similar to: finding the last day of the month

Is there any R function to calculate automatically the last day of a particular month? For example "sep2009" should be converted to last day of September of 2009? Thanks -- View this message in context: http://www.nabble.com/How-to-get-last-day-of-a-month--tp25425645p25425645.html Sent from the R help mailing list archive at Nabble.com.

I have a vector that contains month and year in the format YYYYMM (e.g.“200701”, “200702”) I wish to do to things: 1. I need to convert to a date that is the last calendar day of each month. 2. I need to convert this to a date that is the last U.S. stock-exchange trading day of each month. Any advice is appreciated, mymonths <- c(200701, 200702)

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

I have a vector of dates. I wish to find the month end date for each. Any suggestions? e.g. For 12/15/05, I want 12/31/05, For 10/15/1995, I want 10/31/1995, etc __________________________________________________ [[alternative HTML version deleted]]

Dear Gabor, Thanks for your reply. however: > tail(DJd) ^DJI.Close 2010-04-01 10927.07 2010-04-05 10973.55 2010-04-06 10969.99 2010-04-07 10897.52 2010-04-08 10927.07 *2010-04-09 10997.35* > tail(ag) 2009-11-30 10344.84 2009-12-31 10428.05 2010-01-31 10067.33 2010-02-28 10325.26 2010-03-31 10856.63 *2010-04-30 10997.35 * It seems the script "makes up"

Dear R People: I have a monthly time series which runs from January 1998 to December 2010. When I use tsp I get the following: > tsp(ibm$ts) [1] 1998.000 2010.917 12.000 Is there an easy way to convert this to a seq.Date object, please? I would like to have something to the effect of 1998/01/01 .... 2010/12/01 Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer

Hello, I want to change the day of the month in a date object. What I am doing at the moment is: x=as.POSIXlt(x) x$mday=13 x=as.Date(x) Does anybody know if there is a more "natural" (eficient) way to do this Thank you Felipe Parra [[alternative HTML version deleted]]

Hello, I would like to implement a "turn-of-the-month' trading strategy in R. Given a daily series of stock market return data as a zoo object, the strategy would go long (buy) four trading days before the end of the month, and sell the third trading day of the following month. How can I select these days, particularly the fourth day before and the third day after the turn of the

Hello all, ## I am trying to convert some year and month data into a single variable that has a date format so I can plot a proper x axis. ## I've made a few tries at this and search around but I haven't found anything. I am looking for something of the format "%Y-%m" ## A data.frame df <- data.frame(x=rnorm(36, 1, 10), month=rep(1:12, each = 3), year=c(2000,2001,2002)) ##

Dear R helpers. I am new to plotting time data using R. wonder how to convert character time info into date in R. I searched over the web but did not find answer. the input character string is something like 03_1993 or 03-1993, so the precision is at month level. I tried the following but failed. #R code below. strptime(c("03_1993"),"%m_%Y")

Hello, I have two date strings, say "1972-06-30" and "2012-01-31", and I'd like to get every quarter period end date between those dates? Does anyone know how to do this? Speed is important... Here is a small sample: Two dates: "2007-01-31" "2012-01-31" And I'd like to get this: [1] "2007-03-31" "2007-06-30"

Dear R community, I have the following two zoo objects: MONTHLY CPI > plot(z) > par("usr") [1] 1977.76333 2011.15333 70.39856 227.03744 > z=zooreg(cpius$Value,as.yearmon("1979-11"),frequency=12) > str(z) ?zooreg? series from Nov 1979 to Oct 2010 Data: num [1:372] 76.2 77 77.8 78.5 79.5 80.3 81.1 82 82 82.6 ... Index: Class 'yearmon' num [1:372]

Dear R experts, I have a problem in modifying one column of a dataframe with a datatime format using a datetime operator. Here is my dataframe A: DATACONT PROVINCIA VALORE 1 2007-12-31 MI 1 2 2007-12-31 PV 2 3 2007-12-31 NA 3 4 2007-12-31 MI 4 5 2007-12-31 RM 5 6 2007-12-31 RM 6 7 2007-12-31 MI 7 8

Hi List, The following code returns an "Error in as.POSIXlt.character(x, tz, ...) : character string is not in a standard unambiguous format" require(raster) require(rts) require(stringi) r <- raster(ncol=100, nrow=100) values(r) <- runif(ncell(r)) list(ID=seq(1:24),month=rep(str_pad(1:12, pad = 0,width = 2 , "left"),2),year=sort(rep(2016:2017,12)))->dt

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Is there an elegant way to do operations (+/-/*/ / ) on zoo/xts objects when one serie is monthly (end of month) and the other daily (weekdays only) - typically a monthly economic indicator and a stock index price? Thanks, TDB -- View this message in context: http://r.789695.n4.nabble.com/How-to-do-operations-on-zoo-xts-objects-with-Monthly-and-Daily-periodicities-tp3558081p3558081.html

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 I have a stock market trading values time series. What's the best way to extract the "last day of month" values. I looked at function window() but doesn't appear suitable for this since it expects regular dates. Thank you. lukas - -- Lukas Kubin lukas.kubin at permonik.com phone: 00420603836180 -----BEGIN PGP SIGNATURE-----

Last line in the following (updated) code produces the error require(raster) require(rts) require(stringr) r <- raster(ncol=100, nrow=100) values(r) <- runif(ncell(r)) stack(r)->s r->rs for(i in 1:23){ rs[]<-r[]*i addLayer(s,rs)->s print(nlayers(s)) } dt<-list(ID=seq(1:24),month=rep(formatC(1:12,flag=0,width=2),2), year=sort(rep(2016:2017,12)))

It works if you use as.Date. But this defeates the purpose for the yearmon notion... require(raster) require(rts) require(stringr) r <- raster(ncol=100, nrow=100) values(r) <- runif(ncell(r)) stack(r)->s r->rs for(i in 1:23){ rs[]<-r[]*i addLayer(s,rs)->s print(nlayers(s)) } dt<-list(ID=seq(1:24),month=rep(formatC(1:12,flag=0,width=2),2), year=sort(rep(2016:2017,12)))

Hi, I'd like to make a time series at an annual frequency. > a<-xts(x=c(2,4,5), order.by=c("1991","1992","1993")) Error in xts(x = c(2, 4, 5), order.by = c("1991", "1992", "1993")) : order.by requires an appropriate time-based object > a<-xts(x=c(2,4,5), order.by=1991:1993) Error in xts(x = c(2, 4, 5), order.by =