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is there a way that i can convert this dataframe into a datatype [closed]

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already clean a lot of information into this, and i'm feel stuck now, if anyone can give some ideas and how to proceed i will apreacite so much.
later i need to join with this:
and timezone abbr. ¿there's a library or something that can handle datetypes like this?
my desired output is something like this: 2022-05-04 18:00:00 but i have this:
[]

As mentioned in the comment, you should make sure your dates are strings in order to concatenate them. You can do this like that:
df.column_name = df.column_name.astype(str)
After that you can use python's datetime library to format it as dates.
date = 'May-02-2022'
time = '9:00 AM'
timestamp = a + ' ' + b
timestamp = pd.to_datetime(timestamp, format='%b-%d-%Y %I:%M %p')
print(timstamp)
>>> Timestamp('2022-05-02 09:00:00')

How do I perform the below datetime operation? [closed]

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I have a date with format in 'yyyy-dd-mm'. How do I convert to 'yyyy-mm-dd' format, been bit of a struggle with strftime and strptime!
Assuming the date is '1919-01-12', I want to see it in the format '1919-12-01'.

from datetime import datetime
date = '1919-01-12'
formatted_date = datetime.strptime(date, '%Y-%d-%m').strftime('%Y-%m-%d')
datetime.strptime(date, '%Y-%d-%m') creates a datetime object from a string, .strftime('%Y-%m-%d') formats the datetime as a string accoring to the specified pattern

try this code
import datetime
print (datetime.datetime.strptime("1919-01-12", "%Y-%d-%m").strftime("%Y-%m-%d"))

Create a time object? [closed]

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I have a vast number of files that need to be deleted past a certain date.
I can't use the datestamp of the file because the files are created some time before they are used. A file called OCT21.txt needs to be deleted a few days after October 21st, but it could have been created in May.
My question is: is it possible to convert the "OCT21" string into a format that the time module can use?

This is in the datetime docs
classmethod datetime.strptime(date_string, format)
Return a datetime corresponding to date_string, parsed according to format.
This is equivalent to datetime(*(time.strptime(date_string, format)[0:6])).
ValueError is raised if the date_string and format can’t be parsed by time.strptime() or if it returns a value which isn’t a time tuple.
For a complete list of formatting directives, see strftime() and strptime() Behavior.
You can give it a format (%b%d in your case) and it'll return the datetime.

Try this:
from datetime import datetime
target = datetime.now().strftime('%b%d').upper()
if filename == target + '.txt':
# Do your thing

It is possible, have a look at time.strptime.
import time
filedate = time.strptime("OCT21", "%b%d")
Of course, in the resulting struct_time neither the year nor exact time are valid.

How to convert MS Excel numbers in general format (e.g 43422) to date in python? [closed]

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After fetching data from Excel to Python, the dates have been fetched into python as '43422' format. How do I convert this integer to date format in python now?

You may be able to prevent this from happening in the first place when you read in the data: pd.read_excel('file.xlsx'). Make sure that the data is properly formatted as a date before you read in the file.
If you want to convert this number to a date after you read it in, use the fact that it represents the number of days since January 1, 1900:
import datetime
d = '45422'
date = datetime.date(1900, 1, 1) + datetime.timedelta(int(d))
date
Output:
datetime.date(2024, 5, 12)

Extracting time data from odd Dd HH:MM:SS.mm format [closed]

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I am trying to import experimental time data which is not really a time, date, or datetime according to the python formalism. It is a time-elapsed format as Dd HH:MM:SS.m where the first time point would be 0d 00:00:00.00 and a point 26 hours later would be 1d 02:00:00.00.
Is there a way to use the datetime module to extract the appropriate time information without hard-coding a string search for the day and adding a multiple of 24 to the hour counter?

To convert your 'odd' format to datetime.timedelta object try:
from datetime import timedelta
import re
input = '1d 02:00:00.00'
def to_timedelta(input):
reg = re.search('([0-9])+d ([0-9]){2}:([0-9]){2}:([0-9]){2}\.([0-9]){2}', input)
ints = tuple(int(t) for t in reg.groups())
return timedelta(days=ints[0],
hours=ints[1],
minutes=ints[2],
seconds=ints[3],
milliseconds=ints[4])
td = to_timedelta(input)
print(td)