[python]os模块之路径拼接问题-已解决

这个问题之前遇到过，但是忘了怎么解决的了，现在记录一下，怕自己又忘了

目标：通过遍历文件夹，拼接一个返回文件的路径，然后修改原文件文件名：

import os
import pandas as pd
path = r'D:\test\ori_data'
for root,dir,files in os.walk(path):
    for file in files:
#         print(os.path.join(root,file))
        pth = os.path.join(root,file)
        filename = os.path.splitext(file)[0]
#         print(type(filename))             
        re_path1 = os.path.join('re_',filename,'.txt')
        print(re_path1)


但是得到的结果如下，感觉这条路不大合适，有斜杆，后面用replace试试
re_\id1\.txt
re_\id10\.txt
re_\id11\.txt

直接用replace就可以把id替换，但是这仅限于文件名十分有规律的情况下；
如果文件名没有规律，其实还是需要找别的办法来替代,这里或许是因为用os拼接的所以才会出现有斜杆的情况
import os
import pandas as pd
path = r'D:\test\ori_data'
for root,dir,files in os.walk(path):
    for file in files:
#         print(os.path.join(root,file))
        pth = os.path.join(root,file)
        filename = os.path.splitext(file)[0]
#         print(type(filename))             
#         re_path1 = os.path.join('re_',filename,'.txt')
        re_p = file.replace('id','re_id')
        print(re_p)

 如果文件名不规律的情况下，把filename先放进一个list，然后再进行拼接：

import os
import pandas as pd
path = r'D:\test\ori_data'
filenamelist = []
for root,dir,files in os.walk(path):
    for file in files:
#         print(os.path.join(root,file))
        pth = os.path.join(root,file)
        filename = os.path.splitext(file)[0]
        filenamelist.append(filename)
#         re_path1 = os.path.join('re_',filename,'.txt')
# print(filenamelist)
    filepth1 = []
    for i in filenamelist:
#         直接拼接要+r
        a = r'D:\test\ori_data'+r'\re_'+i+'.txt'
        print(a)
#         os拼接更简单
#         re_path = os.path.join(root,a)
        filepth1.append(re_path)
        print(re_path)
# print(filepth1)

D:\test\ori_data\re_id1.txt
D:\test\ori_data\re_id10.txt
D:\test\ori_data\re_id11.txt
D:\test\ori_data\re_id12.txt

突然想起来，换一个文件夹保存更简单了：

import os
import pandas as pd
path = r'D:\test\ori_data'
filenamelist = []
for root,dir,files in os.walk(path):
    for file in files:
#         print(os.path.join(root,file))
        pth = os.path.join(root,file)
        filename = os.path.splitext(file)[0]
        filenamelist.append(filename)
        此处得加转义的
        re_path1 = os.path.join(r'D:\test\ori',filename)
        print(re_path1)