与Python中的IEEE时间戳格式匹配的最佳正则表达式

Question

I did some searching but didn't find this specifically, and I'm sure it's going to be a quick answer.

I have a python script parsing IEEE date and time stamps out of strings, but I think I'm using python's match objects wrong.

import re
stir = "foo_2015-07-07-17-58-26.log"
timestamp = re.search("([0-9]+-){5}[0-9]+", stir).groups()
print timestamp

Produces

58-

When my intent is to get

2015-07-07-17-58-26

Is there a pre-canned regex that would work better here? Am I getting tripped up on re's capture groups? Why is the length of the groups() tuple only 1?

Edit

I was misinterpreting the way capture groups work in python's re module - there is only one set of parentheses in the statement, so the re module returned the most recently grabbed capture group - the "58-".

The way I ended up doing it was by referencing group(0), as Dawg suggests below.

timestamp = re.search("([0-9]+-){5}[0-9]+", stir)

print timestamp.group(0)
2015-07-07-17-58-26

Answer 1

You need a single capture group or groups:

(\d\d\d\d-\d\d-\d\d-\d\d-\d\d-\d\d)

Demo

Or, use nested capture groups:

>>> re.search(r'(\d{4}(?:-\d{2}){5})', 'foo_2015-07-07-17-58-26.log')
<_sre.SRE_Match object at 0x100b49dc8>
>>> _.group(1)
'2015-07-07-17-58-26'

Or, you can use your pattern and just use group(0) instead of groups() :

>>> re.search("([0-9]+-){5}[0-9]+", "foo_2015-07-07-17-58-26.log").group(0)
'2015-07-07-17-58-26'

Or, use findall with an additional capture group (and the other a non capture group):

>>> re.findall("((?:[0-9]+-){5}[0-9]+)", 'foo_2015-07-07-17-58-26.log')
['2015-07-07-17-58-26']

But that will find the digits that are not part of the timestamp.

Answer 2

if you want the timestamp in one match object, i think this should work

 \\d{4}(?:\\d{2}){5} 

then use group() or group(0)

also, match.groups actually returns the number of group objects, you should try .group() instead (your code would still not work though because you grouped the 5 sets of numbers in and the final -58 would be omitted

Answer 3

I'd use below:

_(\d{4}-\d{2}-\d{2}-\d{2}-\d{2}-\d{2}).

_ and . to mark the starting and the end.

import re
r = r'_(\d{4}-\d{2}-\d{2}-\d{2}-\d{2}-\d{2}).'
s = 'some string'
lst = re.findall(s,r)

link

Answer 4

You might want

re.findall(r"([0-9-]+)", stir)


>>> import re
>>> stir = "foo_2015-07-07-17-58-26.log"
>>> re.findall(r"([0-9-]+)", stir)
['2015-07-07-17-58-26']