-2], result[w/2, h-3], result[w/2, h-4],
result[w/2, h-5], result[w/2, h-6])):
vcode += 'h'
printf('h', debug)
else:
# y的下半部分会出现先遇到7再遇到0的情况
num_70 = 0
for y in range(2*h/3, h):
s = ''
for x in range(w):
s += str(result[x, y])
s = unique(s)
if s.find('070') != -1:
continue
if s.find('70') != -1:
num_70 += 1
if num_70 > 2:
vcode += 'y'
printf('y', debug)
else:
# 倒数第5排出现了7则为u
y = h - 5
u = False
for x in range(w):
if result[x, y] == 7:
u = True
break
if u:
vcode += 'u'
printf('u', debug)
else:
vcode += 'v'
printf('v', debug)
continue
for y in range(h):
for x in range(w):
print result[x, y],
print
print
return vcode
def getCode(filename):
im = Image.open(filename)
im = eraseGridAndArc(im) # 去除背景及弧线并转换为二值图
start_x, end_x, graph = getVerticalProjection(im)
chars = cut(start_x, end_x, graph, im)
return otsu(chars, False)
#chars[0].show()
#for i in chars:i.show()
if __name__ == '__main__':
print getCode('./0.png')
## 下面是用识别的结果重命名那100个图片，遇到重复的字符串，就生成一个随机的后缀避免覆盖已有的图片
## import os, glob
## import random
## for f in glob.glob('*.png'):
## try:
## im = Image.open(f)
## im = eraseGridAndArc(im) # 去除背景及弧线并转换为二值图
## start_x, end_x, graph = getVerticalProjection(im)
## chars = cut(start_x, end_x, graph, im)
## vcode = otsu(chars, False)
## print vcode
## if os.path.exists(vcode + '.png'):
## os.rename(f, vcode + ' - %f.png' % random.random())
## else:
## os.rename(f, vcode + '.png')
## except:
## print 'error occurs when otsu %s' % f最后来几个效果图：
单个字母识别率还是挺高的，单词组合起来就不太理想了，提升空间很大。
第3个好像也是错的
第3篇：http://blog.youkuaiyun.com/bh20077/article/details/7311183
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