Question

我创建了一个脚本来重命名给定目录中的索引文件

例如，如果目录包含以下文件＆gt;＆gt; （bar001.txt，bar004.txt，bar007.txt，foo2.txt，foo5.txt，morty.dat，rick.py）。我的脚本应该能够“仅”重命名索引文件，并缩小像这样的间隙＆gt;＆gt; （bar001.txt，bar002.txt，bar003.txt，foo1.txt，foo2.txt ...）。

我把下面的完整脚本放在哪里不起作用。该错误是合乎逻辑的，因为没有给出错误消息，但目录中的文件保持不变。

#! python3

import os, re

working_dir = os.path.abspath('.')

# A regex pattern that matches files with prefix,numbering and then extension
pattern = re.compile(r'''
    ^(.*?)        # text before the file number
    (\d+)         # file index
    (\.([a-z]+))$ # file extension
''',re.VERBOSE)

# Method that renames the items of an array
def rename(array):
    for i in range(len(array)):
        matchObj = pattern.search(array[i])
        temp = list(matchObj.group(2))
        temp[-1] = str(i+1)
        index = ''.join(temp)
        array[i] = matchObj.group(1) + index + matchObj.group(3)
    return(array)

array = []
directory = sorted(os.listdir('.'))

for item in directory:
    matchObj = pattern.search(item)
    if not matchObj:
        continue
    if len(array) == 0 or matchObj.group(1) in array[0]:
        array.append(item)
    else:
        temp = array
        newNames = rename(temp)
        for i in range(len(temp)):
            os.rename(os.path.join(working_dir,temp[i]),
                        os.path.join(working_dir,newNames[i]))
        array.clear() #reset array for other files
        array.append(item)

Answer 1

总而言之，您希望找到名称以数字和结尾的每个文件为每个具有相同名称的文件填写空白，除了数字后缀。您不想要创建任何新文件;相反，应使用最高数字的那些填补空白。

由于此摘要可以很好地转换为代码，因此我会这样做，而不是使用代码。

import re
import os

from os import path

folder  = 'path/to/folder/'
pattern = re.compile(r'(.*?)(\d+)(\.[a-z]+)$')
summary = {}

for fn in os.listdir(folder):
  m = pattern.match(fn)
  if m and path.isfile(path.join(folder, fn)):
    # Create a key if there isn't one, add the 'index' to the set
    # The first item in the tuple - len(n) - tells use how the numbers should be formatted later on
    name, n, ext = m.groups()
    summary.setdefault((name, ext), (len(n), set()))[1].add(int(n))

for (name, ext), (n, current) in summary.items():
  required = set(range(1, len(current)+1)) # You want these
  gaps     = required - current            # You're missing these
  superfluous = current - required         # You don't need these, so they should be renamed to fill the gaps

  assert(len(gaps) == len(superfluous)), 'Something has gone wrong'

  for old, new in zip(superfluous, gaps):
      oldname = '{name}{n:>0{pad}}{ext}'.format(pad=n, name=name, n=old, ext=ext)
      newname = '{name}{n:>0{pad}}{ext}'.format(pad=n, name=name, n=new, ext=ext)

      print('{old} should be replaced with {new}'.format(old=oldname, new=newname))

我认为这涵盖了它。

Python：检索和重命名目录中的索引文件

1 个答案: