关于重命名文件夹中文件名的问题。我的文件名如下所示:
EPG CRO 24 Kitchen 09.2013.xsl
中间有命名空间,我使用了这样的代码:
#!/usr/bin/python
# -*- coding: utf-8 -*-
# Remove whitespace from files where EPG named with space " " replace with "_"
for filename in os.listdir("."):
if filename.find("2013|09 ") > 0:
newfilename = filename.replace(" ","_")
os.rename(filename, newfilename)
使用此代码,我删除了空格,但是如何从文件名中删除日期,使其看起来像这样:EPG_CRO_24_Kitche.xsl
.你能给我一些解决方案吗?
正则表达式
正如utdemir所回避的那样,正则表达式在这种情况下确实可以提供帮助。如果您从未接触过它们,一开始可能会感到困惑。结帐 https://www.debuggex.com/r/4RR6ZVrLC_nKYs8g 可帮助您构造正则表达式的有用工具。
溶液
更新的解决方案是:
import re
def rename_file(filename):
if filename.startswith('EPG') and ' ' in filename:
# s+ means 1 or more whitespace characters
# [0-9]{2} means exactly 2 characters of 0 through 9
# . means find a '.' character
# [0-9]{4} means exactly 4 characters of 0 through 9
newfilename = re.sub("s+[0-9]{2}.[0-9]{4}", '', filename)
newfilename = newfilename.replace(" ","_")
os.rename(filename, newfilename)
旁注
# Remove whitespace from files where EPG named with space " " replace with "_"
for filename in os.listdir("."):
if filename.find("2013|09 ") > 0:
newfilename = filename.replace(" ","_")
os.rename(filename, newfilename)
除非我弄错了,否则您上面发表的评论filename.find("2013|09 ") > 0
行不通。
鉴于以下情况:
In [76]: filename = "EPG CRO 24 Kitchen 09.2013.xsl"
In [77]: filename.find("2013|09 ")
Out[77]: -1
而您描述的评论,您可能需要更多类似以下内容:
In [80]: if filename.startswith('EPG') and ' ' in filename:
....: print('process this')
....:
process this
如果所有文件名的格式相同:NAME_20XX_XX.xsl
,那么你可以使用python的列表slicing
而不是regex
:
name.replace(' ','_')[:-12] + '.xsl'
如果日期格式始终相同;
>>> s = "EPG CRO 24 Kitchen 09.2013.xsl"
>>> re.sub("s+d{2}.d{4}..{3}$", "", s)
'EPG CRO 24 Kitchen'
小切片怎么样:
newfilename = input1[:input1.rfind(" ")].replace(" ","_")+input1[input1.rfind("."):]