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
Related
I imported date variables as strings from SQL (date1) into Stata and then created a new date variable (date2) like this:
gen double date2 = clock(date1, "YMDhms")
format date2 %tc
However, now I want to calculate the number of days between two dates (date3-date2), formatted as above, but I can't seem to do it.
I don't care about the hms, so perhaps I should strip that out first? And then deconstruct the date into YYYY MM DD as separate variables? Nothing I seems to do is working right now.
It sounds like by dates you actually mean timestamp (aka datetime) variables. In my experience, there's usually no need to cast dates/timestamps as strings since ODBC and Stata will handle the conversion to SIF td/tc formats nicely.
But perhaps you exported to a text file and then read in the data instead. Here are a couple solutions.
tc timestamps are in milliseconds since 01jan1960 00:00:00.000, assuming 1000*60*60*24=86,400 seconds/day (that is, ignoring leap seconds). This means that you need to divide your difference by that number to get elapsed days.
For example, 2016 was a leap year:
. display (tc(01jan2017 00:00:00) - tc(01jan2016 00:00:00))/(1000*60*60*24)
366
You can also use the dofc() function to make dates out of timestamps and omit the division:
. display (dofc(tc(01jan2018 00:00:00)) - dofc(tc(01jan2016 00:00:00)))
731
2017 is not a leap year, so 366 + 365 = 731 days.
You can use generate with all these functions, though display is often easier for debugging initial attempts.
I'm looking for a way to determine the week number (week beginning on Monday) over several years. That means I don't want to have 0-53 but if, let's say I have 2 years of dates, I want them to be numbered with 0-106 in R.
I tried strftime(Datum, format ="%W") but then I only get the annual week number and not as a whole.
Given that you did not provide any data, I took the liberty of creating some:
#create data
Datum<-c("2013-03-01", "2014-06-02", "2013-06-01")
# format data to year-month-day with strptime
Datum<-strptime(Datum, "%Y-%m-%d")
You now need to identify the origin year. As I'm sure you are aware not all years have the same number of weeks 52.29 in a leap year vs. 52.4 in a standard calendar year but as this is unlikely to be a consideration for only 2 years we can use the number of weeks returned through the strftime function.
origin.year=as.numeric(min(substring(Datum,1,4)))
# number of weeks in first year (offset for second year)
n.weeks<-52
Now we can create a vector containing the number of weeks to offset each week in Datum (X).
X<-as.numeric(substring(Datum,1,4)!=origin.year)*n.weeks
We can then simply add this vector to the number of weeks returned by strftime when it is applied to Datum
week.vec<-as.numeric(strftime(Datum, "%W"))+X
This will work for 2 years, but if you have more years than this, you will need to modify the offsets to account for this.
I want to create a single column with a sequence of date/time increasing every hour for one year or one month (for example). I was using a code like this to generate this sequence:
start.date<-"2012-01-15"
start.time<-"00:00:00"
interval<-60 # 60 minutes
increment.mins<-interval*60
x<-paste(start.date,start.time)
for(i in 1:365){
print(strptime(x, "%Y-%m-%d %H:%M:%S")+i*increment.mins)
}
However, I am not sure how to specify the range of the sequence of dates and hours. Also, I have been having problems dealing with the first hour "00:00:00"? Not sure what is the best way to specify the length of the date/time sequence for a month, year, etc? Any suggestion will be appreciated.
I would strongly recommend you to use the POSIXct datatype. This way you can use seq without any problems and use those data however you want.
start <- as.POSIXct("2012-01-15")
interval <- 60
end <- start + as.difftime(1, units="days")
seq(from=start, by=interval*60, to=end)
Now you can do whatever you want with your vector of timestamps.
Try this. mondate is very clever about advancing by a month. For example, it will advance the last day of Jan to last day of Feb whereas other date/time classes tend to overshoot into Mar. chron does not use time zones so you can't get the time zone bugs that code as you can using POSIXct. Here x is from the question.
library(chron)
library(mondate)
start.time.num <- as.numeric(as.chron(x))
# +1 means one month. Use +12 if you want one year.
end.time.num <- as.numeric(as.chron(paste(mondate(x)+1, start.time)))
# 1/24 means one hour. Change as needed.
hours <- as.chron(seq(start.time.num, end.time.num, 1/24))
How can I accurately convert the products (units is in days) of the difftime below to years, months and days?
difftime(Sys.time(),"1931-04-10")
difftime(Sys.time(),"2012-04-10")
This does years and days but how could I include months?
yd.conv<-function(days, print=TRUE){
x<-days*0.00273790700698851
x2<-floor(x)
x3<-x-x2
x4<-floor(x3*365.25)
if (print) cat(x2,"years &",x4,"days\n")
invisible(c(x2, x4))
}
yd.conv(difftime(Sys.time(),"1931-04-10"))
yd.conv(difftime(Sys.time(),"2012-04-10"))
I'm not sure how to even define months either. Would 4 weeks be considered a month or the passing of the same month day. So for the later definition of a month if the initial date was 2012-01-10 and the current 2012-05-31 then we'd have 0 years, 5 months and 21 days. This works well but what if the original date was on the 31st of the month and the end date was on feb 28 would this be considered a month?
As I wrote this question the question itself evolved so I'd better clarify:
What would be the best (most logical approach) to defining months and then how to find diff time in years, months and days?
If you're doing something like
difftime(Sys.time(), someDate)
It comes as implied that you must know what someDate is. In that case, you can convert this to a POSIXct class object that gives you the ability to extract temporal information directly (package chron offers more methods, too). For instance
as.POSIXct(c(difftime(Sys.time(), someDate, units = "sec")), origin = someDate)
This will return your desired date object. If you have a timezone tz to feed into difftime, you can also pass that directly to the tz parameter in as.POSIXct.
Now that you have your date object, you can run things like months(.) and if you have chron you can do years(.) and days(.) (returns ordered factor).
From here, you could do more simple math on the difference of years, months, and days separately (converting to appropriate numeric representations). Of course, convert someDate to POSIXct will be required.
EDIT: On second thought, months(.) returns a character representation of the month, so that may not be efficient. At least, it'll require a little processing (not too difficult) to give a numeric representation.
R has not implemented these features out of ignorance. difftime objects are transitive. A 700 day difference on any arbitrary start-date can yield a differing number of years depending on whether there was a leap year or not. Similarly for months, they take between 28-31 days.
For research purposes, we use these units a lot (months and years) and pragmatically, we define a year as 365.25 days and a month as 365.25/12 = 30.4375 days.
To do arithmetic on a given difftime, you must convert this value to numeric using as.numeric(difftime.obj) which is, in default, days so R stops spouting off the units.
You can not simply convert a difftime to month, since the definition of months depends on the absolute time at which the difftime has started.
You'll need to know the start date or the end date to accurately tell the number of months.
You could then, e.g., calculate the number of months in the first year of your timespan, the number of month in the last your of the timespan, and add the number of years between times 12.
Hmm. I think the most sensible would be to look at the various units themselves. So compare the day of the month first, then compare the month of the year, then compare the year. At each point, you can introduce a carry to avoid negative values.
In other words, don't work with the product of difftime, but recode your own difftime.
Wikipedia gives an example of an ISO 8601 example of a repeating interval:
R5/2008-03-01T13:00:00Z/P1Y2M10DT2H30M
This is what this means:
R5 means that the interval after the slash is repeated 5 times.
2008-03-01T13:00:00Z means that the interval begins at this given datetime.
P1Y2M10DT2H30M means that the interval lasts for
1 year
2 months
10 days
2 hours
30 minutes
My problem is that I do not know exactly what is being repeated here. Does the repetition
occur immediately after the interval ends? Can I specify that every Monday something happens from 13:00 to 14:00?
The standard itself doesn't clarify, but the only obvious interpretation here is that the interval repeats back-to-back. So this recurring interval:
R2/2008-03-01T13:00:00Z/P1Y2M10DT2H30M
Will be equivalent to these non-recurring intervals:
2008-03-01T13:00:00Z/P1Y2M10DT2H30M
2009-05-01T15:30:00Z/P1Y2M10DT2H30M
(Note: my reading is that the number of repetitions does include the first occurrence)
There is no way to represent "every Monday from 13:00 to 14:00" inside of ISO 8601, but it's natural to do for a VEVENT in the iCalendar format. (If you could do that entirely within ISO 8601, then that would give rise to a slew of further feature requests)
Yes, ISO8601 does define a regular repeating interval (or as regular as a "month" can be as one of the units).
R5/2008-03-01T13:00:00Z/P1Y2M10DT2H30M
Should generate these times:
2009-05-11T15:30:00Z
2010-07-21T18:00:00Z
2011-10-01T20:30:00Z
2012-12-11T23:00:00Z
2014-02-22T00:30:00Z
It doesn't define a "start time" and "end time" like RFC5545 (iCalendar) does, or even irregular repetition like RRULE or crontab can.
You should be able to specify a weekly repetition using the ISO Week Date as a starting point, but you'll need separate repetitions for "start" and "end" times:
R/2021-W01-1T13:00:00Z/P1W
R/2021-W01-1T14:00:00Z/P1W
The first interval is for the start times: Mondays at 13:00 (starting in 2021), and the second is for the end times: Mondays at 14:00 (starting in 2021).
I'm probably being an idiot (Long Covid Brain) but isn't the obvious extension to ISO-8601 a second duration part? In the absence of the second duration, the repeats are back to back, in its presence what is actually repeating is a smaller duration event at the start of each period. e.g.
R/2021-W01-1T13:00:00Z/P1W/P1H
indefinite weekly repeat of hour long slots every Monday 1pm starting week 1 2021.
EDIT: Maybe you could even nest them ...
R/2021-W01-1T09:00:00Z/P1W/R5/P1D/P8H
Mon to Fri, 9am to 5pm, every week? Ok I'll get my coat