Datetimes and Timedeltas#
New in version 1.7.0.
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 Python standard library.
Datetime64 Conventions and Assumptions#
Similar to the Python date
class, dates are expressed in the current
Gregorian Calendar, indefinitely extended both in the future and in the past.
[1] Contrary to Python date
, which supports only years in the 1 AD — 9999
AD range, datetime64
allows also for dates BC; years BC follow the Astronomical
year numbering
convention, i.e. year 2 BC is numbered −1, year 1 BC is numbered 0, year 1 AD is
numbered 1.
Time instants, say 16:23:32.234, are represented counting hours, minutes, seconds and fractions from midnight: i.e. 00:00:00.000 is midnight, 12:00:00.000 is noon, etc. Each calendar day has exactly 86400 seconds. This is a “naive” time, with no explicit notion of timezones or specific time scales (UT1, UTC, TAI, etc.). [2]
Basic Datetimes#
The most basic way to create datetimes is from strings in ISO 8601 date
or datetime format. It is also possible to create datetimes from an integer by
offset relative to the Unix epoch (00:00:00 UTC on 1 January 1970).
The unit for internal storage is automatically selected from the
form of the string, and can be either a date unit or a
time unit. 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
some additional SI-prefix seconds-based units. The datetime64
data type
also accepts the string “NAT”, in any combination of lowercase/uppercase
letters, for a “Not A Time” value.
Example
A simple ISO date:
>>> np.datetime64('2005-02-25')
numpy.datetime64('2005-02-25')
From an integer and a date unit, 1 year since the UNIX epoch:
>>> np.datetime64(1, 'Y')
numpy.datetime64('1971')
Using months for the unit:
>>> np.datetime64('2005-02')
numpy.datetime64('2005-02')
Specifying just the month, but forcing a ‘days’ unit:
>>> np.datetime64('2005-02', 'D')
numpy.datetime64('2005-02-01')
From a date and time:
>>> np.datetime64('2005-02-25T03:30')
numpy.datetime64('2005-02-25T03:30')
NAT (not a time):
>>> np.datetime64('nat')
numpy.datetime64('NaT')
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.
Example
>>> 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]')
>>> np.array(['2001-01-01T12:00', '2002-02-03T13:56:03.172'], dtype='datetime64')
array(['2001-01-01T12:00:00.000', '2002-02-03T13:56:03.172'],
dtype='datetime64[ms]')
An array of datetimes can be constructed from integers representing POSIX timestamps with the given unit.
Example
>>> np.array([0, 1577836800], dtype='datetime64[s]')
array(['1970-01-01T00:00:00', '2020-01-01T00:00:00'],
dtype='datetime64[s]')
>>> np.array([0, 1577836800000]).astype('datetime64[ms]')
array(['1970-01-01T00:00:00.000', '2020-01-01T00:00:00.000'],
dtype='datetime64[ms]')
The datetime type works with many common NumPy functions, for
example arange
can be used to generate ranges of dates.
Example
All the dates for one month:
>>> 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]')
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.
Example
>>> np.datetime64('2005') == np.datetime64('2005-01-01')
True
>>> np.datetime64('2010-03-14T15') == np.datetime64('2010-03-14T15:00:00.00')
True
Deprecated since version 1.11.0: NumPy does not store timezone information. For backwards compatibility, datetime64 still parses timezone offsets, which it handles by converting to UTC±00:00 (Zulu time). This behaviour is deprecated and will raise an error in the future.
Datetime and Timedelta Arithmetic#
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
. The arguments for timedelta64
are a number,
to represent the number of units, and a date/time unit, such as
(D)ay, (M)onth, (Y)ear, (h)ours, (m)inutes, or (s)econds. The timedelta64
data type also accepts the string “NAT” in place of the number for a “Not A Time” value.
Example
>>> np.timedelta64(1, 'D')
numpy.timedelta64(1,'D')
>>> np.timedelta64(4, 'h')
numpy.timedelta64(4,'h')
>>> np.timedelta64('nAt')
numpy.timedelta64('NaT')
Datetimes and Timedeltas work together to provide ways for simple datetime calculations.
Example
>>> np.datetime64('2009-01-01') - np.datetime64('2008-01-01')
numpy.timedelta64(366,'D')
>>> np.datetime64('2009') + np.timedelta64(20, 'D')
numpy.datetime64('2009-01-21')
>>> np.datetime64('2011-06-15T00:00') + np.timedelta64(12, 'h')
numpy.datetime64('2011-06-15T12:00')
>>> np.timedelta64(1,'W') / np.timedelta64(1,'D')
7.0
>>> np.timedelta64(1,'W') % np.timedelta64(10,'D')
numpy.timedelta64(7,'D')
>>> np.datetime64('nat') - np.datetime64('2009-01-01')
numpy.timedelta64('NaT','D')
>>> np.datetime64('2009-01-01') + np.timedelta64('nat')
numpy.datetime64('NaT')
There are two Timedelta units (‘Y’, years and ‘M’, 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.
Example
>>> a = np.timedelta64(1, 'Y')
>>> np.timedelta64(a, 'M')
numpy.timedelta64(12,'M')
>>> 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'
Datetime Units#
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.
Datetimes are always stored with an epoch of 1970-01-01T00:00. This means the supported dates are always a symmetric interval around the epoch, called “time span” in the table below.
The length of the span is the range of a 64-bit integer times the length of the date or unit. For example, the time span for ‘W’ (week) is exactly 7 times longer than the time span for ‘D’ (day), and the time span for ‘D’ (day) is exactly 24 times longer than the time span for ‘h’ (hour).
Here are the date units:
Code |
Meaning |
Time span (relative) |
Time span (absolute) |
---|---|---|---|
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] |
And here are the time units:
Code |
Meaning |
Time span (relative) |
Time span (absolute) |
---|---|---|---|
h |
hour |
+/- 1.0e15 years |
[1.0e15 BC, 1.0e15 AD] |
m |
minute |
+/- 1.7e13 years |
[1.7e13 BC, 1.7e13 AD] |
s |
second |
+/- 2.9e11 years |
[2.9e11 BC, 2.9e11 AD] |
ms |
millisecond |
+/- 2.9e8 years |
[ 2.9e8 BC, 2.9e8 AD] |
us / μs |
microsecond |
+/- 2.9e5 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] |
Business Day Functionality#
To allow the datetime to be used in contexts where only certain days of the week are valid, NumPy includes a set of “busday” (business day) functions.
The default for busday functions is that the only valid days are Monday through Friday (the usual business days). The implementation is based on a “weekmask” containing 7 Boolean flags to indicate valid days; custom weekmasks are possible that specify other sets of valid days.
The “busday” functions can additionally check a list of “holiday” dates, specific dates that are not valid days.
The function busday_offset
allows you to apply offsets
specified in business days to datetimes with a unit of ‘D’ (day).
Example
>>> np.busday_offset('2011-06-23', 1)
numpy.datetime64('2011-06-24')
>>> np.busday_offset('2011-06-23', 2)
numpy.datetime64('2011-06-27')
When an input date falls on the weekend or a holiday,
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.
The rules most typically used are ‘forward’ and ‘backward’.
Example
>>> 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
>>> np.busday_offset('2011-06-25', 0, roll='forward')
numpy.datetime64('2011-06-27')
>>> np.busday_offset('2011-06-25', 2, roll='forward')
numpy.datetime64('2011-06-29')
>>> np.busday_offset('2011-06-25', 0, roll='backward')
numpy.datetime64('2011-06-24')
>>> np.busday_offset('2011-06-25', 2, roll='backward')
numpy.datetime64('2011-06-28')
In some cases, an appropriate use of the roll and the offset is necessary to get a desired answer.
Example
The first business day on or after a date:
>>> np.busday_offset('2011-03-20', 0, roll='forward')
numpy.datetime64('2011-03-21')
>>> np.busday_offset('2011-03-22', 0, roll='forward')
numpy.datetime64('2011-03-22')
The first business day strictly after a date:
>>> np.busday_offset('2011-03-20', 1, roll='backward')
numpy.datetime64('2011-03-21')
>>> np.busday_offset('2011-03-22', 1, roll='backward')
numpy.datetime64('2011-03-23')
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 custom weekmask.
Example
>>> np.busday_offset('2012-05', 1, roll='forward', weekmask='Sun')
numpy.datetime64('2012-05-13')
When performance is important for manipulating many business dates
with one particular choice of weekmask and holidays, there is
an object busdaycalendar
which stores the data necessary
in an optimized form.
np.is_busday():#
To test a datetime64
value to see if it is a valid day, use is_busday
.
Example
>>> np.is_busday(np.datetime64('2011-07-15')) # a Friday
True
>>> np.is_busday(np.datetime64('2011-07-16')) # a Saturday
False
>>> np.is_busday(np.datetime64('2011-07-16'), weekmask="Sat Sun")
True
>>> a = np.arange(np.datetime64('2011-07-11'), np.datetime64('2011-07-18'))
>>> np.is_busday(a)
array([ True, True, True, True, True, False, False])
np.busday_count():#
To find how many valid days there are in a specified range of datetime64
dates, use busday_count
:
Example
>>> np.busday_count(np.datetime64('2011-07-11'), np.datetime64('2011-07-18'))
5
>>> np.busday_count(np.datetime64('2011-07-18'), np.datetime64('2011-07-11'))
-5
If you have an array of datetime64 day values, and you want a count of how many of them are valid dates, you can do this:
Example
>>> a = np.arange(np.datetime64('2011-07-11'), np.datetime64('2011-07-18'))
>>> np.count_nonzero(np.is_busday(a))
5
Custom Weekmasks#
Here are several examples of custom weekmask values. These examples specify the “busday” default of Monday through Friday being valid days.
Some examples:
# Positional sequences; positions are Monday through Sunday.
# Length of the sequence must be exactly 7.
weekmask = [1, 1, 1, 1, 1, 0, 0]
# list or other sequence; 0 == invalid day, 1 == valid day
weekmask = "1111100"
# string '0' == invalid day, '1' == valid day
# string abbreviations from this list: Mon Tue Wed Thu Fri Sat Sun
weekmask = "Mon Tue Wed Thu Fri"
# any amount of whitespace is allowed; abbreviations are case-sensitive.
weekmask = "MonTue Wed Thu\tFri"
Datetime64 shortcomings#
The assumption that all days are exactly 86400 seconds long makes datetime64
largely compatible with Python datetime
and “POSIX time” semantics; therefore
they all share the same well known shortcomings with respect to the UTC
timescale and historical time determination. A brief non exhaustive summary is
given below.
It is impossible to parse valid UTC timestamps occurring during a positive leap second.
Example
“2016-12-31 23:59:60 UTC” was a leap second, therefore “2016-12-31 23:59:60.450 UTC” is a valid timestamp which is not parseable by
datetime64
:>>> np.datetime64("2016-12-31 23:59:60.450") Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: Seconds out of range in datetime string "2016-12-31 23:59:60.450"
Timedelta64 computations between two UTC dates can be wrong by an integer number of SI seconds.
Example
Compute the number of SI seconds between “2021-01-01 12:56:23.423 UTC” and “2001-01-01 00:00:00.000 UTC”:
>>> ( ... np.datetime64("2021-01-01 12:56:23.423") ... - np.datetime64("2001-01-01") ... ) / np.timedelta64(1, "s") 631198583.423
however correct answer is 631198588.423 SI seconds because there were 5 leap seconds between 2001 and 2021.
Timedelta64 computations for dates in the past do not return SI seconds, as one would expect.
Example
Compute the number of seconds between “000-01-01 UT” and “1600-01-01 UT”, where UT is universal time:
>>> a = np.datetime64("0000-01-01", "us") >>> b = np.datetime64("1600-01-01", "us") >>> b - a numpy.timedelta64(50491123200000000,'us')
The computed results, 50491123200 seconds, is obtained as the elapsed number of days (584388) times 86400 seconds; this is the number of seconds of a clock in sync with earth rotation. The exact value in SI seconds can only be estimated, e.g using data published in Measurement of the Earth’s rotation: 720 BC to AD 2015, 2016, Royal Society’s Proceedings A 472, by Stephenson et.al.. A sensible estimate is 50491112870 ± 90 seconds, with a difference of 10330 seconds.