我还是初学者,但我想写一个字符识别程序.该计划尚未准备好.我编辑了很多,因此评论可能不完全匹配.我将使用8连接进行连接组件标记.
from PIL import Image
import numpy as np
im = Image.open("D:\\Python26\\PYTHON-PROGRAMME\\bild_schrift.jpg")
w,h = im.size
w = int(w)
h = int(h)
#2D-Array for area
area = []
for x in range(w):
area.append([])
for y in range(h):
area[x].append(2) #number 0 is white, number 1 is black
#2D-Array for letter
letter = []
for x in range(50):
letter.append([])
for y in range(50):
letter[x].append(0)
#2D-Array for label
label = []
for x in range(50):
label.append([])
for y in range(50):
label[x].append(0)
#image to number conversion
pix = im.load()
threshold = 200
for x in range(w):
for y in range(h):
aaa = pix[x, y]
bbb = aaa[0] + aaa[1] + aaa[2] #total value
if bbb<=threshold:
area[x][y] = 1
if bbb>threshold:
area[x][y] = 0
np.set_printoptions(threshold='nan', linewidth=10)
#matrix transponation
ccc = np.array(area)
area = ccc.T #better solution?
#find all black pixel and set temporary label numbers
i=1
for x in range(40): # width (later)
for y in range(40): # heigth (later)
if area[x][y]==1:
letter[x][y]=1
label[x][y]=i
i += 1
#connected components labeling
for x in range(40): # width (later)
for y in range(40): # heigth (later)
if area[x][y]==1:
label[x][y]=i
#if pixel has neighbour:
if area[x][y+1]==1:
#pixel and neighbour get the lowest label
pass # tomorrows work
if area[x+1][y]==1:
#pixel and neighbour get the lowest label
pass # tomorrows work
#should i also compare pixel and left neighbour?
#find width of the letter
#find height of the letter
#find the middle of the letter
#middle = [width/2][height/2] #?
#divide letter into 30 parts --> 5 x 6 array
#model letter
#letter A-Z, a-z, 0-9 (maybe more)
#compare each of the 30 parts of the letter with all model letters
#make a weighting
#print(letter)
im.save("D:\\Python26\\PYTHON-PROGRAMME\\bild2.jpg")
print('done')
jbochi.. 34
OCR确实不是一件容易的事.这就是为什么文本CAPTCHA仍然有效:)
要仅讨论字母提取而不是模式识别,您用来分隔字母的技术称为连接组件标签.由于您要求更有效的方法来执行此操作,请尝试实现本文中描述的两遍算法.可以在文章Blob提取中找到另一种描述.
编辑:这是我建议的算法的实现:
import sys
from PIL import Image, ImageDraw
class Region():
def __init__(self, x, y):
self._pixels = [(x, y)]
self._min_x = x
self._max_x = x
self._min_y = y
self._max_y = y
def add(self, x, y):
self._pixels.append((x, y))
self._min_x = min(self._min_x, x)
self._max_x = max(self._max_x, x)
self._min_y = min(self._min_y, y)
self._max_y = max(self._max_y, y)
def box(self):
return [(self._min_x, self._min_y), (self._max_x, self._max_y)]
def find_regions(im):
width, height = im.size
regions = {}
pixel_region = [[0 for y in range(height)] for x in range(width)]
equivalences = {}
n_regions = 0
#first pass. find regions.
for x in xrange(width):
for y in xrange(height):
#look for a black pixel
if im.getpixel((x, y)) == (0, 0, 0, 255): #BLACK
# get the region number from north or west
# or create new region
region_n = pixel_region[x-1][y] if x > 0 else 0
region_w = pixel_region[x][y-1] if y > 0 else 0
max_region = max(region_n, region_w)
if max_region > 0:
#a neighbour already has a region
#new region is the smallest > 0
new_region = min(filter(lambda i: i > 0, (region_n, region_w)))
#update equivalences
if max_region > new_region:
if max_region in equivalences:
equivalences[max_region].add(new_region)
else:
equivalences[max_region] = set((new_region, ))
else:
n_regions += 1
new_region = n_regions
pixel_region[x][y] = new_region
#Scan image again, assigning all equivalent regions the same region value.
for x in xrange(width):
for y in xrange(height):
r = pixel_region[x][y]
if r > 0:
while r in equivalences:
r = min(equivalences[r])
if not r in regions:
regions[r] = Region(x, y)
else:
regions[r].add(x, y)
return list(regions.itervalues())
def main():
im = Image.open(r"c:\users\personal\py\ocr\test.png")
regions = find_regions(im)
draw = ImageDraw.Draw(im)
for r in regions:
draw.rectangle(r.box(), outline=(255, 0, 0))
del draw
#im.show()
output = file("output.png", "wb")
im.save(output)
output.close()
if __name__ == "__main__":
main()
这是输出文件:
死链接
它并非100%完美,但由于您只是出于学习目的而这样做,这可能是一个很好的起点.使用每个角色的边界框,您现在可以像其他人在这里建议的那样使用神经网络.
OCR确实不是一件容易的事.这就是为什么文本CAPTCHA仍然有效:)
要仅讨论字母提取而不是模式识别,您用来分隔字母的技术称为连接组件标签.由于您要求更有效的方法来执行此操作,请尝试实现本文中描述的两遍算法.可以在文章Blob提取中找到另一种描述.
编辑:这是我建议的算法的实现:
import sys
from PIL import Image, ImageDraw
class Region():
def __init__(self, x, y):
self._pixels = [(x, y)]
self._min_x = x
self._max_x = x
self._min_y = y
self._max_y = y
def add(self, x, y):
self._pixels.append((x, y))
self._min_x = min(self._min_x, x)
self._max_x = max(self._max_x, x)
self._min_y = min(self._min_y, y)
self._max_y = max(self._max_y, y)
def box(self):
return [(self._min_x, self._min_y), (self._max_x, self._max_y)]
def find_regions(im):
width, height = im.size
regions = {}
pixel_region = [[0 for y in range(height)] for x in range(width)]
equivalences = {}
n_regions = 0
#first pass. find regions.
for x in xrange(width):
for y in xrange(height):
#look for a black pixel
if im.getpixel((x, y)) == (0, 0, 0, 255): #BLACK
# get the region number from north or west
# or create new region
region_n = pixel_region[x-1][y] if x > 0 else 0
region_w = pixel_region[x][y-1] if y > 0 else 0
max_region = max(region_n, region_w)
if max_region > 0:
#a neighbour already has a region
#new region is the smallest > 0
new_region = min(filter(lambda i: i > 0, (region_n, region_w)))
#update equivalences
if max_region > new_region:
if max_region in equivalences:
equivalences[max_region].add(new_region)
else:
equivalences[max_region] = set((new_region, ))
else:
n_regions += 1
new_region = n_regions
pixel_region[x][y] = new_region
#Scan image again, assigning all equivalent regions the same region value.
for x in xrange(width):
for y in xrange(height):
r = pixel_region[x][y]
if r > 0:
while r in equivalences:
r = min(equivalences[r])
if not r in regions:
regions[r] = Region(x, y)
else:
regions[r].add(x, y)
return list(regions.itervalues())
def main():
im = Image.open(r"c:\users\personal\py\ocr\test.png")
regions = find_regions(im)
draw = ImageDraw.Draw(im)
for r in regions:
draw.rectangle(r.box(), outline=(255, 0, 0))
del draw
#im.show()
output = file("output.png", "wb")
im.save(output)
output.close()
if __name__ == "__main__":
main()
这是输出文件:
死链接
它并非100%完美,但由于您只是出于学习目的而这样做,这可能是一个很好的起点.使用每个角色的边界框,您现在可以像其他人在这里建议的那样使用神经网络.
OCR非常非常困难.即使使用计算机生成的字符,如果您事先不知道字体和字体大小,也会非常具有挑战性.即使你完全匹配字符,我也不会把它称为"开始"编程项目; 这很微妙.
如果你想识别扫描或手写的字符,那就更难了 - 你需要使用高级数学,算法和机器学习.有很多书和数千篇关于这个主题的文章,所以你不需要重新发明轮子.
我很佩服你的努力,但我认为你还没有达到任何实际困难.到目前为止,您只是随机探索像素并将它们从一个阵列复制到另一个阵列.你还没有真正做过任何比较,我不确定你的"随机游走"的目的.
为什么随意？编写正确的随机算法非常困难.我建议首先从确定性算法开始.
你为什么要从一个阵列复制到另一个阵列？为什么不直接比较？
当你得到比较时,你将不得不处理这个图像与"原型"不完全相同的事实,并且不清楚你将如何处理它.
根据你到目前为止所编写的代码,我有一个想法:尝试编写一个程序,通过图像中的"迷宫"找到它.输入将是图像,加上起始像素和目标像素.输出是从开始到目标的迷宫路径.这是一个比OCR更容易解决的问题 - 解决迷宫是计算机非常适合的事情 - 但它仍然充满乐趣和挑战性.
目前大多数OCR算法都基于神经网络算法. Hopfield网络是一个很好的起点.基于C语言中提供的Hopfield模型,我在python中构建了一个非常基本的图像识别算法,与您描述的类似.我在这里发布了完整的来源.这是一个玩具项目,不适合真正的OCR,但可以让你开始朝着正确的方向前进.
Hopfield模型用作自动关联存储器来存储和调用一组位图图像.通过计算相应的权重矩阵来存储图像.此后,从任意配置开始,存储器将精确地定位在存储的图像上,该存储的图像在汉明距离方面最接近起始配置.因此,给定存储图像的不完整或损坏版本,网络能够调用相应的原始图像.
可以在这里找到一个带有示例玩具的Java小程序; 使用数字0-9的示例输入训练网络.在右侧的框中绘制,单击测试并查看网络结果.
不要让数学符号吓到你,一旦你得到源代码,算法就很简单了.
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