由于上面的验证码是24位的jpeg图像，并且包含了噪点，所以我们要做的就是去噪和去色，我拿PS找了张验证码试了试，使用PS滤镜中的去噪效果还行， 但是没有在PIL找到去噪的函数，后来发现中值过滤后可以去掉大部分的噪点，而且PIL里有现成的函数，接下来我试着直接把图像转换为单色，结果发现还是 会有不过的噪点留了下来，因为中值过滤时把不少噪点淡化了，但转换为音色时这些噪点又被强化显示了，于是在中值过滤后对图像亮度进行加强处理，然后再转换 为单色，这样验证码图片就变得比较容易识别了:
上面这些处理使用Python才几行：
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im
=
Image.
open
(image_name)
im
=
im.
filter
(ImageFilter.MedianFilter())
enhancer
=
ImageEnhance.Contrast(im)
im
=
enhancer.enhance(
2
)
im
=
im.convert(
'1'
)
im.show()
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接下来就是提取这些数字的字模，使用shell脚本下载100幅图片，抽出三张图片获取字模：
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#!/usr/bin/env python
#encoding=utf-8
import
Image,ImageEnhance,ImageFilter
import
sys
image_name
=
"./images/81.jpeg"
im
=
Image.
open
(image_name)
im
=
im.
filter
(ImageFilter.MedianFilter())
enhancer
=
ImageEnhance.Contrast(im)
im
=
enhancer.enhance(
2
)
im
=
im.convert(
'1'
)
#im.show()
#all by pixel
s
=
12
#start postion of first number
w
=
10
#width of each number
h
=
15
#end postion from top
t
=
2
#start postion of top
im_new
=
[]
#split four numbers in the picture
for
i
in
range
(
4
):
im1
=
im.crop((s
+
w
*
i
+
i
*
2
,t,s
+
w
*
(i
+
1
)
+
i
*
2
,h))
im_new.append(im1)
f
=
file
(
"data.txt"
,
"a"
)
for
k
in
range
(
4
):
l
=
[]
#im_new[k].show()
for
i
in
range
(
13
):
for
j
in
range
(
10
):
if
(im_new[k].getpixel((j,i))
=
=
255
):
l.append(
0
)
else
:
l.append(
1
)
f.write(
"l=["
)
n
=
0
for
i
in
l:
if
(n
%
10
=
=
0
):
f.write(
"/n"
)
f.write(
str
(i)
+
","
)
n
+
=
1
f.write(
"]/n"
)
|
把字模保存为list，用于接下来的匹配；
提取完字模后剩下来的就是对需要处理的图片进行与数据库中的字模进行匹配了，基本的思路就是看相应点的重合率，但是由于噪点的影响在对(6,8) (8,3)(5,9)的匹配时容易出错，俺自己针对已有的100幅图片数据采集进行分析，采用了双向匹配(图片与字模分别作为基点)，做了半天的测试终于 可以实现100%的识别率。
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#!/usr/bin/env python
#encoding=utf-8
import
Image,ImageEnhance,ImageFilter
import
Data
DEBUG
=
False
def
d_print(
*
msg):
global
DEBUG
if
DEBUG:
for
i
in
msg:
print
i,
print
else
:
pass
def
Get_Num(l
=
[]):
min1
=
[]
min2
=
[]
for
n
in
Data.N:
count1
=
count2
=
count3
=
count4
=
0
if
(
len
(l) !
=
len
(n)):
print
"Wrong pic"
exit()
for
i
in
range
(
len
(l)):
if
(l[i]
=
=
1
):
count1
+
=
1
if
(n[i]
=
=
1
):
count2
+
=
1
for
i
in
range
(
len
(l)):
if
(n[i]
=
=
1
):
count3
+
=
1
if
(l[i]
=
=
1
):
count4
+
=
1
d_print(count1,count2,count3,count4)
min1.append(count1
-
count2)
min2.append(count3
-
count4)
d_print(min1,
"/n"
,min2)
for
i
in
range
(
10
):
if
(min1[i] <
=
2
or
min2[i] <
=
2
):
if
((
abs
(min1[i]
-
min2[i])) <
10
):
return
i
for
i
in
range
(
10
):
if
(min1[i] <
=
4
or
min2[i] <
=
4
):
if
(
abs
(min1[i]
-
min2[i]) <
=
2
):
return
i
for
i
in
range
(
10
):
flag
=
False
if
(min1[i] <
=
3
or
min2[i] <
=
3
):
for
j
in
range
(
10
):
if
(j !
=
i
and
(min1[j] <
5
or
min2[j] <
5
)):
flag
=
True
else
:
pass
if
(
not
flag):
return
i
for
i
in
range
(
10
):
if
(min1[i] <
=
5
or
min2[i] <
=
5
):
if
(
abs
(min1[i]
-
min2[i]) <
=
10
):
return
i
for
i
in
range
(
10
):
if
(min1[i] <
=
10
or
min2[i] <
=
10
):
if
(
abs
(min1[i]
-
min2[i]) <
=
3
):
return
i
#end of function Get_Num
def
Pic_Reg(image_name
=
None
):
im
=
Image.
open
(image_name)
im
=
im.
filter
(ImageFilter.MedianFilter())
enhancer
=
ImageEnhance.Contrast(im)
im
=
enhancer.enhance(
2
)
im
=
im.convert(
'1'
)
im.show()
#all by pixel
s
=
12
#start postion of first number
w
=
10
#width of each number
h
=
15
#end postion from top
t
=
2
#start postion of top
im_new
=
[]
#split four numbers in the picture
for
i
in
range
(
4
):
im1
=
im.crop((s
+
w
*
i
+
i
*
2
,t,s
+
w
*
(i
+
1
)
+
i
*
2
,h))
im_new.append(im1)
s
=
""
for
k
in
range
(
4
):
l
=
[]
#im_new[k].show()
for
i
in
range
(
13
):
for
j
in
range
(
10
):
if
(im_new[k].getpixel((j,i))
=
=
255
):
l.append(
0
)
else
:
l.append(
1
)
s
+
=
str
(Get_Num(l))
return
s
print
Pic_Reg(
"./images/22.jpeg"
)
|
这里再提一下验证码识别的基本方法：截图，二值化、中值滤波去噪、分割、紧缩重排（让高矮统一）、字库特征匹配识别。
http://www.jb51.net/article/78598.htm