Pandas – Alter week number based on the day of the week derived from the date column

You can use pandas Timedelta objects.

Using just your Date column, you can convert it to a pandas datetime object (and in fact use it to create your WeekDay column).

>>> import pandas as pd
>>> df = pd.DataFrame( 
    data = {"Date":["1/1/23", "2/1/23", "4/1/23", "5/1/23", "7/1/23" ]}
)
>>> df.Date = pd.to_datetime( df.Date, dayfirst=True )
>>> df

    Date
0   2023-01-02
1   2023-01-03
2   2023-01-04
3   2023-01-05

Create the WeekDay column:

>>> dayOfWeekMap = { 0: "Monday", 1: "Tuesday", 2: "Wednesday", 3: "Thursday", 4: "Friday",
                5: "Saturday", 6: "Sunday" }
>>> df["WeekDay"] = df.Date.dt.dayofweek.map( dayOfWeekMap )
>>> df

    Date    WeekDay
0   2023-01-01  Sunday
1   2023-01-02  Monday
2   2023-01-04  Wednesday
3   2023-01-05  Thursday
4   2023-01-07  Saturday

Finally, get the week number with your custom cutoff. First, define the day you want to start. For your test data, I would start at Wednesday 28/12/2023 so that 3/1/2023 is has Week_Num = 1, but 4/1/2023 has Week_Num = 2.

>>> start_date = pd.to_datetime( "28/12/2022", dayfirst=True )
>>> df["Week_Num"] = ( ( df.Date - start_date ).dt.days // 7 ).astype( int ) + 1
>>> df

    Date        WeekDay Week_Num
0   2023-01-01  Sunday      1
1   2023-01-02  Monday      1
2   2023-01-04  Wednesday   2
3   2023-01-05  Thursday    2
4   2023-01-07  Saturday    2

What’s happening here: We take the difference in the number of days from the date in the observation from the start date, floor division by 7 (so 6 days after start date is 0, 7 days from start date is 1, 8 days from start date is 1), and then add 1 so that our counter starts at 1.

There is no simple, non-iterative solution using the current data frame. If the WeekDay column was expressed as a number rather than text, a simple df.loc[] statement would give you the desired result.

df.loc[df["WeekDay"] > 4, "Week_Num"] += 1

This is saying to locate the rows where the Weekday > 4 and increment the Week_Num value by 1.

Using the data frame you have posted, a slower, iterative solution (not recommended) can be used along with the weekday() function from the datetime library.

#loop through each data frame row
for i, row in df.iterrows():
    #if the date is past wednesday, increment week_num
    if(row["dates"].weekday() > 4):
        df.at[i, "week_num"] += 1                                                    

The code iterates through each row item and increments the Week_Num based on an if statement. Using this method also means the WeekDay column is obsolete since datetime allows you to get the weekday number from the date.

"Shortly" (one-liner), you can use cumsum():

df['Week_Num'] = df['WeekDay'].eq('Wednesday').cumsum().add(1).shift(1).fillna(1).astype(int)

Example input:

df = pd.DataFrame.from_dict({
    'Date': ['01/01/23', '02/01/23', '04/01/23', '05/01/23', '07/01/23', '07/01/23', '07/01/23', '07/01/23', '07/01/23', '07/01/23'],
    'WeekDay': ['Sunday', 'Monday', 'Wednesday', 'Thursday', 'Saturday', 'Wednesday', 'Thursday', 'Saturday', 'Wednesday', 'Thursday']
})

Outputs:

       Date    WeekDay  Week_Num
0  01/01/23     Sunday         1
1  02/01/23     Monday         1
2  04/01/23  Wednesday         1
3  05/01/23   Thursday         2
4  07/01/23   Saturday         2
5  07/01/23  Wednesday         2
6  07/01/23   Thursday         3
7  07/01/23   Saturday         3
8  07/01/23  Wednesday         3
9  07/01/23   Thursday         4

Explanations:

1. We introduce a new column Week_Num to df
2. We use cumsum() which accumulates a value when we meet a specific requirement notated by .eq
3. The .eq part is if we meet a WeekDay ‘Wednesday’
4. We add one to each row because cumsum starts from 0 and we want to start from 1.
5. We shift downwards each row in the dataframe by 1 to change only the rows not including ‘Wednesday’
6. Because we shifted each row downwards then the first row will have NaN – so we fill it with the week_num of 1
7. We convert the row to have integer values with the astype(int)