Python: Convert Timedelta to Int in a Dataframe

The Series class has a pandas.Series.dt accessor object with several useful datetime attributes, including dt.days. Access this attribute via:

timedelta_series.dt.days

You can also get the seconds and microseconds attributes in the same way.

You could do this, where td is your series of timedeltas. The division converts the nanosecond deltas into day deltas, and the conversion to int drops to whole days.

import numpy as np

(td / np.timedelta64(1, 'D')).astype(int)

Timedelta objects have read-only instance attributes .days, .seconds, and .microseconds.

If the question isn't just "how to access an integer form of the timedelta?" but "how to convert the timedelta column in the dataframe to an int?" the answer might be a little different. In addition to the .dt.days accessor you need either df.astype or pd.to_numeric

Either of these options should help:

df['tdColumn'] = pd.to_numeric(df['tdColumn'].dt.days, downcast='integer')

or

df['tdColumn'] = df['tdColumn'].dt.days.astype('int16')

I think it's much easier way to it with this (where dif is the difference between dates):

dif_In_Days = dif.days

The simplest way to do this is by

df["DateColumn"] = (df["DateColumn"]).dt.days