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I need to add business days to the dates listed in an existing column in a dataframe. The data in the column is in the datetime64[ns] type. Using the suggestions from Business days in Python, I tried the code listed below.

import datetime from pandas.tseries.offsets import BDay df['Math Admin Date'] + BDay(1)

I got the following error message:

TypeError: cannot use a non-absolute DateOffset in datetime/timedelta operations [<BusinessDay>]

How can I add business days to my datetime column?

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  • We need to see where df['Math Admin Date'] is coming from. Commented Jul 23, 2015 at 13:07

2 Answers 2

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Unfortunately offsets don't support operations using array like objects so you have to apply the offset for each element:

In [208]: import datetime as dt from pandas.tseries.offsets import BDay ​ df = pd.DataFrame({'Math Admin Date':pd.date_range(start=dt.datetime(2015,6,1), end = dt.datetime(2015,6,30))}) df['Math Admin Date'].apply(lambda x: x + BDay(1)) Out[208]: 0 2015-06-02 1 2015-06-03 2 2015-06-04 3 2015-06-05 4 2015-06-08 5 2015-06-08 6 2015-06-08 7 2015-06-09 8 2015-06-10 9 2015-06-11 10 2015-06-12 11 2015-06-15 12 2015-06-15 13 2015-06-15 14 2015-06-16 15 2015-06-17 16 2015-06-18 17 2015-06-19 18 2015-06-22 19 2015-06-22 20 2015-06-22 21 2015-06-23 22 2015-06-24 23 2015-06-25 24 2015-06-26 25 2015-06-29 26 2015-06-29 27 2015-06-29 28 2015-06-30 29 2015-07-01 Name: Math Admin Date, dtype: datetime64[ns]

Using a Timedelta would work but this doesn't support business days at the moment:

df['Math Admin Date'] + pd.Timedelta(1,'D')
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2 Comments

Thank you for the suggestion! This works, but I'm planning to add different numbers of business days to different columns, and I'm wondering if it's possible to make this a function with multiple arguments rather than continuously use an anonymous function. For example def add_busdays(df.column, number of business days to add).
Yes it would work just change to def func(x, num_days=1): x + BDay(num_days) and then call it on your column: df['column'].apply(lambda x: func(x, some_num))
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For versions:

In [1140]: pd.__version__ Out[1140]: '1.1.0'

No need of using apply which is loops under the hood. You can simply do:

In [1138]: df['Math Admin Date'] = df['Math Admin Date'] + BDay(1) In [1139]: df Out[1139]: Math Admin Date 0 2015-06-02 1 2015-06-03 2 2015-06-04 3 2015-06-05 4 2015-06-08 5 2015-06-08 6 2015-06-08 7 2015-06-09 8 2015-06-10 9 2015-06-11 10 2015-06-12 11 2015-06-15 12 2015-06-15 13 2015-06-15 14 2015-06-16 15 2015-06-17 16 2015-06-18 17 2015-06-19 18 2015-06-22 19 2015-06-22 20 2015-06-22 21 2015-06-23 22 2015-06-24 23 2015-06-25 24 2015-06-26 25 2015-06-29 26 2015-06-29 27 2015-06-29 28 2015-06-30 29 2015-07-01
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