Question

我需要从图像中提取URL和错误消息。图像的DPI为300，尺寸为W 1536像素H 1068像素。

import glob
list_f = glob.glob('/home/agnihotri/img/*.jpg')
if len(list_f) != 0:
     res_final = []
     for f in list_f:

          from PIL import Image
          import pytesseract
          import argparse
          import cv2
          import os, glob
          from matplotlib import pyplot as plt
          import re

          if not os.path.exists('processed/images'):
              os.makedirs('processed/images')

          image = cv2.imread(f)
          if image.shape[0]< 715 or image.shape[1]<1020:
              print('Resolution too low')
          else:
              image = cv2.resize(image, None, fx = 10, fy = 10)
              config='-c tessedit_char_whitelist=0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ:.// -psm 3'

              text = pytesseract.image_to_string(image, config=config, nice=2)
          try:
              url = (''.join(
                   re.findall('(http|ftp|https):(//|H)([\w_-]+(?:(?:\.[\w_-]+)+))([\w.,@?^=%&:/~+#-]*[\w@?^=%&/~+#-])?',
                               text)[0]).replace('H', '://').replace('::', ':').replace('I', '/').replace('BBBB', '8888'))
           except:
               url = 'No url found in image'
               print(url)

           err_message = []
           for l in text.split(sep='\n'):
               if any(i in l.lower() for i in ['error', 'cannot', 'warning', 'problem', "can't", 'unable', 'could not', 'invalid']):

                    err_message.append(l)

           if len(err_message) == 0:
               err_message=['No error message found']
           err_message = '\n'.join(err_message)
           print (err_message)

但是，在执行上述代码后，我确实收到了错误消息：

No url found in image
Traceback (most recent call last): File "./oc_script.py", line 37, in <module>
TypeError: split() takes no keyword arguments

输入图片的链接是https://drive.google.com/open?id=1drdqJjR06EIkt_YbBcVLDZW1s8FRj1J6

有谁知道为什么会这样？

Answer 1

更改

Not_Passed

到

for l in text.split(sep='\n'):

从图像中提取URL和错误消息

1 个答案: