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.. currentmodule:: numpy | ||
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.. _arrays.datetime: | ||
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************************ | ||
Datetimes and Timedeltas | ||
************************ | ||
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.. versionadded:: 1.7.0 | ||
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Starting in NumPy 1.7, there are core array data types which natively | ||
support datetime functionality. The data type is called "datetime64", | ||
so named because "datetime" is already taken by the datetime library | ||
included in Python. | ||
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Basic Datetimes | ||
=============== | ||
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The most basic way to create datetimes is from strings in | ||
ISO 8601 date or datetime format. The unit for internal storage | ||
is automatically selected from the form of the string, and can | ||
be either a :ref:`date unit <arrays.dtypes.dateunits>` or a | ||
:ref:`time unit <arrays.dtypes.timeunits>`. The date units are years ('Y'), | ||
months ('M'), weeks ('W'), and days ('D'), while the time units are | ||
hours ('h'), minutes ('m'), seconds ('s'), milliseconds ('ms'), and | ||
more SI-prefix seconds-based units. | ||
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.. admonition:: Example | ||
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A simple ISO date: | ||
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>>> np.datetime64('2005-02-25') | ||
numpy.datetime64('2005-02-25') | ||
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Using months for the unit: | ||
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>>> np.datetime64('2005-02') | ||
numpy.datetime64('2005-02') | ||
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Specifying just the month, but forcing a 'days' unit: | ||
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>>> np.datetime64('2005-02', 'D') | ||
numpy.datetime64('2005-02-01') | ||
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Using UTC "Zulu" time: | ||
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>>> np.datetime64('2005-02-25T03:30Z') | ||
numpy.datetime64('2005-02-24T21:30-0600') | ||
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ISO 8601 specifies to use the local time zone | ||
if none is explicitly given: | ||
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>>> np.datetime64('2005-02-25T03:30') | ||
numpy.datetime64('2005-02-25T03:30-0600') | ||
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When creating an array of datetimes from a string, it is still possible | ||
to automatically select the unit from the inputs, by using the | ||
datetime type with generic units. | ||
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.. admonition:: Example | ||
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>>> np.array(['2007-07-13', '2006-01-13', '2010-08-13'], dtype='datetime64') | ||
array(['2007-07-13', '2006-01-13', '2010-08-13'], dtype='datetime64[D]') | ||
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>>> np.array(['2001-01-01T12:00', '2002-02-03T13:56:03.172'], dtype='datetime64') | ||
array(['2001-01-01T12:00:00.000-0600', '2002-02-03T13:56:03.172-0600'], dtype='datetime64[ms]') | ||
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The datetime type works with many common NumPy functions, for | ||
example :meth:`arange` can be used to generate ranges of dates. | ||
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.. admonition:: Example | ||
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All the dates for one month: | ||
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>>> np.arange('2005-02', '2005-03', dtype='datetime64[D]') | ||
array(['2005-02-01', '2005-02-02', '2005-02-03', '2005-02-04', | ||
'2005-02-05', '2005-02-06', '2005-02-07', '2005-02-08', | ||
'2005-02-09', '2005-02-10', '2005-02-11', '2005-02-12', | ||
'2005-02-13', '2005-02-14', '2005-02-15', '2005-02-16', | ||
'2005-02-17', '2005-02-18', '2005-02-19', '2005-02-20', | ||
'2005-02-21', '2005-02-22', '2005-02-23', '2005-02-24', | ||
'2005-02-25', '2005-02-26', '2005-02-27', '2005-02-28'], | ||
dtype='datetime64[D]') | ||
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The datetime object represents a single moment in time. If two | ||
datetimes have different units, they may still be representing | ||
the same moment of time, and converting from a bigger unit like | ||
months to a smaller unit like days is considered a 'safe' cast | ||
because the moment of time is still being represented exactly. | ||
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.. admonition:: Example | ||
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>>> np.datetime64('2005') == np.datetime64('2005-01-01') | ||
True | ||
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>>> np.datetime64('2010-03-14T15Z') == np.datetime64('2010-03-14T15:00:00.00Z') | ||
True | ||
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An important exception to this rule is between datetimes with | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. The 'date units' and 'time units' should probably be introduced up near the beginning and the distinction explained. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Sure There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. What about just splitting up the table of units in the section below into two tables, one for each type? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. That would make one table rather small. Maybe star the lower case types, or explain why they are lower case. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Looking at the change to the table, I think your idea of two tables would look better. |
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:ref:`date units <arrays.dtypes.dateunits>` and datetimes with | ||
:ref:`time units <arrays.dtypes.timeunits>`. This is because this kind | ||
of conversion generally requires a choice of timezone and | ||
particular time of day on the given date. | ||
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.. admonition:: Example | ||
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>>> np.datetime64('2003-12-25', 's') | ||
Traceback (most recent call last): | ||
File "<stdin>", line 1, in <module> | ||
TypeError: Cannot parse "2003-12-25" as unit 's' using casting rule 'same_kind' | ||
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>>> np.datetime64('2003-12-25') == np.datetime64('2003-12-25T00Z') | ||
False | ||
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Datetime and Timedelta Arithmetic | ||
================================= | ||
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NumPy allows the subtraction of two Datetime values, an operation which | ||
produces a number with a time unit. Because NumPy doesn't have a physical | ||
quantities system in its core, the timedelta64 data type was created | ||
to complement datetime64. | ||
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Datetimes and Timedeltas work together to provide ways for | ||
simple datetime calculations. | ||
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.. admonition:: Example | ||
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>>> np.datetime64('2009-01-01') - np.datetime64('2008-01-01') | ||
numpy.timedelta64(366,'D') | ||
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>>> np.datetime64('2009') + np.timedelta64(20, 'D') | ||
numpy.datetime64('2009-01-21') | ||
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>>> np.datetime64('2011-06-15T00:00') + np.timedelta64(12, 'h') | ||
numpy.datetime64('2011-06-15T12:00-0500') | ||
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>>> np.timedelta64(1,'W') / np.timedelta64(1,'D') | ||
7.0 | ||
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There are two Timedelta units, years and months, which are treated | ||
specially, because how much time they represent changes depending | ||
on when they are used. While a timedelta day unit is equivalent to | ||
24 hours, there is no way to convert a month unit into days, because | ||
different months have different numbers of days. | ||
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.. admonition:: Example | ||
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>>> a = np.timedelta64(1, 'Y') | ||
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>>> np.timedelta64(a, 'M') | ||
numpy.timedelta64(12,'M') | ||
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>>> np.timedelta64(a, 'D') | ||
Traceback (most recent call last): | ||
File "<stdin>", line 1, in <module> | ||
TypeError: Cannot cast NumPy timedelta64 scalar from metadata [Y] to [D] according to the rule 'same_kind' | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. line break needed? There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Maybe, I guess the 'doctest' thing isn't taken very seriously anyway... |
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Datetime Units | ||
============== | ||
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The Datetime and Timedelta data types support a large number of time | ||
units, as well as generic units which can be coerced into any of the | ||
other units based on input data. | ||
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Datetimes are always stored based on POSIX time (though having a TAI | ||
mode which allows for accounting of leap-seconds is proposed), with | ||
a epoch of 1970-01-01T00:00Z. This means the supported dates are | ||
always a symmetric interval around 1970. | ||
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Here are the date units: | ||
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.. _arrays.dtypes.dateunits: | ||
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======== ================ ======================= ========================== | ||
Date unit Time span Time span (years) | ||
------------------------- ----------------------- -------------------------- | ||
Code Meaning Relative Time Absolute Time | ||
======== ================ ======================= ========================== | ||
Y year +- 9.2e18 years [9.2e18 BC, 9.2e18 AD] | ||
M month +- 7.6e17 years [7.6e17 BC, 7.6e17 AD] | ||
W week +- 1.7e17 years [1.7e17 BC, 1.7e17 AD] | ||
D day +- 2.5e16 years [2.5e16 BC, 2.5e16 AD] | ||
======== ================ ======================= ========================== | ||
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And here are the time units: | ||
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.. _arrays.dtypes.timeunits: | ||
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======== ================ ======================= ========================== | ||
Time unit Time span Time span (years) | ||
------------------------- ----------------------- -------------------------- | ||
Code Meaning Relative Time Absolute Time | ||
======== ================ ======================= ========================== | ||
h hour +- 1.0e15 years [1.0e15 BC, 1.0e15 AD] | ||
m minute +- 1.7e13 years [1.7e13 BC, 1.7e13 AD] | ||
s second +- 2.9e12 years [ 2.9e9 BC, 2.9e9 AD] | ||
ms millisecond +- 2.9e9 years [ 2.9e6 BC, 2.9e6 AD] | ||
us microsecond +- 2.9e6 years [290301 BC, 294241 AD] | ||
ns nanosecond +- 292 years [ 1678 AD, 2262 AD] | ||
ps picosecond +- 106 days [ 1969 AD, 1970 AD] | ||
fs femtosecond +- 2.6 hours [ 1969 AD, 1970 AD] | ||
as attosecond +- 9.2 seconds [ 1969 AD, 1970 AD] | ||
======== ================ ======================= ========================== | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. So if you add 1as to "2011-6-30' you get an overflow error. This unit doesn't seem that useful except for elapsed time from some separately specified epoch. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Silent overflow, fully compatible with the silent overflow in all of NumPy's integer types. ;) There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Hmm, silent overflow seems a bit dangerous for business. Actually, I did get an error:
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. But then this
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. True, I don't think business stuff will accidentally use attoseconds, though. I've added it to my TODO list though |
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Business Day Functionality | ||
========================== | ||
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To allow the datetime to be used in contexts where accounting for weekends | ||
and holidays is important, NumPy includes a set of functions for | ||
working with business days. | ||
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The function :meth:`busday_offset` allows you to apply offsets | ||
specified in business days to datetimes with a unit of 'day'. By default, | ||
a business date is defined to be any date which falls on Monday through | ||
Friday, but this can be customized with a weekmask and a list of holidays. | ||
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.. admonition:: Example | ||
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>>> np.busday_offset('2011-06-23', 1) | ||
numpy.datetime64('2011-06-24') | ||
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>>> np.busday_offset('2011-06-23', 2) | ||
numpy.datetime64('2011-06-27') | ||
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When an input date falls on the weekend or a holiday, | ||
:meth:`busday_offset` first applies a rule to roll the | ||
date to a valid business day, then applies the offset. The | ||
default rule is 'raise', which simply raises an exception. | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. By the example, the default rule looks like 'forward' There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Which example? I see an example raising an exception. Maybe an offset of 0 instead of 2 in that example would help. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Where's the exception? I assume that the exception is only raised if the start date isn't a business day. |
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The rules most typically used are 'forward' and 'backward'. | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. You have to think of it in terms of what the function does:
I tried to explain it in the paragraph, but maybe you have ideas to make it more clear. Negative offsets work as one would expect. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Yeah, I saw that further down. That's why I deleted my comment. |
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.. admonition:: Example | ||
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>>> np.busday_offset('2011-06-25', 2) | ||
Traceback (most recent call last): | ||
File "<stdin>", line 1, in <module> | ||
ValueError: Non-business day date in busday_offset | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Here's the exception |
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>>> np.busday_offset('2011-06-25', 0, roll='forward') | ||
numpy.datetime64('2011-06-27') | ||
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>>> np.busday_offset('2011-06-25', 2, roll='forward') | ||
numpy.datetime64('2011-06-29') | ||
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>>> np.busday_offset('2011-06-25', 0, roll='backward') | ||
numpy.datetime64('2011-06-24') | ||
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>>> np.busday_offset('2011-06-25', 2, roll='backward') | ||
numpy.datetime64('2011-06-28') | ||
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In some cases, an appropriate use of the roll and the offset | ||
is necessary to get a desired answer. | ||
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.. admonition:: Example | ||
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The first business day on or after a date: | ||
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>>> np.busday_offset('2011-03-20', 0, roll='forward') | ||
numpy.datetime64('2011-03-21','D') | ||
>>> np.busday_offset('2011-03-22', 0, roll='forward') | ||
numpy.datetime64('2011-03-22','D') | ||
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The first business day strictly after a date: | ||
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>>> np.busday_offset('2011-03-20', 1, roll='backward') | ||
numpy.datetime64('2011-03-21','D') | ||
>>> np.busday_offset('2011-03-22', 1, roll='backward') | ||
numpy.datetime64('2011-03-23','D') | ||
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The function is also useful for computing some kinds of days | ||
like holidays. In Canada and the U.S., Mother's day is on | ||
the second Sunday in May, which can be computed with a special | ||
weekmask. | ||
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.. admonition:: Example | ||
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>>> np.busday_offset('2012-05', 1, roll='forward', weekmask='Sun') | ||
numpy.datetime64('2012-05-13','D') | ||
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When performance is important for manipulating many business date | ||
with one particular choice of weekmask and holidays, there is | ||
an object :class:`busdaycalendar` which stores the data necessary | ||
in an optimized form. | ||
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The other two functions for business days are :meth:`is_busday` | ||
and :meth:`busday_count`, which are more straightforward and | ||
not explained here. |
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arrays.classes | ||
maskedarray | ||
arrays.interface | ||
arrays.datetime |
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I assume that the year in this and similar example corresponds to the earliest time compatible with both sides of the comparison.
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Not really 'compatible with', more 'equal by definition'.