问题
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. There are 3 columns: date, day of week (Mon, Tue, Wed etc.) and sales. I would like to match nth weekday of Year 1 with Year 2 and compare sales that way.
- Example1: 01/06/2020 matches with 01/04/2021, both are the 1st Monday of that year.
- Example2: 11/02/2019 matches with 10/31/2020, both are the 44th Saturday of that year.
As you can see, I can't simply do a "nth weekday of the MONTH" because sometimes the matched nth weekday would fall in different months (as seen in 11/02/2019 & 10/31/2020).
I am manipulating the table in pandas. I am wondering if there's a quick way for me to create a column that helps me to calculate the "nth weekday of the year" for me, so that I could later match based on that value?
Thanks for your help.
回答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]
})
回答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"])
回答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.
来源:https://stackoverflow.com/questions/61106105/python-how-to-find-the-nth-weekday-of-the-year