Subtract datetime now from a dataframe column without Saturday and Sunday

Currently my script is subtracting my current time with the times that i have in a Dataframe column called "Creation", generating a new column with the days of the difference. I get the difference days with this code:

df['Creation']= pandas.to_datetime(df["Creation"],dayfirst="True")

#Generates new column with the days.
df['Difference'] = df.to_datetime('now') - df['Creation']

What i want to now is for it to give me the days like hes giving me but dont count the Saturdays and Sundays. How can i do that ?

Solution

you can make use of numpy's busday_count, Ex:

import pandas as pd
import numpy as np

# some dummy data
df = pd.DataFrame({'Creation': ['2021-03-29', '2021-03-30']})

# make sure we have datetime
df['Creation'] = pd.to_datetime(df['Creation'])

# set now to a fixed date
now = pd.Timestamp('2021-04-05')

# difference in business days, excluding weekends
# need to cast to datetime64[D] dtype so that np.busday_count works
df['busday_diff'] = np.busday_count(df['Creation'].values.astype('datetime64[D]'),
                                    np.repeat(now, df['Creation'].size).astype('datetime64[D]'))

df['busday_diff'] # since I didn't define holidays, potential Easter holiday is excluded:
0    5
1    4
Name: busday_diff, dtype: int64

If you need the output to be of dtype timedelta, you can easily cast to that via

df['busday_diff'] = pd.to_timedelta(df['busday_diff'], unit='d')

df['busday_diff']
0   5 days
1   4 days
Name: busday_diff, dtype: timedelta64[ns]

Note: np.busday_count also allows you to set a custom weekmask (exclude days other than Saturday and Sunday) or a list of holidays. See the docs I linked on top.