Python替换不使用list

Question

我的代码读取目录并将带有扩展名的文件存储到列表中.我想要做的是用替换摆脱扩展.但是,它没有保存到列表中.

 print projectFilenames

 for f in projectFilenames:
  print f
  f = f.replace('.txt','')
  f = f.replace('.mdown','')
  f = f.replace('.markdown','')
  print f

 print projectFilenames

这是我的输出

['2010-10-30-markdown-example.txt','2010-12-29-hello-world.mdown','2011-1-1-tester.markdown']
2010-10-30-markdown-example .txt
2010-10-30-markdown-example
2010-12-29-hello-world.mdown
2010-12-29-hello-world
2011-1-1-tester.markdown
2011-1-1-tester
['2010 -10-30-markdown-example.txt','2010-12-29-hello-world.mdown','2011-1-1-tester.markdown']

我究竟做错了什么？

Answer 1

列表不会更改,因为您没有更新它.您没有以任何方式触及projectFilenames列表(也不是列表中的字符串.Python变量不是指针)以下是一种方法:

newlist = []
for f in projectFilenames:
    f = f.replace('.txt','')
    f = f.replace('.mdown','')
    f = f.replace('.markdown','')
    newlist.append(f)

projectFilenames = newlist

另外,看一下os.path模块,那里有一些函数可以切断文件扩展名.os.path.splitext()具体来说.所以另一种方法是:

newlist = []
for f in projectFilenames:
    f = os.path.splitext(f)[0]
    newlist.append(f)

projectFilenames = newlist

这反过来可以简化为(并符合PEP 8):

>>> import os
>>> project_filenames = ['2010-10-30-markdown-example.txt', '2010-12-29-hello-world.mdown', '2011-1-1-tester.markdown']
>>> project_filenames = [os.path.splitext(f)[0] for f in project_filenames]
>>> project_filenames
['2010-10-30-markdown-example', '2010-12-29-hello-world', '2011-1-1-tester']