Python: How To Find The Nth Weekday Of The Year?

I have seen a lot of similar posts on 'nth weekday of the month', but my question pertains to 'nth weekday of the year'. Background: I have a table that has daily sales data. The

Solution 1:

The pandas package has some good time/date functions.

For example

import pandas as pd
s = pd.date_range('2020-01-01', '2020-12-31', freq='D').to_series()
print(s.dt.dayofweek)

gives you the weekdays as integers.

2020-01-01    2
2020-01-02    3
2020-01-03    4
2020-01-04    5
2020-01-05    6
2020-01-06    0
2020-01-07    1
2020-01-08    2
2020-01-09    3
2020-01-10    4

(Monday=0)

Then you can do

mondays = s.dt.dayofweek.eq(0) 

If you want to find the first Monday of the year use.

print(mondays.idxmax())                                                    
Timestamp('2020-01-06 00:00:00', freq='D')

Or the 5th Monday:

n = 4                                                                           
print(s[mondays].iloc[n])                                                                
Timestamp('2020-02-03 00:00:00')

If your sales dataframe is df then to compare sales on the first 5 Mondays of two different years you could do something like this:

mondays = df['Date'].dt.dayofweek.eq(0)
mondays_in_y1 = (df['Year'] == 2019) & mondays
mondays_in_y2 = (df['Year'] == 2020) & mondays 

pd.DataFrame({
    2019: df.loc[mondays_in_y1, 'Sales'].values[:5],
    2020: df.loc[mondays_in_y2, 'Sales'].values[:5]
})

Solution 2:

IIUC you can play from

import pandas as pd
import numpy as np

df = pd.DataFrame({"date":pd.date_range(start="2020-01-01",
                                        end="2020-12-31")})
# weekday number Monday is 0
df["dow"] = df["date"].dt.weekday

# is weekday as int
df["is_weekday"] =  (df["dow"]<5).astype(int)

df["n"] = df["is_weekday"].cumsum()

# remove weekends
df["n"] = np.where(df["n"]==df["n"].shift(), np.nan, df["n"])

df[df["n"]==100]["date"]

Edit In two lines only

df["n"] = (df["date"].dt.weekday<5).astype(int).cumsum()
df["n"] = np.where(df["n"]==df["n"].shift(), np.nan, df["n"])

Solution 3:

You can try using dt.week. It returns a series, but you can simply define a new column with these values.

For example:

import pandas as pd
rng = pd.date_range('2015-02-24', periods=5, freq='D')
df = pd.DataFrame({ 'Date': rng, 'Val' : np.random.randn(len(rng))})

Output:

       Date        Val

0   2015-02-24  -0.977278
1   2015-02-25  0.950088
2   2015-02-26  -0.151357
3   2015-02-27  -0.103219
4   2015-02-28  0.410599

The you should input df['Week_Number'] = df['Date'].dt.week, so you will make a new column with the week number:

       Date        Val     Week_Number

0   2015-02-24  -0.977278   9
1   2015-02-25   0.950088   9
2   2015-02-26  -0.151357   9
3   2015-02-27  -0.103219   9
4   2015-02-28   0.410599   9

Hope it helps. It's my first contribution.