Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using> x$Date <-as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried> as.Date(as.yearmon(x$Date,frac=0))But I don't get the last day of the month here. (Tried frac=1 too.) I then add a month to the date, substract one day from the resultant date. But this wouldn't work for December.> x$YearEnd[1] 200203 200303 200403 200503 200603 200603 200312 200503 200603 200203 200303 [12] 200403 200503 200512 200612 200203 200303 200403 200503 200603> > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),+ as.integer(substr(x$YearEnd,5,6))+1, + 1))-1> x$Date[1] "2002-03-31" "2003-03-31" "2004-03-31" "2005-03-31" "2006-03-31" "2006-03-31" [7] NA "2005-03-31" "2006-03-31" "2002-03-31" "2003-03-31" "2004-03-31" [13] "2005-03-31" NA NA "2002-03-31" "2003-03-31" "2004-03-31" [19] "2005-03-31" "2006-03-31" So I add a year, and set the month to 1 in a quick function.> GetEOM <- function(yyyymm=200406){year <- as.integer(substr(yyyymm,1,4)) month <- as.integer(substr(yyyymm,5,6)) if (month==12){ date <- as.Date(ISOdate(year+1,1,1))-1 }else{ date <- as.Date(ISOdate(year,month+1,1))-1 } print(date) } x$Date <- as.vector(sapply(x$YearEnd,GetEOM)) str(x$Date) Is there a simpler way to do this please? TIA and best, -Tir Tirthankar Patnaik India Strategy Citigroup Investment Research +91-22-6631 9887
Try this: library(zoo) x <- "200212" as.Date(as.yearmon(paste(x, "01", sep = ""), "%Y%m%d"), frac = 1) On 5/10/07, Patnaik, Tirthankar <tirthankar.patnaik at citi.com> wrote:> Hi, > Given a date, how do I get the last date of that month? I have > data in the form YYYYMM, that I've read as a date using > > > x$Date <- > as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) > > But this gives the first day of the month. To get the last day of the > month, I tried > > > as.Date(as.yearmon(x$Date,frac=0)) > > But I don't get the last day of the month here. (Tried frac=1 too.) > > I then add a month to the date, substract one day from the resultant > date. But this wouldn't work for December. > > > x$YearEnd > [1] 200203 200303 200403 200503 200603 200603 200312 200503 200603 > 200203 200303 > [12] 200403 200503 200512 200612 200203 200303 200403 200503 200603 > > > > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4), > + as.integer(substr(x$YearEnd,5,6))+1, > + 1))-1 > > x$Date > [1] "2002-03-31" "2003-03-31" "2004-03-31" "2005-03-31" "2006-03-31" > "2006-03-31" > [7] NA "2005-03-31" "2006-03-31" "2002-03-31" "2003-03-31" > "2004-03-31" > [13] "2005-03-31" NA NA "2002-03-31" "2003-03-31" > "2004-03-31" > [19] "2005-03-31" "2006-03-31" > > So I add a year, and set the month to 1 in a quick function. > > > GetEOM <- function(yyyymm=200406){ > year <- as.integer(substr(yyyymm,1,4)) > month <- as.integer(substr(yyyymm,5,6)) > if (month==12){ > date <- as.Date(ISOdate(year+1,1,1))-1 > }else{ > date <- as.Date(ISOdate(year,month+1,1))-1 > } > print(date) > } > > x$Date <- as.vector(sapply(x$YearEnd,GetEOM)) > > str(x$Date) > > > Is there a simpler way to do this please? > > > TIA and best, > -Tir > > Tirthankar Patnaik > India Strategy > Citigroup Investment Research > +91-22-6631 9887 > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >
http://finzi.psych.upenn.edu/R/library/fCalendar/html/3D-TimeDateSpecDates.html
try this also RSiteSearch("last day of the month") to get more
pointers
- Regards,
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"Patnaik, Tirthankar " <tirthankar.patnaik@citi.com>
Sent by: r-help-bounces@stat.math.ethz.ch
05/10/2007 07:12 PM
To
<r-help@stat.math.ethz.ch>
cc
Subject
[R] Getting the last day of the month.
Hi,
Given a date, how do I get the last date of that month? I
have
data in the form YYYYMM, that I've read as a date using
> x$Date <-
as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1))
But this gives the first day of the month. To get the last day of the
month, I tried
> as.Date(as.yearmon(x$Date,frac=0))
But I don't get the last day of the month here. (Tried frac=1 too.)
I then add a month to the date, substract one day from the resultant
date. But this wouldn't work for December.
> x$YearEnd
[1] 200203 200303 200403 200503 200603 200603 200312 200503 200603
200203 200303
[12] 200403 200503 200512 200612 200203 200303 200403 200503
200603>
> x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),
+ as.integer(substr(x$YearEnd,5,6))+1,
+ 1))-1> x$Date
[1] "2002-03-31" "2003-03-31" "2004-03-31"
"2005-03-31" "2006-03-31"
"2006-03-31"
[7] NA "2005-03-31" "2006-03-31"
"2002-03-31" "2003-03-31"
"2004-03-31"
[13] "2005-03-31" NA NA "2002-03-31"
"2003-03-31"
"2004-03-31"
[19] "2005-03-31" "2006-03-31"
So I add a year, and set the month to 1 in a quick function.
> GetEOM <- function(yyyymm=200406){
year <- as.integer(substr(yyyymm,1,4))
month <- as.integer(substr(yyyymm,5,6))
if (month==12){
date <- as.Date(ISOdate(year+1,1,1))-1
}else{
date <-
as.Date(ISOdate(year,month+1,1))-1
}
print(date)
}
x$Date <- as.vector(sapply(x$YearEnd,GetEOM))
str(x$Date)
Is there a simpler way to do this please?
TIA and best,
-Tir
Tirthankar Patnaik
India Strategy
Citigroup Investment Research
+91-22-6631 9887
______________________________________________
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
===========================================================================================DISCLAIMER
AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}
For 2007:
seq(as.Date('2007-02-01'), length = 12, by = "mon") - 1
Current month:
seq( as.Date( format( Sys.Date(), "%Y-%m-01")), length = 2, by =
"mon")[2] - 1
Eric
Patnaik, Tirthankar wrote:> Hi,
> Given a date, how do I get the last date of that month? I have
> data in the form YYYYMM, that I've read as a date using
>
>
>> x$Date <-
>>
> as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1))
>
> But this gives the first day of the month. To get the last day of the
> month, I tried
>
>
>> as.Date(as.yearmon(x$Date,frac=0))
>>
>
> But I don't get the last day of the month here. (Tried frac=1 too.)
>
> I then add a month to the date, substract one day from the resultant
> date. But this wouldn't work for December.
>
>
>> x$YearEnd
>>
> [1] 200203 200303 200403 200503 200603 200603 200312 200503 200603
> 200203 200303
> [12] 200403 200503 200512 200612 200203 200303 200403 200503 200603
>
>> x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),
>>
> + as.integer(substr(x$YearEnd,5,6))+1,
> + 1))-1
>
>> x$Date
>>
> [1] "2002-03-31" "2003-03-31" "2004-03-31"
"2005-03-31" "2006-03-31"
> "2006-03-31"
> [7] NA "2005-03-31" "2006-03-31"
"2002-03-31" "2003-03-31"
> "2004-03-31"
> [13] "2005-03-31" NA NA
"2002-03-31" "2003-03-31"
> "2004-03-31"
> [19] "2005-03-31" "2006-03-31"
>
> So I add a year, and set the month to 1 in a quick function.
>
>
>> GetEOM <- function(yyyymm=200406){
>>
> year <- as.integer(substr(yyyymm,1,4))
> month <- as.integer(substr(yyyymm,5,6))
> if (month==12){
> date <- as.Date(ISOdate(year+1,1,1))-1
> }else{
> date <- as.Date(ISOdate(year,month+1,1))-1
> }
> print(date)
> }
>
> x$Date <- as.vector(sapply(x$YearEnd,GetEOM))
>
> str(x$Date)
>
>
> Is there a simpler way to do this please?
>
>
> TIA and best,
> -Tir
>
> Tirthankar Patnaik
> India Strategy
> Citigroup Investment Research
> +91-22-6631 9887
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>