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How to change to datetime format date with dashes?

I have a date in format 2022-12-16T16-48-47" and I would like to change it to datetime using function pd.to_datetime.
My first idea was to create split the string to have it in more readable way:
string = "2022-12-16T16-48-47"
date, hour = string.split("T")
string = date + " " + hour
string
And now to use:
import pandas as pd
pd.to_datetime(string, format = "%Y-%M-%D %h-%m-%S")
But I have error:
ValueError: 'D' is a bad directive in format '%Y-%M-%D %h-%m-%S'
Do you know how it should be done properly?

Use Y, m, d and H, M, S instead:
>>> pd.to_datetime(string, format = "%Y-%m-%d %H-%M-%S")
Timestamp('2022-12-16 16:48:47')
>>>
You should check out the strftime format codes documentation for better understanding.

Python strptime errors on parsing strftime output

I am trying to store a date in a human readable format. For that I save and read back a string containing a date. I am using date.min to denote a date before any other.
from datetime import datetime, date
d = date.min
s = datetime.strftime(d, "%Y-%m-%d")
print(s)
# 1-01-01
d2 = datetime.strptime(s, "%Y-%m-%d")
# ValueError: time data '1-01-01' does not match format '%Y-%m-%d'
However, when I try to parse the date using strptime that was output by strftime, I only get an error. It seems that strptime is expecting leading zeros like 0001, which strftime is not outputting.
It might be possible to use None. Are there any other ways to work around what seems like a bug to me?

You need to add leading zeros:
try replacing:
s = {datetime.strftime(d, "%Y-%m-%d")}
with:
s = f'{d.year:04d}-{datetime.strftime(d, "%m-%d")}'
If you want to work with dates easily, I can really suggest the 'Arrow' library.
https://pypi.org/project/arrow/

Python 3.9 on Linux exhibits the problem as expected in the many comments on the question. One workaround that should work on all platforms is to use ISO format for the date string with date.isoformat():
>>> from datetime import date, datetime
>>> s = date.min.isoformat()
>>> s
'0001-01-01'
>>> d = datetime.strptime(s, "%Y-%m-%d")
>>> d
datetime.datetime(1, 1, 1, 0, 0)
>>> assert d.date() == date.min
You can also use date.fromisoformat() instead of strptime():
>>> date.fromisoformat(date.min.isoformat())
datetime.date(1, 1, 1)

The strftime can't add leading zeros to the year. It calls the underlying C function and its behavior on adding leading zeros to the year is platform specific. You can work around this by formatting the date object by yourself. Just do a check if d.year is less than 1000 and add how many leading zeros needed:
d = date.min
year_str = ''.join(['0' for _ in range(4 - len(str(d.year)))]) + str(d.year)
s = '{year}-{md}'.format(year=year_str, md=datetime.strftime(d, "%m-%d"))
d2 = datetime.strptime(s, "%Y-%m-%d")
# now d2 is equal to datetime.datetime(1, 1, 1, 0, 0)

How to change a datetime format in python?

How can one make 2020/09/06 15:59:04 out of 06-09-202015u59m04s.
This is my code:
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%YT%H:%M:%S')
print(date_object)
This is the error I receive:
ValueError: time data '06-09-202014u59m04s' does not match format '%d-%m-%YT%H:%M:%S'

>>> from datetime import datetime
>>> my_time = '06-09-202014u59m04s'
>>> dt_obj = datetime.strptime(my_time,'%d-%m-%Y%Hu%Mm%Ss')
Now you need to do some format changes to get the answer as the datetime object always prints itself with : so you can do any one of the following:
Either get a new format using strftime:
>>> dt_obj.strftime('%Y/%m/%d %H:%M:%S')
'2020/09/06 14:59:04'
Or you can simply use .replace() by converting datetime object to str:
>>> str(dt_obj).replace('-','/')
'2020/09/06 14:59:04'

As your error says what you give does not match format - %d-%m-%YT%H:%M:%S - means you are expecting after year: letter T hour:minutes:seconds when in example show it is houruminutesmsecondss without T, so you should do:
import datetime
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%Y%Hu%Mm%Ss')
print(date_object)
Output:
2020-09-06 14:59:04

You need to always make sure that your desired date format should match up with your required format.
from datetime import datetime
date_object = datetime.strptime("06-09-202015u59m04s", '%d-%m-%Y%Hu%Mm%Ss')
print(date_object.strftime('%Y/%m/%d %H:%M:%S'))
Output
2020/09/06 15:59:04

seconds part of string representing date time is dropped when reading from time from CSV file [duplicate]

Right now I have:
timestamp = datetime.strptime(date_string, '%Y-%m-%d %H:%M:%S.%f')
This works great unless I'm converting a string that doesn't have the microseconds. How can I specify that the microseconds are optional (and should be considered 0 if they aren't in the string)?

You could use a try/except block:
try:
timestamp = datetime.strptime(date_string, '%Y-%m-%d %H:%M:%S.%f')
except ValueError:
timestamp = datetime.strptime(date_string, '%Y-%m-%d %H:%M:%S')

What about just appending it if it doesn't exist?
if '.' not in date_string:
date_string = date_string + '.0'
timestamp = datetime.strptime(date_string, '%Y-%m-%d %H:%M:%S.%f')

I'm late to the party but I found if you don't care about the optional bits this will lop off the .%f for you.
datestring.split('.')[0]

I prefer using regex matches instead of try and except. This allows for many fallbacks of acceptable formats.
# full timestamp with milliseconds
match = re.match(r"\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}\.\d+Z", date_string)
if match:
return datetime.strptime(date_string, "%Y-%m-%dT%H:%M:%S.%fZ")
# timestamp missing milliseconds
match = re.match(r"\d{4}-\d{2}-\d{2}T\d{2}:\d{2}:\d{2}Z", date_string)
if match:
return datetime.strptime(date_string, "%Y-%m-%dT%H:%M:%SZ")
# timestamp missing milliseconds & seconds
match = re.match(r"\d{4}-\d{2}-\d{2}T\d{2}:\d{2}Z", date_string)
if match:
return datetime.strptime(date_string, "%Y-%m-%dT%H:%MZ")
# unknown timestamp format
return false
Don't forget to import "re" as well as "datetime" for this method.

datetime(*map(int, re.findall('\d+', date_string)))
can parse both '%Y-%m-%d %H:%M:%S.%f' and '%Y-%m-%d %H:%M:%S'. It is too permissive if your input is not filtered.
It is quick-and-dirty but sometimes strptime() is too slow. It can be used if you know that the input has the expected date format.

If you are using Pandas you can also filter the the Series and concatenate it. The index is automatically joined.
import pandas as pd
# Every other row has a different format
df = pd.DataFrame({"datetime_string": ["21-06-08 14:36:09", "21-06-08 14:36:09.50", "21-06-08 14:36:10", "21-06-08 14:36:10.50"]})
df["datetime"] = pd.concat([
pd.to_datetime(df["datetime_string"].iloc[1::2], format="%y-%m-%d %H:%M:%S.%f"),
pd.to_datetime(df["datetime_string"].iloc[::2], format="%y-%m-%d %H:%M:%S"),
])
datetime_string
datetime
0
21-06-08 14:36:09
2021-06-08 14:36:09
1
21-06-08 14:36:09.50
2021-06-08 14:36:09.500000
2
21-06-08 14:36:10
2021-06-08 14:36:10
3
21-06-08 14:36:10.50
2021-06-08 14:36:10.500000

using one regular expression and some list expressions
time_str = "12:34.567"
# time format is [HH:]MM:SS[.FFF]
sum([a*b for a,b in zip(map(lambda x: int(x) if x else 0, re.match(r"(?:(\d{2}):)?(\d{2}):(\d{2})(?:\.(\d{3}))?", time_str).groups()), [3600, 60, 1, 1/1000])])
# result = 754.567

For my similar problem using jq I used the following:
|split("Z")[0]|split(".")[0]|strptime("%Y-%m-%dT%H:%M:%S")|mktime
As the solution to sort my list by time properly.

'020-06-08 17:11:02+00:00' I want to extract the date from this time format in Python using datetime

This is what I have
date = datetime.datetime.strptime('2020-06-08 17:11:02+00:00', '%Y-%m-%d %H:%M:%S+%f')
d= date.strftime('%Y-%m-%d')
print(d);

This is one approach. Since you need only the date I have stripped :00 from the original date string.
Demo:
import datetime
sData = '2020-06-08 17:11:02+00:00'
date = datetime.datetime.strptime(sData[:-3], '%Y-%m-%d %H:%M:%S+%f')
d = date.strftime('%Y-%m-%d')
print(d)
Output:
2020-06-08