I have column long_term_remaining_days, and I want to create a Date field which
will be calculate TODAY - long_term_remaining_days and in this way will display a Date, for example 1 Jun 2020.
I tried to do as in Excel and used the formula below, but it doesn't work:
TODAY() - long_term_remaining_days
One way it can be achieved is by using the new PARSE_DATE, UNIX_DATE and CURRENT_DATE functions (introduced in the 17 Sep 2020 update to Dates and Times).
1) Calculated Date
Copy-paste the Calculated Field below which converts CURRENT_DATE to UNIX_DATE (the number of days since epoch, 01 Jan 1970) and subtracts long_term_remaining_days (where long_term_remaining_days represents the respective Number field), after which it's converted to a UNIX_TIMESTAMP by multiplying with 86,400 (seconds in a day) before then being recognised as a Google Data Studio date using the PARSE_DATE function:
PARSE_DATE(
"%s",
CAST(((UNIX_DATE(CURRENT_DATE())-long_term_remaining_days)*86400)AS TEXT))
Google Data Studio Report and a GIF to elaborate:
Related
In Stata I have a variable yearmonth which is formatted as 201201, 201202 etc. for the years 2012 - 2019, monthly with no gaps. When I format the variable as
format yearmonth %tm
The results look like: 2.0e+05 for all periods, with the exact same number each time. A Dickey-Fuller test tells me I have gaps in my data (I don't) and a tsfill command generates dozens of empty observations between each period.
How do I properly format my yearmonth variable so I can set it as a monthly date?
You do have gaps — between 201212 and 201301, for example. Consider a statement like
gen wanted = ym(floor(yearmonth/100), mod(yearmonth, 100))
which parses your integers like 201201 into year and month components. So floor(201201/100) is floor(2012.01) and so 2012 while mod(201201, 100) is 1. The two components are then the arguments of ym() which expects a year and a month argument.
Then and only then will your format statement do you want. That command won’t create date variables.
See help datetime in Stata for more information and Problem with displaying reformatted string into a four-digit year in Stata 17 for an explanation of the difference between a date value and a date display format.
I searched on Stackoverflow and was able to write this formula by reading a cell's Month Year Date Time to give me this (Column A from reading Column H):
Column A Cell A2 (A2 indicates 4/1/2021) =DATE(YEAR(H2),MONTH(H2)+1,1) which reads this 3/30/2021 5:09:55 PM.
Then I wrote a formula in Column I for giving me the Month Year format for I2 reading from A2:
=TEXT(A2,"mmmm yyyy") which is April 2021 then I copy and paste as value into Column B (B2 as April 2021)
Is there an IF statement or a formula I can write when scanning say Column I (I2 = "April 2021) to give me the Month~MonthEnd-Date~Year format? ie "April 2021" to "April 30th 2021"?
This might require a new thread but how do I "turn-on" the auto feature whenever I add a new row to keep the new rows as part of a Table?
If you are just looking for the start of the following month in column A, you formula works fine. An alternative formula for column A could be:
=EOMONTH(I2,0)+1
As an alternative to converting the date to a text value, you can format the cell to display date in the same format. The advantage to this is the cell is still an excel date that formulas can easily work with.
Again to get the last day of the month use EOMONTH formula. If you want to keep your text method going, use the following formula:
=TEXT(EOMONTH(A2,0),"mmmm dd yyyy")
Alternatively if you format the cell appropriately, you can just use:
=EOMONTH(A2,0)
Note if you really need the st for days ending in 1 and nd for days end 2, etc, it gets more complicated. I am assuming that they are not really needed.
As you can see the question above, I was wondering if IDL is able to add or subtract days / months / years to a given date.
For example:
given_date = anytim('01-jan-2000')
print, given_date
1-Jan-2000 00:00:00.000
When I would add 2 weeks to the given_date, then this date should appear:
15-Jan-2000 00:00:00.000
I was already looking for a solution for this problem, but I unfortunately couldn't find any solution.
Note:
I am using a normal calendar date, not the julian date.
Are you only concerned with dates after 1582? Is accuracy to the second important?
The ANYTIM routine is not part of the IDL distribution. Possibly there are third party routines to handle time increments, but I don't know of any builtin to the IDL library.
By default, which you are using, ANYTIM returns seconds from Jan 1, 1979. So to add/subtract some number of days, weeks, or years, you could calculate the number of seconds in the time interval. Of course, this does not take into account leap seconds/years (but leap years are fairly easy to take into account, leap seconds requires a database of when they were added). And adding months is going to require determining which month so to determine the number of days in it.
IDL can convert to and from Julian dates using JULDAY and CALDAT.
You may also read and write Julian dates (which are doubles or long integers) to and from strings using the format keyword to PRINT, STRING, and READS.
You'll want to use the (C()) calendar date format code.
format='(c(cdi0,"-",cMoa,"-"cyi04," ",cHi02,":",cmi02,":",csf06.3))'
date = julday(1, 1, 2000)
print, date, format=format
; 1-Jan-2000 00:00:00.000
date = date + 14
print, date, format=format
; 15-Jan-2000 00:00:00.000
I tried to group my daily data by week (given a reference date) to generate a smaller panel data set.
I used postgres before and there it was quite easy:
CREATE TABLE videos_weekly AS SELECT channel_id,
CEIL(DATE_PART('day', observation_date - '2016-02-10')/7) AS week
FROM videos GROUP BY channel_id, week;
But it seems like it is not possible to subtract a timestamp with a date string in Drill. I found the AGE function, which returns an interval between two dates, but how to convert this into an integer (number of days or weeks)?
DATE_SUB may help you here. Following is an example:
SELECT extract(day from date_sub('2016-11-13', cast('2015-01-01' as timestamp)))/7 FROM (VALUES(1));
This will return number of weeks between 2015-01-01 and 2016-11-13.
Click here for documentation
The %tw format in Stata has the form: 1960w1 which has no equivalent in R.
Therefore %tw dates must be post-processed.
Importing a .dta file into R, the date is an integer like 1304 (instead of 1985w5) or 1426 (instead of 1987w23). If it was a simple time series you could set a starting date as follows:
ts(df, start= c(1985,5), frequency=52)
Another possibility would be:
as.Date(Camp$date, format= "%Yw%W" , origin = "1985w5")
But if each row is not a single date, then you must convert it.
The package ISOweek is based on ISO-8601 with the form "1985-W05" and does not process the Stata %tw.
The Lubridate package does not work with this format. The week() returns the number of complete seven day periods that have occurred between the date and January 1st, plus one. week function
In Stata week 1 of any year starts on 1 January, whatever day of the week that is. Stata Documentation on Dates
In the format %W of Date in R the week starts as Monday as first day of the week.
From strptime %V is
the Week of the year as decimal number (00--53) as defined in ISO
8601. If the week (starting on Monday) containing 1 January has four or more days in the new year, then it is considered week 1. Otherwise,
it is the last week of the previous year, and the next week is week 1.
(Accepted but ignored on input.) Strptime
Larmarange noted on Github that Haven doesn't interpret dates properly:
months, week, quarter and halfyear are specific format from Stata,
respectively %tm, %tw, %tq and %th. I'm not sure that there are
corresponding formats available in R. So far they are imported as
integers.
Is there a way to convert Stata %tw to a date format R understands?
Here is an Stata file with dates
This won't be an answer in terms of R code, but it is commentary on Stata weeks that can't be fitted into a comment.
Strictly, dates in Stata are not defined by the display formats that make them intelligible to people. A date in Stata is always a numeric variable or scalar or macro defined with origin the first instance in 1960. Thus it is at best a shorthand to talk about %tw dates, etc. We can use display to see the effects of different date display formats:
. di %td 0
01jan1960
. di %tw 0
1960w1
. di %tq 0
1960q1
. di %td 42
12feb1960
. di %tw 42
1960w43
. di %tq 42
1970q3
A subtle point made explicit above is that changing the display format will not change what is stored, i.e. the numeric value.
Otherwise put, dates in Stata are not distinct data types; they are just integers made intelligible as dates by a pertinent display format.
The question presupposes that it was correct to describe some weekly dates in terms of Stata weeks. This seems unlikely, as I know no instance in which a body outside StataCorp uses the week rules of Stata, not only that week 1 always starts on 1 January, but also that week 52 always includes either 8 or 9 days and hence that there is never a week 53 in a calendar year.
So, you need to go upstream and find out what the data should have been. Failing some explanation, my best advice is to map the 52 weeks of each year to the days that start them, namely days 1(7)358 of each calendar year.
Stata weeks won't map one-to-one to any other scheme for defining weeks.
More in this article on Stata weeks
It's not completely clear what the question is but the year and week corresponding to 1304 are:
wk <- 1304
1960 + wk %/% 52
## [1] 1985
wk %% 52 + 1
## [1] 5
so assuming that the first week of the year is week 1 and starts on Jan 1st, the beginning of the above week is this date:
as.Date(paste(1960 + wk %/% 52, 1, 1, sep = "-")) + 7 * (wk %% 52)
## [1] "1985-01-29"