Python regular expression find and output a part of the pattern in multiple times

Question

Assume I have a string as follows:

2021/12/23 13:00 14:00 2021/12/24 13:00 14:00 15:00

Where a date comes with time several times. Is it possible that regular expression can find all time after each date such as follows?

[('2021/12/23', '13:00','14:00'), ('2021/12/24', '13:00','14:00','15:00')]

I tried the following code in Python, but it returns only the first time:

re.findall(r'(\d+/\d+/\d+)(\s\d+\:\d+)+','2021/12/23 13:00 14:00 2021/12/24 13:00 14:00 15:00')

>>>[('2021/12/23', ' 14:00'), ('2021/12/24', ' 15:00')]

Appendix: The original problem is in fact a more complicated case, where there are several texts between time, and it is difficult to replace them to '' directly:

2021/12/31 14:00 start 15:00 end 17:00 pending 18:00 ok 2021/12/31 14:00 begin 15:00 end  17:00 start 18:00 suspend

So the robust method here should be this answer.

pattern = regex.compile(r'(?P<date>\d{4}/\d+/\d+)(?:\s?(?P<time>(\d+:\d+))\s+([^\d]+))+')
for m in pattern.finditer('2021/12/31 14:00 start 15:00 end 17:00 pending 18:00 ok 2021/12/30 14:00 begin 15:00 end  17:00 ggh 18:00 suspend'):
    print(m.capturesdict())
>>> {'date': ['2021/12/31'], 'time': ['14:00', '15:00', '17:00', '18:00']}
{'date': ['2021/12/30'], 'time': ['14:00', '15:00', '17:00', '18:00']}

Tim Biegeleisen · Accepted Answer · 2022-01-13 09:53:21Z

2

Use re.findall:

inp = '2021/12/23 13:00 14:00 2021/12/24 13:00 14:00 15:00'
matches = re.findall(r'\d{4}/\d{2}/\d{2}(?: \d{1,2}:\d{2})*', inp)
print(matches)

This prints:

['2021/12/23 13:00 14:00', '2021/12/24 13:00 14:00 15:00']

Explanation of regex:

\d{4}/\d{2}/\d{2}    match a date in YYYY/MM/DD format
(?: \d{1,2}:\d{2})*  match a space followed by hh:mm time, 0 or more times

answered Jan 13, 2022 at 9:53

Tim Biegeleisen

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anubhava · Accepted Answer · 2022-01-13 09:56:34Z

2

You can use this findall + split solution:

import re

s = '2021/12/23 13:00 14:00 2021/12/24 13:00 14:00 15:00'

for i in re.findall(r'\d+/\d+/\d+(?:\s\d+\:\d+)+', s): print (i.split())

Output:

['2021/12/23', '13:00', '14:00']
['2021/12/24', '13:00', '14:00', '15:00']

Code Demo

\d+/\d+/\d+(?:\s\d+\:\d+)+ matches a date string followed by 1 or more time strings.

You. could also use:

print ([i.split() for i in re.findall(r'\d+/\d+/\d+(?:\s\d+\:\d+)+', s)])

To get output:

[['2021/12/23', '13:00', '14:00'], ['2021/12/24', '13:00', '14:00', '15:00']]

answered Jan 13, 2022 at 9:56

anubhava

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2 Comments

natsuapo Over a year ago

Sorry, the case is just an example, it will be much more complex than using split.

anubhava Over a year ago

Provide details or more examples in question. An answer is based on the question asked.

Wiktor Stribiżew · Accepted Answer · 2022-01-13 09:58:56Z

1

You can use PyPi regex library to get the following to work:

import regex
pattern = regex.compile(r'(?P<date>\d+/\d+/\d+)(?:\s+(?P<time>\d+:\d+))+')
for m in pattern.finditer('2021/12/23 13:00 14:00 2021/12/24 13:00 14:00 15:00'):
    print(m.capturesdict())

Output:

{'date': ['2021/12/23'], 'time': ['13:00', '14:00']}
{'date': ['2021/12/24'], 'time': ['13:00', '14:00', '15:00']}

See the Python demo.

Since PyPi regex library does not "forget" all captures inside a group, and provided the groups are named, the match.capturesdict() returns the dictionary of all groups with their captures.

answered Jan 13, 2022 at 9:58

Wiktor Stribiżew

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1 Comment

Wiktor Stribiżew Over a year ago

FYI: one can install PyPi regex module by running pip install regex (or pip3 install regex depending on environment) in the terminal/console window.

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