乍一看验证码还挺吓人的，背景杂点线条都有，字符也是变形的。
不过经过仔细观察，字符虽然是经过变形的，但是每个字符都是一样的....也就是可以通过字模匹配来完成识别了..
第一步：二值化
原始图：
灰度图：
二值化图：
二值化要先选择一个阈值，然后按照灰度进行分类
python
#二值化
def calcThreshold(img):
im=Image.open(img)
L = im.convert('L').histogram()
sum = 0
threshold = 0
for i in xrange(len(L)):
sum += L[i]
if sum >= 530:
threshold = i
break
return threshold
def binaryzation(img,threshold = 90):
table = []
for i in range(256):
if i < threshold:
table.append(0)
else:
table.append(1)
im=Image.open(img)
imgry = im.convert('L')
imout = imgry.point(table,'1')
return imout第二步：抽取字符
先用 floodfill 把连片的点抽取出来，去除太小的片(杂点)，然后对其到(0,0)
python
'''
抽取出字符矩阵 列表
'''
def extractChar(im):
OFFSETLIST = [(1,0),(0,1),(-1,0),(0,-1),(1,1),(-1,1),(1,-1),(-1,-1)]
pixelAccess = im.load()
num = 1
queue = []
ff = [[0]*im.size[1] for i in xrange(im.size[0])]
'''
floodfill 提出块
'''
for i in xrange(im.size[0]):
for j in xrange(im.size[1]):
'''
pixelAccess[i,j] == 0 表示是黑点
'''
if pixelAccess[i,j] == 0 and ff[i][j] == 0:
ff[i][j] = num
queue.append((i,j))
while len(queue) > 0 :
a,b = queue[0]
queue = queue[1:]
for offset1,offset2 in OFFSETLIST:
x,y = a + offset1, b + offset2
if x < 0 or x >= im.size[0]:continue
if y < 0 or y >= im.size[1]:continue
if pixelAccess[x,y] == 0 and ff[x][y] == 0:
ff[x][y] = num
queue.append((x,y))
num += 1
'''
字符点阵的坐标列表，对齐到 (0,0)
eg: [(1,2),(3,24),(54,23)]
'''
#初始化字符数组
info = {
"x_min":im.size[0],
"y_min":im.size[1],
"x_max":0,
"y_max":0,
"width":0,
"height":0,
"number":0,
"points":[]
}
charList = [copy.deepcopy(info) for i in xrange(num)]
#统计
for i in xrange(im.size[0]):
for j in xrange(im.size[1]):
if ff[i][j] == 0:
continue
id = ff[i][j]
if i < charList[id]['x_min']:charList[id]['x_min'] = i
if j < charList[id]['y_min']:charList[id]['y_min'] = j
if i > charList[id]['x_max']:charList[id]['x_max'] = i
if j > charList[id]['y_max']:charList[id]['y_max'] = j
charList[id]['number'] += 1
charList[id]['points'].append((i,j))
for i in xrange(num):
charList[i]['width'] = charList[i]['x_max'] - charList[i]['x_min'] + 1
charList[i]['height'] = charList[i]['y_max'] - charList[i]['y_min'] + 1
#修正偏移
charList[i]['points'] = [(x-charList[i]['x_min'], y-charList[i]['y_min']) for x,y in charList[i]['points'] ]
#过滤杂点
ret = [one for one in charList if one['number'] > 4]
#排序
ret.sort(lambda a,b:a['x_min'] < b['x_min'])
return ret第三步：识别字符
用预先整理好的字模数据跟每个 抽取出来的字符，进行比较。 由于可能抽取出来的字符和字模的长宽不太一致，所以需要进行“晃动”全匹配，取最大相似度的
python
'''
识别字符
'''
def charSimilarity(charA,charB):
s2 = set([(one[0],one[1]) for one in charB['points']])
sumlen = len(charA['points']) + len(charB['points'])
max = 0
# 晃动匹配
i_adjust = 1 if charB['width'] - charA['width'] >= 0 else -1
j_adjust = 1 if charB['height'] - charA['height'] >= 0 else -1
for i in xrange(0,charB['width'] - charA['width'] + i_adjust,i_adjust):
for j in xrange(0,charB['height'] - charA['height'] + j_adjust,j_adjust):
s1 = set([(one[0]+i,one[1]+j) for one in charA['points']])
sim = len(s1&s2) *2.0 / sumlen
if sim > max:
max = sim
return max
def recognise(one):
max = 0
ret = None
for char in CharMatrix:
s = charSimilarity(one,CharMatrix[char])
#print s * 100,"%"
if s > max:
ret = char
max = s
return ret整个步骤的顺序就是这样的：
python
'''
识别验证码
'''
def DoWork(img):
ans = []
threshold = calcThreshold(img)
print 'threshold:',threshold
im = binaryzation(img,threshold)
chars = extractChar(im)
for one in chars:
ans.append(recognise(one))
return ans完整代码：https://github.com/dingyaguang117/hack58/blob/master/hack58/src/ocr58.py
识别正确率应该在 70%左右，二值化没有做太多优化。。 识别效果图：
[图片因故障丢失]