Get the first day of the week for a Pandas series
Question:
I have the following df :
import pandas as pd
from datetime import datetime, timedelta
df = pd.DataFrame([
["A", "2018-08-03"],
["B", "2018-08-20"]
])
df.columns = ["Item", "Date"]
I want to get the first day of the week for every line of my df. I tried to do this :
df['Date'] = pd.to_datetime(df['Date'], format='%Y-%m-%d')
df["Day_of_Week"] = df.Date.dt.weekday
df["First_day_of_the_week"] = df.Date - timedelta(days=df.Day_of_Week)
But I got that error message :
TypeError: unsupported type for timedelta days component: Series
How can I get the first day of the week for a Series ?
My expected result is that :
- “A”, “2018-08-03”, “2018-07-30”
- “B”, “2018-08-20”, “2018-08-20”
Answers:
Unfortunately timedelta doesn’t support a vectorized form so I would go for an apply
df["First_day_of_the_week"] = df.apply(lambda x: x['Date'] - timedelta(days=x['Day_of_Week']), axis=1)
EDIT
timedelta doesn’t support vectorized arguments but can be multiplied by a vector 🙂
df["First_day_of_the_week"] = df.Date - df.Day_of_Week * timedelta(days=1)
Leave out your ‘Day of week” calculation and do this.
df["First_day_of_the_week"] = df['Date'].apply(lambda x: (x - timedelta(days=x.dayofweek)))
print(df)
giving
Item Date First_day_of_the_week
0 A 2018-08-03 2018-07-30
1 B 2018-08-20 2018-08-20
A vectorised solution is possible with NumPy:
df['First_day'] = df['Date'] - df['Date'].dt.weekday * np.timedelta64(1, 'D')
print(df)
Item Date First_day
0 A 2018-08-03 2018-07-30
1 B 2018-08-20 2018-08-20
You can stay in Pandas and use its DateOffset objects:
>>> from pandas.tseries.offsets import Week
>>> df.Date.where(df.Date.dt.weekday == 0, df.Date - Week(weekday=0))
0 2018-07-30
1 2018-08-20
Name: Date, dtype: datetime64[ns]
The trick being that you need to not do the subtraction where the weekday is already Monday (weekday == 0). This says, “in cases where weekday is already zero, do nothing; else, return Monday of that week.”
pandas version
df = pd.DataFrame({
'Item': ['A', 'B'],
'Date': ['2018-08-03', '2018-08-20']
})
df['Date'] = pd.to_datetime(df.Date) #Use pd.Timestamp
df.Date - pd.TimedeltaIndex(df.Date.dt.dayofweek,unit='d')
Output:
0 2018-07-30
1 2018-08-20
dtype: datetime64[ns]
Docs on used functions: pd.TimedeltaIndex, pd.to_datetime
Working with date and time: Time Series / Date functionality
Here’s a solution that doesn’t require timedelta or lambda functions with subtractions:
df['Date'].dt.to_period('W').dt.to_timestamp()
I have the following df :
import pandas as pd
from datetime import datetime, timedelta
df = pd.DataFrame([
["A", "2018-08-03"],
["B", "2018-08-20"]
])
df.columns = ["Item", "Date"]
I want to get the first day of the week for every line of my df. I tried to do this :
df['Date'] = pd.to_datetime(df['Date'], format='%Y-%m-%d')
df["Day_of_Week"] = df.Date.dt.weekday
df["First_day_of_the_week"] = df.Date - timedelta(days=df.Day_of_Week)
But I got that error message :
TypeError: unsupported type for timedelta days component: Series
How can I get the first day of the week for a Series ?
My expected result is that :
- “A”, “2018-08-03”, “2018-07-30”
- “B”, “2018-08-20”, “2018-08-20”
Unfortunately timedelta doesn’t support a vectorized form so I would go for an apply
df["First_day_of_the_week"] = df.apply(lambda x: x['Date'] - timedelta(days=x['Day_of_Week']), axis=1)
EDIT
timedelta doesn’t support vectorized arguments but can be multiplied by a vector 🙂
df["First_day_of_the_week"] = df.Date - df.Day_of_Week * timedelta(days=1)
Leave out your ‘Day of week” calculation and do this.
df["First_day_of_the_week"] = df['Date'].apply(lambda x: (x - timedelta(days=x.dayofweek)))
print(df)
giving
Item Date First_day_of_the_week
0 A 2018-08-03 2018-07-30
1 B 2018-08-20 2018-08-20
A vectorised solution is possible with NumPy:
df['First_day'] = df['Date'] - df['Date'].dt.weekday * np.timedelta64(1, 'D')
print(df)
Item Date First_day
0 A 2018-08-03 2018-07-30
1 B 2018-08-20 2018-08-20
You can stay in Pandas and use its DateOffset objects:
>>> from pandas.tseries.offsets import Week
>>> df.Date.where(df.Date.dt.weekday == 0, df.Date - Week(weekday=0))
0 2018-07-30
1 2018-08-20
Name: Date, dtype: datetime64[ns]
The trick being that you need to not do the subtraction where the weekday is already Monday (weekday == 0). This says, “in cases where weekday is already zero, do nothing; else, return Monday of that week.”
pandas version
df = pd.DataFrame({
'Item': ['A', 'B'],
'Date': ['2018-08-03', '2018-08-20']
})
df['Date'] = pd.to_datetime(df.Date) #Use pd.Timestamp
df.Date - pd.TimedeltaIndex(df.Date.dt.dayofweek,unit='d')
Output:
0 2018-07-30
1 2018-08-20
dtype: datetime64[ns]
Docs on used functions: pd.TimedeltaIndex, pd.to_datetime
Working with date and time: Time Series / Date functionality
Here’s a solution that doesn’t require timedelta or lambda functions with subtractions:
df['Date'].dt.to_period('W').dt.to_timestamp()