Python基礎 - 重命名文件

Question

例如，根據序列，有5個文件要重命名，1到1。

我可以通過將名稱放入Excel電子表格並將其重命名為1來實現。但我希望從列表方式中學習它。

我嘗試了以下方法：

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，只重命名第一個文件。

從列表中做到這一點的正確方法是什么？ 是否存在重命名文件但風格不正確的風險？

Answer 1

你可以使用zip

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

Answer 2

問題是嵌套循環，看看它做了什么：

>>> 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

你的程序可以通過迭代zip(l,ll)來修復：

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

Answer 3

如果你想重命名一對一試試這個：

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)])