RegEx for matching the month, day and year - regex

Question

I'm trying to find a regular expression to extract the month, day and year from a datetime stamp in this format:
01/20/2019 12:34:54
It should return a list:
['01', '20', '2019']
I know this can be solved using:
dt.split(' ')[0].split('/')
But, I'm trying to find a regex to do it:
[^\/\s]+
But, I need it to exclude everything after the space.

Answer 1

As you are expecting the date month and year to be returned as a list, you can use this Python code,
import re
s = '01/20/2019 12:34:54'
print(re.findall(r'\d+(?=[ /])', s))
Prints,
['01', '20', '2019']
Otherwise, you can better write your regex as,
(\d{2})/(\d{2})/(\d{4})
And get date, month and year from group1, group2 and group3
Regex Demo
Python code in this way should be,
import re
s = '01/20/2019 12:34:54'
m = re.search(r'(\d{2})/(\d{2})/(\d{4})', s)
if m:
print([m.group(1), m.group(2), m.group(3)])
Prints,
['01', '20', '2019']

Answer 2

You should absolutely be taking advantage of Python's date/time API here. Use strptime to parse your input datetime string to a bona fide Python datetime. Then, just build a list, accessing the various components you need.
dt = "01/20/2019 12:34:54"
dto = datetime.strptime(dt, '%m/%d/%Y %H:%M:%S')
list = [dto.month, dto.day, dto.year]
print(list)
[1, 20, 2019]
If you really want/need to work with the original datetime string, then split provides an option, without even formally using a regex:
dt = "01/20/2019 12:34:54"
dt = dt.split()[0].split('/')
print(dt)
['01', '20', '2019']

Answer 3

This RegEx might help you to do so.
([0-9]+)\/([0-9]+)\/([0-9]+)\s[0-9]+:[0-9]+:[0-9]+
Code:
import re
string = '01/20/2019 12:34:54'
matches = re.search(r'([0-9]+)/([0-9]+)/([0-9]+)', string)
if matches:
print([matches.group(1), matches.group(2), matches.group(3)])
else:
print('Sorry! No matches! Something is not right! Call 911')
Output
['01', '20', '2019']