Get week start date (Monday) from a date column in Python (pandas)?
Question:
I have seen a lot of posts about how you can do it with a date string but I am trying something for a dataframe column and haven’t got any luck so far.
My current method is : Get the weekday from ‘myday’ and then offset to get monday.
df['myday'] is column of dates.
mydays = pd.DatetimeIndex(df['myday']).weekday
df['week_start'] = pd.DatetimeIndex(df['myday']) - pd.DateOffset(days=mydays)
But I get
TypeError: unsupported type for timedelta days component: numpy.ndarray
How can I get week start date from a df column?
Answers:
it fails because pd.DateOffset expects a single integer as a parameter (and you are feeding it an array). You can only use DateOffset to change a date column by the same offset.
try this :
import datetime as dt
# Change 'myday' to contains dates as datetime objects
df['myday'] = pd.to_datetime(df['myday'])
# 'daysoffset' will container the weekday, as integers
df['daysoffset'] = df['myday'].apply(lambda x: x.weekday())
# We apply, row by row (axis=1) a timedelta operation
df['week_start'] = df.apply(lambda x: x['myday'] - dt.TimeDelta(days=x['daysoffset']), axis=1)
I haven’t actually tested this code, (there was no sample data), but that should work for what you have described.
However, you might want to look at pandas.Resample, which might provide a better solution – depending on exactly what you are looking for.
Another alternative:
df['week_start'] = df['myday'].dt.to_period('W').apply(lambda r: r.start_time)
This will set ‘week_start’ to be the first Monday before the time in ‘myday’.
You can choose different week starts via anchored offsets e.g. ’W-THU’ to start the week on Thursday instead. (Thanks @Henry Ecker for that suggestion)
While both @knightofni’s and @Paul’s solutions work I tend to try to stay away from using apply in Pandas because it is usually quite slow compared to array-based methods. In order to avoid this, after casting to a datetime column (via pd.to_datetime) we can modify the weekday based method and simply cast the day of the week to be a numpy timedelta64[D] by either casting it directly:
df['week_start'] = df['myday'] - df['myday'].dt.weekday.astype('timedelta64[D]')
or by using to_timedelta as @ribitskiyb suggested:
df['week_start'] = df['myday'] - pd.to_timedelta(df['myday'].dt.weekday, unit='D').
Using test data with 60,000 datetimes I got the following times using the suggested answers using the newly released Pandas 1.0.1.
%timeit df.apply(lambda x: x['myday'] - datetime.timedelta(days=x['myday'].weekday()), axis=1)
>>> 1.33 s ± 28.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df['myday'].dt.to_period('W').apply(lambda r: r.start_time)
>>> 5.59 ms ± 138 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df['myday'] - df['myday'].dt.weekday.astype('timedelta64[D]')
>>> 3.44 ms ± 106 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df['myday'] - pd.to_timedelta(df['myday'].dt.weekday, unit='D')
>>> 3.47 ms ± 170 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
These results show that Pandas 1.0.1 has dramatically improved the speed of the to_period apply based method (vs Pandas <= 0.25) but show that converting directly to a timedelta (by either casting the type directly .astype('timedelta64[D]') or using pd.to_timedelta is still superior. Based on these results I would suggest using pd.to_timedelta going forward.
(Just adding to n8yoder‘s answer)
Using .astype('timedelta64[D]') seems not so readable to me — found an alternative using just the functionality of pandas:
df['myday'] - pd.to_timedelta(arg=df['myday'].dt.weekday, unit='D')
from datetime import datetime, timedelta
# Convert column to pandas datetime equivalent
df['myday'] = pd.to_datetime(df['myday'])
# Create function to calculate Start Week date
week_start_date = lambda date: date - timedelta(days=date.weekday())
# Apply above function on DataFrame column
df['week_start_date'] = df['myday'].apply(week_start_date)
I have seen a lot of posts about how you can do it with a date string but I am trying something for a dataframe column and haven’t got any luck so far.
My current method is : Get the weekday from ‘myday’ and then offset to get monday.
df['myday'] is column of dates.
mydays = pd.DatetimeIndex(df['myday']).weekday
df['week_start'] = pd.DatetimeIndex(df['myday']) - pd.DateOffset(days=mydays)
But I get
TypeError: unsupported type for timedelta days component: numpy.ndarray
How can I get week start date from a df column?
it fails because pd.DateOffset expects a single integer as a parameter (and you are feeding it an array). You can only use DateOffset to change a date column by the same offset.
try this :
import datetime as dt
# Change 'myday' to contains dates as datetime objects
df['myday'] = pd.to_datetime(df['myday'])
# 'daysoffset' will container the weekday, as integers
df['daysoffset'] = df['myday'].apply(lambda x: x.weekday())
# We apply, row by row (axis=1) a timedelta operation
df['week_start'] = df.apply(lambda x: x['myday'] - dt.TimeDelta(days=x['daysoffset']), axis=1)
I haven’t actually tested this code, (there was no sample data), but that should work for what you have described.
However, you might want to look at pandas.Resample, which might provide a better solution – depending on exactly what you are looking for.
Another alternative:
df['week_start'] = df['myday'].dt.to_period('W').apply(lambda r: r.start_time)
This will set ‘week_start’ to be the first Monday before the time in ‘myday’.
You can choose different week starts via anchored offsets e.g. ’W-THU’ to start the week on Thursday instead. (Thanks @Henry Ecker for that suggestion)
While both @knightofni’s and @Paul’s solutions work I tend to try to stay away from using apply in Pandas because it is usually quite slow compared to array-based methods. In order to avoid this, after casting to a datetime column (via pd.to_datetime) we can modify the weekday based method and simply cast the day of the week to be a numpy timedelta64[D] by either casting it directly:
df['week_start'] = df['myday'] - df['myday'].dt.weekday.astype('timedelta64[D]')
or by using to_timedelta as @ribitskiyb suggested:
df['week_start'] = df['myday'] - pd.to_timedelta(df['myday'].dt.weekday, unit='D').
Using test data with 60,000 datetimes I got the following times using the suggested answers using the newly released Pandas 1.0.1.
%timeit df.apply(lambda x: x['myday'] - datetime.timedelta(days=x['myday'].weekday()), axis=1)
>>> 1.33 s ± 28.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df['myday'].dt.to_period('W').apply(lambda r: r.start_time)
>>> 5.59 ms ± 138 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df['myday'] - df['myday'].dt.weekday.astype('timedelta64[D]')
>>> 3.44 ms ± 106 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df['myday'] - pd.to_timedelta(df['myday'].dt.weekday, unit='D')
>>> 3.47 ms ± 170 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
These results show that Pandas 1.0.1 has dramatically improved the speed of the to_period apply based method (vs Pandas <= 0.25) but show that converting directly to a timedelta (by either casting the type directly .astype('timedelta64[D]') or using pd.to_timedelta is still superior. Based on these results I would suggest using pd.to_timedelta going forward.
(Just adding to n8yoder‘s answer)
Using .astype('timedelta64[D]') seems not so readable to me — found an alternative using just the functionality of pandas:
df['myday'] - pd.to_timedelta(arg=df['myday'].dt.weekday, unit='D')
from datetime import datetime, timedelta
# Convert column to pandas datetime equivalent
df['myday'] = pd.to_datetime(df['myday'])
# Create function to calculate Start Week date
week_start_date = lambda date: date - timedelta(days=date.weekday())
# Apply above function on DataFrame column
df['week_start_date'] = df['myday'].apply(week_start_date)