Finding the third Friday for an expiration date using pandas datetime-CodePudding

I have a simple definition which finds the third friday of the month. I use this function to populate the dataframe for the third fridays and that part works fine.

The trouble I'm having is finding the third friday for an expiration_date that doesn't fall on a third friday.

This is my code simplified:

import pandas as pd

def is_third_friday(d):
    return d.weekday() == 4 and 15 <= d.day <= 21

x = ['09/23/2022','09/26/2022','09/28/2022','09/30/2022','10/3/2022','10/5/2022',
'10/7/2022','10/10/2022','10/12/2022','10/14/2022','10/17/2022','10/19/2022','10/21/2022',
'10/24/2022','10/26/2022','10/28/2022','11/4/2022','11/18/2022','12/16/2022','12/30/2022',
'01/20/2023','03/17/2023','03/31/2023','06/16/2023','06/30/2023','09/15/2023','12/15/2023',
'01/19/2024','06/21/2024','12/20/2024','01/17/2025']

df = pd.DataFrame(x)
df.rename( columns={0 :'expiration_date'}, inplace=True )
df['expiration_date']= pd.to_datetime(df['expiration_date'])
expiration_date = df['expiration_date']
df["is_third_friday"] = [is_third_friday(x) for x in expiration_date]
third_fridays = df.loc[df['is_third_friday'] == True]
df["current_monthly_exp"] = third_fridays['expiration_date'].min() 
df["monthly_exp"] = third_fridays[['expiration_date']]
df.to_csv(path_or_buf = f'C:/Data/Date Dataframe.csv',index=False)

What I'm looking for is any expiration_date that is prior to the monthly expire, I want to populate the dataframe with that monthly expire. If it's past the monthly expire date I want to populate the dataframe with the following monthly expire.

I thought I'd be able to use a new dataframe with only the monthly expires as a lookup table and do a timedelta, but when you look at 4/21/2023 and 7/21/2023 these dates don't exist in that dataframe.

This is my current output:

This is the output I'm seeking:

I was thinking I could handle this problem with something like:

date_df["monthly_exp"][0][::-1].expanding().min()[::-1]

But, it wouldn't solve for the 4/21/2023 and 7/21/2023 problem. Additionally, Pandas won't let you do this in a datetime dataframe.

>>> df = pd.DataFrame([1, nan,2,nan,nan,nan,4])
>>> df
     0
0  1.0
1  NaN
2  2.0
3  NaN
4  NaN
5  NaN
6  4.0
>>> df["b"] = df[0][::-1].expanding().min()[::-1]
>>> df
     0    b
0  1.0  1.0
1  NaN  2.0
2  2.0  2.0
3  NaN  4.0
4  NaN  4.0
5  NaN  4.0
6  4.0  4.0

I've also tried something like the following in many different forms with little luck:

if df['is_third_friday'].any() == True:
    df["monthly_exp"] = third_fridays[['expiration_date']]
else:
    df["monthly_exp"] = third_fridays[['expiration_date']].shift(third_fridays)

Any suggestions to get me in the right direction would be appreciated. I've been stuck on this problem for sometime.

CodePudding user response：

You could add these additional lines of code (to replace df["monthly_exp"] = third_fridays[['expiration_date']]:

# DataFrame of fridays from minimum expiration_date to 30 days after last
fri_3s = pd.DataFrame(pd.date_range(df["expiration_date"].min(),
                                    df["expiration_date"].max() pd.tseries.offsets.Day(30),
                                    freq="W-FRI"),
                      columns=["monthly_exp"])
# only keep those that are between 15th and 21st (as your function did)
fri_3s = fri_3s[fri_3s.monthly_exp.dt.day.between(15, 21)]

# merge_asof to get next third friday
df = pd.merge_asof(df, fri_3s, left_on="expiration_date", right_on="monthly_exp", direction="forward")

This creates a second DataFrame with the 3rd Fridays, and then by merging with merge_asof returns the next of these from the expiration_date.

And to simplify your date_df["monthly_exp"][0][::-1].expanding().min()[::-1] and use it for datetime, you could instead write df["monthly_exp"].bfill() (which backward fills). As you mentioned, this will only include Fridays that exist in your DataFrame already, so creating a list of the possible Fridays might be the easiest way.