Python: 检索和重命名目录中的索引文件

Question

我创建了一个脚本来重命名给定目录中的索引文件

例如如果该目录有以下文件>> (bar001.txt, bar004.txt, bar007.txt, foo2.txt, foo5.txt, morty.dat, rick.py).我的脚本应该能够重命名 'only' 索引文件并像这样缩小差距 >> (bar001.txt, bar002.txt, bar003.txt, foo1.txt, foo2.txt...）。

我把完整的脚本放在下面，但它不起作用。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，您想找到名称以数字结尾的每个文件并且为每组具有相同名称的文件填充空白，除了数字后缀。您不想创建任何新文件；相反，应使用最高数字的数字来填补空白。

由于此摘要可以很好地转换为代码，因此我将这样做而不是处理您的代码。

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))

我想差不多就涵盖了。

Python: 检索和重命名目录中的索引文件

Python: Retrieving and renaming indexed files in a directory

python

filesystems

file-management

python-3.x