6.3. Date Shifts

6.3.1. SetUp

>>> import pandas as pd

6.3.2. Between Time

• https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.between_time.html

6.3.3. Timedelta

• Represents a duration, the difference between two dates or times

• Difference expressed in: days, hours, minutes, seconds

• Similar to datetime.timedelta from the standard library

• Can be both positive and negative

• https://pandas.pydata.org/pandas-docs/stable/reference/arrays.html#timedelta-data

>>> pd.Timedelta('1 day')
Timedelta('1 days 00:00:00')

>>> pd.Timedelta(days=1)
Timedelta('1 days 00:00:00')

>>> feb2000 = pd.Timestamp('2000-02-28')
>>> mar2000 = pd.Timestamp('2000-03-01')
>>>
>>> feb2000 + pd.Timedelta(days=1)
Timestamp('2000-02-29 00:00:00')
>>>
>>> feb2000 + pd.Timedelta(days=2)
Timestamp('2000-03-01 00:00:00')
>>>
>>> mar2000 - pd.Timedelta(days=1)
Timestamp('2000-02-29 00:00:00')

>>> feb2001 = pd.Timestamp('2001-02-28')
>>> mar2001 = pd.Timestamp('2001-03-01')
>>>
>>> feb2001 + pd.Timedelta(days=1)
Timestamp('2001-03-01 00:00:00')
>>>
>>> mar2001 - pd.Timedelta(days=1)
Timestamp('2001-02-28 00:00:00')

Leap second has not been added:

>>> leap = pd.Timestamp('2016-12-31 23:59:59')
>>>
>>> leap + pd.Timedelta(seconds=1)
Timestamp('2017-01-01 00:00:00')

6.3.4. DateOffset

• A relative time duration that respects calendar arithmetic

• If a date is Sat then adding a Bday will return the next Monday (next Business day) instead of a Saturday

• Test if a date is in the DateOffset().onOffset(date)

>>> first_step = pd.Timestamp('1969-07-21 02:56:15')
>>>
>>> first_step + pd.DateOffset(months=3)
Timestamp('1969-10-21 02:56:15')

>>> epoch = pd.Timestamp('1970-01-01 00:00:00')
>>>
>>> epoch + pd.DateOffset(month=3)
Timestamp('1970-03-01 00:00:00')

>>> mar = pd.Timestamp('1970-03-01 00:00:00')
>>>
>>> mar - pd.DateOffset(days=1)
Timestamp('1970-02-28 00:00:00')