Python: Convert timedelta to int in a dataframe

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
    print("This prints once a minute.")
    time.sleep(60) # Delay for 1 minute (60 seconds).

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

• os.remove() removes a file.

• os.rmdir() removes an empty directory.

• shutil.rmtree() deletes a directory and all its contents.

Path objects from the Python 3.4+ pathlib module also expose these instance methods:

• pathlib.Path.unlink() removes a file or symbolic link.

• pathlib.Path.rmdir() removes an empty directory.

I found this solution to be good. (This uses the python-dateutil extension)

from datetime import date
from dateutil.relativedelta import relativedelta

six_months = date.today() + relativedelta(months=+6)

The advantage of this approach is that it takes care of issues with 28, 30, 31 days etc. This becomes very useful in handling business rules and scenarios (say invoice generation etc.)

$ date(2010,12,31)+relativedelta(months=+1)
  datetime.date(2011, 1, 31)

$ date(2010,12,31)+relativedelta(months=+2)
  datetime.date(2011, 2, 28)

I was preparing for my coding interview, thanks for clarifying this - Python: Convert timedelta to int in a dataframe in Python is not the simplest one. Will get back tomorrow with feedback

Python functions is always a bit confusing 😭 Python: Convert timedelta to int in a dataframe is not the only problem I encountered. Checked yesterday, it works!

Thanks for explaining! I was stuck with Python: Convert timedelta to int in a dataframe for some hours, finally got it done 🤗. I just hope that will not emerge anymore