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Patnaik, Tirthankar

2007-May-10 13:42 UTC

[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

Gabor Grothendieck

2007-May-10 14:12 UTC

head link

[R] Getting the last day of the month.

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.
>

gyadav at ccilindia.co.in

2007-May-11 05:33 UTC

head link

[R] Getting the last day of the month.

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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        (  o   o  )
oOOo-(_)-oOOo--------
|
| Gaurav Yadav
| Assistant Manager, CCIL, Mumbai (India)
| Mob: +919821286118 Email: emailtogauravyadav@gmail.com
| Man is made by his belief, as He believes, so He is.
|                                           --- Bhagavad Gita 
|_______Oooo________
         oooO(      )
         (      )   )   /
          \   (    (_/
            \_ )




"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}}

Eric

2007-May-11 05:55 UTC

head link

[R] Getting the last day of the month.

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.
>

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