在python中从反向切片一个字符串

Question

I had a string for example as below

i = 'backup_1.2012-12-19.tar.gz'

I need to fetch only the 2012-12-19.tar.gz from the string from the reverse(Because sometimes the string alters like 'backup_2.2012-12-20.tar.gz','backup_3.2012-12-21.tar.gz' )

So what ever the string is, i need to slice the characters from 2012 to .gz from reverse in python

Can anyone let me know how to do this python

Answer 1

Can't you just take everything after the first . ?

>>> s.split('.', 1)[1]
'2012-12-19.tar.gz'

Answer 2

Check out str.partition :

In [160]: i = 'backup_1.2012-12-19.tar.gz'

In [161]: i.partition('.')
Out[161]: ('backup_1', '.', '2012-12-19.tar.gz')

In [162]: i.partition('.')[-1]
Out[162]: '2012-12-19.tar.gz'

EDIT : If you want to "reverse" the date:

In [163]: i.partition('.')[-1].partition('.')[0]
Out[163]: '2012-12-19'

In [164]: i.partition('.')[-1].partition('.')[0].split('-')[::-1]
Out[164]: ['19', '12', '2012']

In [165]: '-'.join(i.partition('.')[-1].partition('.')[0].split('-')[::-1])
Out[165]: '19-12-2012'

Answer 3

您可以使用带有负索引的切片：

s[-17:]

Answer 4

 ".".join(i.split(".")[1:]) 

this will also work. But Jon's ans is more apt as there you will be doing only two operations.

or do it by following way. This will always return you the string from last even if the string contains more '.' in beginning.

".".join(i.split(".")[-3:])

Answer 5

i = "backup_1.2012-12-19.tar.gz"
z = i[9:26]
print(z)