I'm having a DataFrame with a date column. How can I map each date d
to the start day of the week containing d
?
The day of the week with Monday=0, Sunday=6. Return the day of the week. It is assumed the week starts on Monday, which is denoted by 0 and ends on Sunday which is denoted by 6. This method is available on both Series with datetime values (using the dt accessor) or DatetimeIndex.
round() function is used to round a DataFrame to a variable number of decimal places. This function provides the flexibility to round different columns by different places. Parameters : decimals : Number of decimal places to round each column to.
import pandas as pd
df['Date'] - pd.to_timedelta(df['Date'].dt.dayofweek, unit='d')
Here is an alternative approach for calculating beginning of the week series by using convenience method pd.DateOffset(weekday=0,weeks=1)
:
import pandas as pd, numpy as np
df=pd.DataFrame({'date':pd.date_range('2016-10-01','2016-10-31')})
df['BeginWeek']=np.where(df.date.dt.weekday==0, # offset on Non Mondays only
df['date'],
df['date']-np.timedelta64(1,'W')),
)
Thanks to ribitskyib np.where
was added to pick current Monday when date is already Monday. Confirmation that the above works now:
Some additional ideas provided by others:
Here is a quick list of days of the week:
df['BeginWeek'].dt.strftime("%a").unique()
array(['Mon'], dtype=object)
and the days in the original column are:
df['date'].dt.strftime("%a").unique()
array(['Sat', 'Sun', 'Mon', 'Tue', 'Wed', 'Thu', 'Fri'], dtype=object)
If you don't have dates, the accepted answer won't work (at least for me). If you have pandas._libs.tslibs.timestamps.Timestamp
objects use this instead:
df["week_start"] = pd.to_datetime(["timestamp_col"]).dt.to_period('W-SUN').dt.start_time
GH discussion on it here
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