python get part of file name-CodePudding

I have a file name:

filename = 'Review Report - 2020-3.2021081716552'

Now I used 2 lines to just pick '2020-3', '2020-3' is dynamic ,can be other value:

report_name, ext = os.path.basename(filename).split(".")

a, b = report_name.split(" - ")

b will equal to '2020-3'

Can I just use 1 line code to get '2020-3' ?

CodePudding user response：

Use Maxsplit: One Liner:

date = filename.split(".")[0].split(" - ",maxsplit=1)[1]

Using one Liner is not recommended in Development of Projects, I recommend you to use the following:

filename = 'Review Report - 2020-3.2021081716552'
report_name, ext = filename.split(".")
a, b = report_name.split(" - ",maxsplit=1)

print(b)

OUTPUT:

2020-3

CodePudding user response：

The split() function returns a list, so you can use the indices to get the respective string instead of declaring two variables to hold the values.

import os

filename = 'Review Report - 2020-3.2021081716552'

the_thing_you_want = os.path.basename(filename).split(".")[0].split(" - ")[1]

print(the_thing_you_want)

It will return 2020-3.

If that helped, it'd be great if you could upvote and/or mark mine as the correct answer :)