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Convert custom string date to date

Is there a way to convert a string date that is stored in some non-traditional custom manner into a date using datetime (or something equivalent)? The dates I am dealing with are S3 partitions that look like this:
year=2023/month=2/dayofmonth=3
I can accomplish this with several replaces but im hoping to find a clean single operation to do this.

You might provide datetime.datetime.strptime with format string holding text, in this case
import datetime
dt = datetime.datetime.strptime("year=2023/month=2/dayofmonth=3","year=%Y/month=%m/dayofmonth=%d")
d = dt.date()
print(d) # 2023-02-03

you can do that converting your string into a date object using "datetime" combined with strptime() method.
The strtime() takes two arguments, the first is the string to be parsed, and the second a string with the format.
Here's an example:
from datetime import datetime
# your string
date_string = "year=2023/month=2/dayofmonth=3"
# parse the string into a datetime object
date = datetime.strptime(date_string, "year=%Y/month=%m/dayofmonth=%d")
# print the datetime object
print(date)

Can't produce a certain format of date when I use a customized date instead of now

I'm trying to format a date to a customized one. When I use datetime.datetime.now(), I get the right format of date I'm after. However, my intention is to get the same format when I use 1980-01-22 instead of now.
import datetime
date_string = "1980-01-22"
item = datetime.datetime.now(datetime.timezone.utc).isoformat(timespec="milliseconds").replace("+00:00", "Z")
print(item)
Output I get:
2021-05-04T09:52:04.010Z
How can I get the same format of date when I use a customized date, as in 1980-01-22 instead of now?

MrFuppes suggestion in the comments is the shortest way to accomplish your date conversion and formatting use case.
Another way is to use the Python module dateutil. This module has a lot of flexibility and I use it all the time.
Using dateutil.parser.parse:
from dateutil.parser import parse
# ISO FORMAT
ISO_FORMAT_MICROS = "%Y-%m-%dT%H:%M:%S.%f%z"
# note the format of these strings
date_strings = ["1980-01-22",
"01-22-1980",
"January 22, 1980",
"1980 January 22"]
for date_string in date_strings:
dt = parse(date_string).strftime(ISO_FORMAT_MICROS)
# strip 3 milliseconds for the output and add the ZULU time zone designator
iso_formatted_date = f'{dt[:-3]}Z'
print(iso_formatted_date)
# output
1980-01-22T00:00:00.000Z
1980-01-22T00:00:00.000Z
1980-01-22T00:00:00.000Z
1980-01-22T00:00:00.000Z
Using dateutil.parser.isoparse:
from dateutil.parser import isoparse
from dateutil.tz import *
dt = isoparse("1980-01-22").isoformat(timespec="milliseconds")
iso_formatted_date = f'{dt}Z'
print(iso_formatted_date)
# output
1980-01-22T00:00:00.000Z

Is this what your trying to achieve?
date_string = "1980-01-22"
datetime.datetime.strptime(date_string, "%Y-%m-%d").isoformat(timespec="milliseconds")
Output
'1980-01-22T00:00:00.000'

Python Regular expression to replace date with timezone

I have CSV file which contains various timezone dates, but before feeding those data to tests, I want to replace all the dates with unify value.
date column contains values like below,
2019-01-01 00:00:00+05:30
2018-12-31 18:30:00+00
2018-02-02 00:00:00-04:00
I want replace them like
2019-01-01 00:00:00+00
2018-12-31 00:00:00+00
2018-02-02 00:00:00+00
How do I write Regex to cover all possible timezones?
I wrote:
([0-9]){4}(-:?)([0-9]){2}(-:?)([0-9]){2} ([0-9]){2}:([0-9]){2}:([0-9]){2}(+-?)([0-9]){2}:([0-9]){2}
but it fails when it encounter 2018-12-31 18:30:00+00, How can I handle this case?

Tim Biegeleisen is very right, you should not be using regex for this, you should use a datetime API provided by Python. I have sourced my answer from an excellent post on this by jfs here
The below is for Python 3.3+ (since you have tagged your question with Python 3.0
time_string = "2019-01-01 00:00:00+05:30"
# Parses a datetime instance from a string
dt = datetime.datetime.strptime(time_string,'%Y-%m-%d %H:%M:%S%z')
# Changes the timezone to UTC by setting tzinfo
timestamp = dt.replace(tzinfo=datetime.timezone.utc).timestamp()
# Converts back to a datetime object
dt = datetime.datetime.fromtimestamp(timestamp)
# Formats and prints it out.
print(dt.strftime('%Y-%m-%d %H:%M:%S %Z'))
For Python versions < 3.3, for an aware datetime
time_string = "2019-01-01 00:00:00+05:30"
# Parses a datetime instance from a string
dt = datetime.datetime.strptime(time_string,'%Y-%m-%d %H:%M:%S%z')
# Changes the timezone to UTC by setting tzinfo
timestamp = (dt - datetime(1970,1,1, tzinfo=timezone.utc)) / timedelta(seconds=1)
# Converts back to a datetime object
dt = datetime.datetime.fromtimestamp(timestamp)
# Formats and prints it out.
print(dt.strftime('%Y-%m-%d %H:%M:%S %Z'))
Terminology
An aware object is used to represent a specific moment in time that is
not open to interpretation
For our case, timezone information is known.

The best way to solve this problem is using **python datetime **(strp and strf)
If you want to solve it using regex then as per python doc https://docs.python.org/2/library/re.html
you can do something like this
def dashrepl(matchobj):
return "{0} 00:00:00+00".format(matchobj.group(1))
import re
k="(\d{4}(-\d{2}){2})\s(\d{2}:?){3}.[\d:]+"
ab = re.sub(k, dashrepl, "2019-01-01 00:00:00+05:30")

You don't need to use regex for this as it seems to be straight forward. You can use the below snippet
ts = ["2019-01-01 00:00:00+05:30", "2018-12-31 18:30:00+00", "2018-02-02 00:00:00-04:00"]
l = [x.split()[0] + " 00:00:00+00" for x in ts]
OR
l = [x[:11] + "00:00:00+00" for x in ts]

Python Change how string is printed

I have a string output from another program that shows the date as
16/05/03 # (YY/MM/DD)
and I wish to change it to
03/05/16 #(DD/MM/YY)
and here is how the date is supplied
(date = info[4].replace('"', '')
i have tried
dates = str(date)[::-1]
but that gave me an output of
40/50/61
not quite what I wanted
any ideas using a minimal code as possible?

>>> '/'.join('16/05/03'.split('/')[::-1])
'03/05/16'
or
>>> '/'.join(reversed('16/05/03'.split('/')))
'03/05/16'
or using datetime library:
>> from datetime import datetime
>>> datetime.strftime(datetime.strptime('16/05/03', '%y/%m/%d'), '%d/%m/%y')
'03/05/16'

Using datetime give you alot more control with changing the format to suite what you want.
import datetime
d = datetime.strptime('16/05/03', '%y/%m/%d')
print d.strftime('%d/%m/%y')

calculating the next day from a "YYYYMMDD" formatted string

How can I calculate the next day from a string like 20110531 in the same YYYYMMDD format? In this particular case, I like to have 20110601 as the result. Calculating "tomorrow" or next day in static way is not that tough, like this:
>>> from datetime import date, timedelta
>>> (date.today() + timedelta(1)).strftime('%Y%m%d')
'20110512'
>>>
>>> (date(2011,05,31) + timedelta(1)).strftime('%Y%m%d')
'20110601'
But how can I use a string like dt = "20110531" to get the same result as above?

Here is an example of how to do it:
import time
from datetime import date, timedelta
t=time.strptime('20110531','%Y%m%d')
newdate=date(t.tm_year,t.tm_mon,t.tm_mday)+timedelta(1)
print newdate.strftime('%Y%m%d')

>>> from datetime import datetime
>>> print datetime.strptime('20110531', '%Y%m%d')
2011-05-31 00:00:00
And then do math on that date object as you show in your question.
The datetime library docs.

You are most of the way there! along with the strftime function which converts a date to a formatted string, there is also a strptime function which converts back the other way.
To solve your problem you can just replace date.today() with strptime(yourDateString, '%Y%m%d').
ED: and of course you will also have to add strptime to the end of your from datetime import line.