Python: Extract two dates from string

Question

I have a string s which contains two dates in it and I am trying to extract these two dates in order to subtract them from each other to count the number of days in between. In the end I am aiming to get a string like this: s = "o4_24d_20170708_20170801"

At the company I work we can't install additional packages so I am looking for a solution using native python. Below is what I have so far by using the datetime package which only extracts one date: How can I get both dates out of the string?

import re, datetime
s = "o4_20170708_20170801"
match = re.search('\d{4}\d{2}\d{2}', s)
date = datetime.datetime.strptime(match.group(), '%Y%m%d').date()
print date

How exactly do you arrive at "o4_24d_20170708_20170801"? In your input ("o4_20170708_20170801") 24 does not exist anywhere. — Ajax1234
– Ajax1234, Commented May 3, 2018 at 14:01
regex seems overly complicated for this. Why don't you use s.split('_')? — Mike Scotty
– Mike Scotty, Commented May 3, 2018 at 14:02
Could you post a list of possible inputs? If all of them are the same length, you can just access by index, or split by _ — Susensio
– Susensio, Commented May 3, 2018 at 14:03
@Ajax1234 sorry maybe it was not clear; I am planning to subtract these two dates and then indicate it in the string. — kurdtc
– kurdtc, Commented May 3, 2018 at 14:04
@Susenio: unfortunately the input is not always the same length, the strings are taken from filenames with various extensions such as e.g. "o5_20170808_20160801_test.tv14.tif" etc. — kurdtc
– kurdtc, Commented May 3, 2018 at 14:06

Susensio · Accepted Answer · 2018-05-03 14:21:30Z

3

from datetime import datetime
import re

s = "o4_20170708_20170801"
pattern = re.compile(r'(\d{8})_(\d{8})')
dates = pattern.search(s)
# dates[0] is full match, dates[1] and dates[2] are captured groups
start = datetime.strptime(dates[1], '%Y%m%d')
end = datetime.strptime(dates[2], '%Y%m%d')
difference = end - start

print(difference.days)

will print

then, you could do something like:

days = 'd{}_'.format(difference.days)
match_index = dates.start()
new_name = s[:match_index] + days + s[match_index:]
print(new_name)

to get

o4_d24_20170708_20170801

edited May 3, 2018 at 14:21

answered May 3, 2018 at 14:15

Susensio

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Comments

NBlaine · Accepted Answer · 2018-05-03 14:07:45Z

0

import re, datetime
s = "o4_20170708_20170801"
match = re.findall('\d{4}\d{2}\d{2}', s)
for a_date in match:
  date = datetime.datetime.strptime(a_date, '%Y%m%d').date()
  print date

This will print:

2017-07-08
2017-08-01

Your regex was working correctly at regexpal

edited May 3, 2018 at 14:07

answered May 3, 2018 at 14:06

NBlaine

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4 Comments

jwoos Over a year ago

The regex can just be simplified to \d{8} as well

NBlaine Over a year ago

sure, I think the issue was with re.search not doing what he wanted rather than the reg itself

kurdtc Over a year ago

@NathanBlaine Thank you, this looks very promising. However I would like to subtract the dates from each other in order to calculate the days inbetween. When I type 'date' into the console I get 'datetime.date(2017,7,8)'. Is it possible to write it into two different variables in order to calculate the days inbetween the two dates?

NBlaine Over a year ago

@kurdtc it sure is possible! You seem like a capable dude though I bet you can figure it out ;)

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