Question

我正试图创建一个＆＃39;审查员＆＃39;读取“检查表”的程序。用语言来审查censorfile＆＃39;。我觉得我很接近，但是当我运行该程序时，它只会审查一行。任何帮助将不胜感激！

#!/usr/bin/python

import sys, os

censorlist = raw_input("File with words to censor? ")
if os.path.isfile(censorlist):
  print "The file", censorlist, "exists!"
else:
  print "The file", censorlist, "doesn't exist!"
  sys.exit()
  print ""

filterword = {}
for currentline in censorlist:
  line = currentline.strip()
  filterword.append(line)

censorfile = raw_input("File to censor using provided list? ")
script = []
for currentline in censorfile:
  line = currentline.strip()
  script.append(line)

for lines in script:
  results = []
  words = lines.split()
  for word in words:
    if word in filter:
      results.append("****")
    else:
      results.append(word)

resultfile = open(censorfile, 'w')
for items in results:
    resultfile.write(items)

Answer 1

您只需使用字符串的replace方法替换子字符串的每个实例...

'foo bar spam foo'.replace('foo', '****') # '**** bar spam ****'

您只需要遍历单词列表进行审核，并为每个单词调用replace。

for word in censorlist:
    string.replace(word, '****')

使用CensorFile从文件中检查单词 - Python

1 个答案: