Regular expressions for pattern matching time format in python

Question

I am looking to match following time formats using regular expressions in python and mark True or False whenever the match is found/ not found in a line. Sample text as below. How can I achieve this task using only regular expressions?

2am-8pm
2:00am - 8:00pm
08:00am-05:00pm
5:30am - 8:59pm

One can observe '_am - _pm' and '_am-_pm' patterns that are consistent in every notation.The colons and number format with spaces matching is what I have been attempting to do. below is what I found from here

HH:MM 12-hour format, optional leading 0, mandatory meridiems (AM/PM)
/((1[0-2]|0?[1-9]):([0-5][0-9]) ?([AaPp][Mm]))/

Sample text:

Lorem Ipsum is dummy text of the printing and typesetting industry between 2am-8pm. 
Contrary to popular belief, Lorem Ipsum is not simply random text. : False
Lorem has been the industry between 2:00am - 8:00pm standard dummy text since the 1500s. 
It has survived not only five centuries, but also between 08:00am-05:00pm 
It was popularised from 5:30am - 8:59pm with the release of Letraset sheets. 
More recently with desktop publishing software like Aldus PageMaker 983-765-0976.

Desired output:

Lorem Ipsum is dummy text of the printing and typesetting industry between 2am-8pm. : True
Contrary to popular belief, Lorem Ipsum is not simply random text. : False
Lorem has been the industry between 2:00am - 8:00pm standard dummy text since the 1500s. : True
It has survived not only five centuries, but also between 08:00am-05:00pm : True
It was popularised from 5:30am - 8:59pm with the release of Letraset sheets. : True
More recently with desktop publishing software like Aldus PageMaker 983-765-0976. : False

You only need yet to match the - with or without spaces around and then the same time format again. You should show what you have tried (edit the question to show it properly formatted). — Michael Butscher
– Michael Butscher, Commented Jun 25, 2020 at 21:42

Wiktor Stribiżew · Accepted Answer · 2020-06-25 21:55:19Z

You may use

(?i)(?<!\d)(?:1[0-2]|0?[1-9])(?::(?:[0-5][0-9]))?\s?[ap]m\s*-\s*(?:1[0-2]|0?[1-9])(?::(?:[0-5][0-9]))?\s?[ap]m\b

See the regex demo

Details

(?i) - case insensitive mode on
(?<!\d) - no digit before is allowed
(?:1[0-2]|0?[1-9])(?::(?:[0-5][0-9]))? - time pattern:
- (?:1[0-2]|0?[1-9]) - 0 to 12 with an optional leading 0 before 1-9 digits
- (?::(?:[0-5][0-9]))? - an optional minut sequence with : separator
\s? - an optional whitespace
[ap]m - a or p and then m
\s*-\s* - a hyphen enclosed with 0+ whitespaces
(?:1[0-2]|0?[1-9])(?::(?:[0-5][0-9]))?\s?[ap]m - the same time pattern as above
\b - word boundary.

Python demo:

import re
time = r'(?:1[0-2]|0?[1-9])(?::(?:[0-5][0-9]))?\s?[ap]m'
pattern = re.compile(r'(?i)(?<!\d){0}\s*-\s*{0}\b'.format(time))
texts = ['Lorem Ipsum is dummy text of the printing and typesetting industry between 2am-8pm.',
'Contrary to popular belief, Lorem Ipsum is not simply random text.',
'Lorem has been the industry between 2:00am - 8:00pm standard dummy text since the 1500s.',
'It has survived not only five centuries, but also between 08:00am-05:00pm',
'It was popularised from 5:30am - 8:59pm with the release of Letraset sheets.',
'More recently with desktop publishing software like Aldus PageMaker 983-765-0976.']
for text in texts:
    print (text, bool(pattern.search(text)), sep=" : ")

Output:

Lorem Ipsum is dummy text of the printing and typesetting industry between 2am-8pm. : True
Contrary to popular belief, Lorem Ipsum is not simply random text. : False
Lorem has been the industry between 2:00am - 8:00pm standard dummy text since the 1500s. : True
It has survived not only five centuries, but also between 08:00am-05:00pm : True
It was popularised from 5:30am - 8:59pm with the release of Letraset sheets. : True
More recently with desktop publishing software like Aldus PageMaker 983-765-0976. : False

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