where is a pattern?

[snippsat]

That's pattern is way longer than it need to be,have you look into the basic regex before?
Quick test.

>>> import re
>>> 
>>> text = "foo 2019-06-12.123 bar"
>>> r = re.search(r"\d{4}-\d{2}-\d{2}\.\d{3}", text).group()
>>> r
'2019-06-12.123'

If it's a common valid date format can parse with dateutil as mention bye @Gribouillis
I like pendulum the best(and most correct) date tool that's is made for Python in the latest years.

>>> import pendulum
>>> 
>>> d = '2019-06-12'
>>> pendulum.parse(d)
DateTime(2019, 6, 12, 0, 0, 0, tzinfo=Timezone('UTC')

It will fail on auto parse with 2019-06-12.123,but can write a own with formatter.

>>> import pendulum
>>> 
>>> dt = pendulum.from_format('2019-06-12.123', 'YYYY-DD-MM.hms')
>>> dt
DateTime(2019, 12, 6, 1, 2, 3, tzinfo=Timezone('UTC'))

As you see pendulum dos this way better than strptime().