Question

例如，根据顺序，有5个文件要重命名，1比1。

我可以通过将名称放入Excel电子表格并将其重命名为1来实现。但我希望从列表方式中学习它。

我尝试了以下内容：

import os

l = ['c:\\3536 OK-LKF.txt', 'c:\\2532 PK-HHY.txt', 'c:\\1256 OK-ASR.txt', 'c:\\521 OL-MRA.txt', 'c:\\2514 LP-GRW.txt']

ll = ['c:\\aa.txt', 'c:\\bb.txt', 'c:\\cc.txt', 'c:\\dd.txt', 'c:\\ee.txt']

for a in l:
    for b in ll:
        os.rename(a, b)

它不起作用，只重命名第一个文件。

从列表中执行此操作的正确方法是什么？是否存在重命名文件但风格不正确的风险？

Answer 1

您可以使用zip

for a,b in zip(l,ll):
    os.rename(a, b)

Answer 2

问题是嵌套循环，看看它的作用：

>>> l = ['c:\\3536 OK-LKF.txt', 'c:\\2532 PK-HHY.txt', 'c:\\1256 OK-ASR.txt', 'c:\\521 OL-MRA.txt', 'c:\\2514 LP-GRW.txt']
>>> 
>>> ll = ['c:\\aa.txt', 'c:\\bb.txt', 'c:\\cc.txt', 'c:\\dd.txt', 'c:\\ee.txt']
>>> for a in l:
...     for b in ll:
...         print('renaming {} to {}'.format(a,b))
... 
renaming c:\3536 OK-LKF.txt to c:\aa.txt
renaming c:\3536 OK-LKF.txt to c:\bb.txt
renaming c:\3536 OK-LKF.txt to c:\cc.txt
renaming c:\3536 OK-LKF.txt to c:\dd.txt
renaming c:\3536 OK-LKF.txt to c:\ee.txt
renaming c:\2532 PK-HHY.txt to c:\aa.txt
renaming c:\2532 PK-HHY.txt to c:\bb.txt
renaming c:\2532 PK-HHY.txt to c:\cc.txt
renaming c:\2532 PK-HHY.txt to c:\dd.txt
renaming c:\2532 PK-HHY.txt to c:\ee.txt
renaming c:\1256 OK-ASR.txt to c:\aa.txt
renaming c:\1256 OK-ASR.txt to c:\bb.txt
renaming c:\1256 OK-ASR.txt to c:\cc.txt
renaming c:\1256 OK-ASR.txt to c:\dd.txt
renaming c:\1256 OK-ASR.txt to c:\ee.txt
renaming c:\521 OL-MRA.txt to c:\aa.txt
renaming c:\521 OL-MRA.txt to c:\bb.txt
renaming c:\521 OL-MRA.txt to c:\cc.txt
renaming c:\521 OL-MRA.txt to c:\dd.txt
renaming c:\521 OL-MRA.txt to c:\ee.txt
renaming c:\2514 LP-GRW.txt to c:\aa.txt
renaming c:\2514 LP-GRW.txt to c:\bb.txt
renaming c:\2514 LP-GRW.txt to c:\cc.txt
renaming c:\2514 LP-GRW.txt to c:\dd.txt
renaming c:\2514 LP-GRW.txt to c:\ee.txt

您的程序可以通过迭代zip(l,ll)：

来修复

for old, new in zip(l,ll):
    os.rename(old,new)

Answer 3

如果你想一对一地重命名，试试这个：

import os

l = ['c:\\3536 OK-LKF.txt', 'c:\\2532 PK-HHY.txt', 'c:\\1256 OK-ASR.txt', 'c:\\521 OL-MRA.txt', 'c:\\2514 LP-GRW.txt']

ll = ['c:\\aa.txt', 'c:\\bb.txt', 'c:\\cc.txt', 'c:\\dd.txt', 'c:\\ee.txt']

for a in l:
    os.rename(a, ll[l.index(a)])

Python基础 - 重命名文件

3 个答案: