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python - 如何从验证码中完全删除行

转载 作者:塔克拉玛干 更新时间:2023-11-03 03:16:31 26 4
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我编写了一个程序来从这个验证码中删除该行:

captcha with line

首先,我通过中值滤波器提高图像可见性

def apply_median_filter(self,img):
img_gray=img.convert('L')
img_gray=cv2.medianBlur(np.asarray(img_gray),3)
img_bw=(img_gray>np.mean(img_gray))*255
return img_bw

然后我尝试删除行:

def eliminate_zeros(self,vector):
return [(dex,v) for (dex,v) in enumerate(vector) if v!=0 ]

def get_line_position(self,img):
sumx=img.sum(axis=0)
list_without_zeros=self.eliminate_zeros(sumx)
min1,min2=heapq.nsmallest(2,list_without_zeros,key=itemgetter(1))
l=[dex for [dex,val] in enumerate(sumx) if val==min1[1] or val==min2[1]]
mindex=[l[0],l[len(l)-1]]
cols=img[:,mindex[:]]
col1=cols[:,0]
col2=cols[:,1]
col1_without_0=self.eliminate_zeros(col1)
col2_without_0=self.eliminate_zeros(col2)
line_length=len(col1_without_0)
dex1=col1_without_0[round(len(col1_without_0)/2)][0]
dex2=col2_without_0[round(len(col2_without_0)/2)][0]
p1=[dex1,mindex[0]]
p2=[dex2,mindex[1]]
return p1,p2,line_length

最后我按位置删除了一行:

def remove_line(self,p1,p2,LL,img):
m=(p2[0]-p1[0])/(p2[1]-p1[1]) if p2[1]!=p1[1] else np.inf
w,h=len(img),len(img[0])
x=[x for x in range(w)]
y=[p1[0]+k for k in [m*t for t in [v-p1[1] for v in x]]]
img_removed_line=img
for dex in range(w):
i,j=np.round([y[dex],x[dex]])
i=int(i)
j=int(j)
rlist=[]
while True:
f1=i
if img_removed_line[i,j]==0 and img_removed_line[i-1,j]==0:
break
rlist.append(i)
i=i-1

i,j=np.round([y[dex],x[dex]])
i=int(i)
j=int(j)
while True:
f2=i
if img_removed_line[i,j]==0 and img_removed_line[i+1,j]==0:
break
rlist.append(i)
i=i+1
print([np.abs(f2-f1),[LL+1,LL,LL-1]])
if np.abs(f2-f1) in [LL+1,LL,LL-1]:
rlist=list(set(rlist))
img_removed_line[rlist,j]=0

return img_removed_line

但在某些情况下线条并没有完全删除,我得到的验证码图像有一些噪音:

captcha with noise and less line

非常感谢您的帮助!

最佳答案

问题解决了!这是我编辑的 python 代码。这从验证码中删除了行。希望对您有所帮助:

from PIL import Image,ImageFilter
from scipy.misc import toimage
from operator import itemgetter
from skimage import measure
import numpy as np
import copy
import heapq
import cv2
import matplotlib.pyplot as plt
from scipy.ndimage.filters import median_filter

#----------------------------------------------------------------
class preprocessing:
def pre_proc_image(self,img):
#img_removed_noise=self.apply_median_filter(img)
img_removed_noise=self.remove_noise(img)
p1,p2,LL=self.get_line_position(img_removed_noise)
img=self.remove_line(p1,p2,LL,img_removed_noise)
img=median_filter(np.asarray(img),1)
return img

def remove_noise(self,img):
img_gray=img.convert('L')
w,h=img_gray.size
max_color=np.asarray(img_gray).max()
pix_access_img=img_gray.load()
row_img=list(map(lambda x:255 if x in range(max_color-15,max_color+1) else 0,np.asarray(img_gray.getdata())))
img=np.reshape(row_img,[h,w])
return img

def apply_median_filter(self,img):
img_gray=img.convert('L')
img_gray=cv2.medianBlur(np.asarray(img_gray),3)
img_bw=(img_gray>np.mean(img_gray))*255
return img_bw

def eliminate_zeros(self,vector):
return [(dex,v) for (dex,v) in enumerate(vector) if v!=0 ]

def get_line_position(self,img):
sumx=img.sum(axis=0)
list_without_zeros=self.eliminate_zeros(sumx)
min1,min2=heapq.nsmallest(2,list_without_zeros,key=itemgetter(1))
l=[dex for [dex,val] in enumerate(sumx) if val==min1[1] or val==min2[1]]
mindex=[l[0],l[len(l)-1]]
cols=img[:,mindex[:]]
col1=cols[:,0]
col2=cols[:,1]
col1_without_0=self.eliminate_zeros(col1)
col2_without_0=self.eliminate_zeros(col2)
line_length=len(col1_without_0)
dex1=col1_without_0[round(len(col1_without_0)/2)][0]
dex2=col2_without_0[round(len(col2_without_0)/2)][0]
p1=[dex1,mindex[0]]
p2=[dex2,mindex[1]]
return p1,p2,line_length

def remove_line(self,p1,p2,LL,img):
m=(p2[0]-p1[0])/(p2[1]-p1[1]) if p2[1]!=p1[1] else np.inf
w,h=len(img),len(img[0])
x=list(range(h))
y=list(map(lambda z : int(np.round(p1[0]+m*(z-p1[1]))),x))
img_removed_line=list(img)
for dex in range(h):
i,j=y[dex],x[dex]
i=int(i)
j=int(j)
rlist=[]
while True:
f1=i
if img_removed_line[i][j]==0 and img_removed_line[i-1][j]==0:
break
rlist.append(i)
i=i-1

i,j=y[dex],x[dex]
i=int(i)
j=int(j)
while True:
f2=i
if img_removed_line[i][j]==0 and img_removed_line[i+1][j]==0:
break
rlist.append(i)
i=i+1
if np.abs(f2-f1) in [LL+1,LL,LL-1]:
rlist=list(set(rlist))
for k in rlist:
img_removed_line[k][j]=0

return img_removed_line

关于python - 如何从验证码中完全删除行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42736616/

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