I try to understand why
print(pd.Timestamp("2015-01-01") - pd.DateOffset(day=1))
does not result in
pd.Timestamp("2014-12-31")
I am using Pandas 0.18. I run within the CET timezone.
You can check pandas.tseries.offsets.DateOffset:
*kwds
Temporal parameter that add to or replace the offset value.
Parameters that add to the offset (like Timedelta):
years
months
weeks
days
hours
minutes
seconds
microseconds
nanoseconds
Parameters that replace the offset value:
year
month
day
weekday
hour
minute
second
microsecond
nanosecond
print(pd.Timestamp("2015-01-01") - pd.DateOffset(days=1))
2014-12-31 00:00:00
Another solution:
print(pd.Timestamp("2015-01-01") - pd.offsets.Day(1))
2014-12-31 00:00:00
Also it is possible to subtract Timedelta:
print(pd.Timestamp("2015-01-01") - pd.Timedelta(1, unit='d'))
pd.DateOffset(day=1) works (ie no error is raised) because "day" is a valid parameter, as is "days".
Look at the below one: "day" resets the actual day, "days" adds to the original day.
pd.Timestamp("2019-12-25") + pd.DateOffset(day=1)
Timestamp('2019-12-01 00:00:00')
pd.Timestamp("2019-12-25") + pd.DateOffset(days=1)
Timestamp('2019-12-26 00:00:00')
Day(d) and DateOffset(days=d) do not behave exactly the same when used on timestamps with timezone information (at least on pandas 0.18.0). It looks like DateOffset add 1 day while keeping the hour information while Day adds just 24 hours of elapsed time.
>>> # 30/10/2016 02:00+02:00 is the hour before the DST change
>>> print(pd.Timestamp("2016-10-30 02:00+02:00", tz="Europe/Brussels") + pd.offsets.Day(1))
2016-10-31 01:00:00+01:00
>>> print(pd.Timestamp("2016-10-30 02:00+02:00", tz="Europe/Brussels") + pd.DateOffset(days=1))
2016-10-31 02:00:00+01:00
Related
How can I get the number of days passed in the current quarter?
For example, if today is 1/2/2021, it will return 2.
If today is 2/5, it will return 36.
If today is 4/2, it will return 2.
Use date_trunc() to get the start of the quarter and subtract dates:
WITH cte(day) AS (
VALUES
(date '2021-01-02')
, (date '2021-02-05')
, (date '2021-04-02')
)
SELECT day
, day - date_trunc('quarter', day)::date + 1 AS days_passed_in_quarter
FROM cte;
day | days_passed_in_quarter
------------+------------------------
2021-01-02 | 2
2021-02-05 | 36
2021-04-02 | 2
+ 1 to fix off-by-one error as you clearly want to include the current day as "passed".
Always use unambiguous ISO 8601 date format (YYYY-MM-DD - 2021-02-05), which is the default in Postgres and always unambiguous, or you depend on the current datestyle setting (and may be in for surprises). Also avoids misunderstandings general communication.
Related:
PostgreSQL: between with datetime
I'm trying to manipulate a date value to go back in time exactly 1 ISO-8601 year.
The following does not work, but best describes what I want to accomplish:
date_add(date '2018-01-03', interval -1 isoyear)
I tried string conversion as an intermediate step, but that doesn't work either:
select parse_date('%G%V%u',safe_cast(safe_cast(format_date('%G%V%u',date '2018-01-03') as int64)-1000 as string))
The error provided for the last one is "Failed to parse input string "2017013"". I don't understand why, this should always resolve to a unique date value.
Is there another way in which I can subtract an ISO year from a date?
This gives the corresponding day of the previous ISO year by subtracting the appropriate number of weeks from the date. I based the calculation on the description of weeks per year from the Wikipedia page:
CREATE TEMP FUNCTION IsLongYear(d DATE) AS (
-- Year starting on Thursday
EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 5 OR
-- Leap year starting on Wednesday
(EXTRACT(DAY FROM DATE_ADD(DATE(EXTRACT(YEAR FROM d), 2, 28), INTERVAL 1 DAY)) = 29
AND EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 4)
);
CREATE TEMP FUNCTION PreviousIsoYear(d DATE) AS (
DATE_SUB(d, INTERVAL IF(IsLongYear(d), 53, 52) WEEK)
);
SELECT PreviousIsoYear('2018-01-03');
This returns 2017-01-04, which is the third day of the 2017 ISO year. 2018-01-03 is the third day of the 2018 ISO year.
I have a table with records, where each record has a date column, then a start time column and end time column.
I am trying to do a datediff to get the duration in hours from start to end date with DateDiff('s',[Start Date[,[End Date])/3600.
This works perfectly for End dates that are on same day as date column, but sometimes the end date would be the next day like 12:45 AM. The date diff will give me a large negative number, how do I let it know its next day?
I dont own the data, so not much I can do with the table
Thanks!
Try something like this:
DateDiff('s',[Start Date],DateAdd('d',IIF([End date]<[Start Date],1,0),[End Date]))/3600
It can be done with pure math:
TotalHours = TimeValue(CDate([End Date] - [Start Date] + 1)) * 24
SELECT cu.user_id, cu.last_activity, cu.updated_time,
DATE_PART('day', cu.last_activity - cu.updated_time), to_char(end_date - start_date, 'DD.HH24')
FROM stats.core_users cu
WHERE cu.user_id = '117132014' or cu.user_id = '117132012';
Get the result like:
117132014 2017-12-11 10:34:51.349905 2017-12-09 12:00:38.503518 1 01.22
117132012 2017-12-11 05:18:20.312283 2017-12-08 15:46:51.914085 2 02.13
Is is feasible to get the day difference with fractions like 1.91 days in the first case, instead of 1 days and 22 hours, to be more precise and easier to fit in a machine learning model?
date_part() does what it's name says: it returns one part of several elements from a date, interval or timestamp. In your case it's one part of an interval (because timestamp - timestamp returns an interval).
If you want the result as a fraction, you need to extract the seconds of the interval and then divide that by 86400 (which is the number of seconds in a day)
extract(epoch from cu.last_activity - cu.updated_time) / 86400
How do I subtract 30 days from a date in PowerBuilder?
I have the following code that returns today's date in a parameter, but I need today - 30 days:
dw_1.setitem(1, "begin_datetime",
datetime(today(), Now()))
you're probably looking for the RelativeDate function. Unfortunately, it takes a Date and not a DateTime as a parameter, but we can get around that:
dw_1.setItem(1, "begin_datetime", DateTime( RelativeDate( today(), -30), Now() )