search for: date1

...7/04/2009 usa 40 12 27/04/2009 Mexique 26 13 27/04/2009 Canada 6 14 27/04/2009 Spain 1 15 28/04/2009 Canada 6 I want to create something like that: ? When entering two dates date1,date2 in the fuction extraction. The result must be: a new subdata with one line per date , per PAYS,per nb_pays.ILI (by summing all the number in variable nb_pays.ILI per date,per country) and the date must be between date1 and date2. I sart to do somethings like that extraction=function(date...

...up as 2020 after conversion. That's just wrong. 2) Sometimes I get NAs. That I don't understand. Can anyone see what I'm doing wrong here? As I say, I think I'd like to convert 01/03/2004 into 1040103 as this numeric format seems very good for doing comparisons. The code: MyDate1 = read.csv("C:\\Date1.txt",header=TRUE) MyDate2 = read.csv("C:\\Date2.txt",header=TRUE) MyDate1 Date1 = MyDate1$Date class(Date1) mode(Date1) Date1 Date1 = as.Date(Date1, "%m/%d/%y") class(Date1) mode(Date1) Date1 MyDate2 Date2 = MyDate2$EnDate class(Date2) mode(Dat...

Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev date1 1 lev1 2007...

...a frame consist of columns of class Date. Below are my test data and best attempt at using apply. I didn't see a solution via Google or the Baron search site. I'd be grateful for any suggestions or solutions. I'm using R 2.0.0 on Mac OS X. Thank you, Stephen Weigand ### Test data date1 <- c(1000, 2000, 3000,4000) date2 <- date1 + 100 date3 <- date2 + 100 class(date1) <- class(date2) <- class(date3) <- "Date" test <- data.frame(date1, date2, date3) print(test) ### create a function for apply() medDate <- function(x){ obj <- unclass(unlis...

Hi All, I have an script in R which accepts user inputs for certain parameters, particularly dates, which the user inputs as character strings. eg: > date1 <- "2009-01-21" The script later parses the input via the as.Date function: > as.Date(date1) However, as.Date encounters an error when the string does not represent an actual date. eg: > date1 <- "2009-02-29" # Note: 2009 not a leap year > as.Date(date1) Error...

Hi Elisa, Try this: date1<-format(seq.Date(as.Date("1991.1.1",format="%Y.%m.%d"),as.Date("1996.12.31",format="%Y.%m.%d"),by="day"),"%Y.%m.%d") ?length(date1) #[1] 2192 mat1<-matrix(c(.314,.314,.273,.273,.236,.236,.236,.236,.273,.314,.403,.314),ncol=1) res1&lt...

...omebody point me to references or provide some code on dealing with this date issue. Basically, I have two vectors of values that represent dates. I want to convert these values into a date format and subtract the differences to show elapsed time in days. More specifically, here is the example: Date1 Date2 032398 061585 032398 061585 111694 101994 111694 101994 062695 021595 051898 111597 072495 040195 072495 040195 The dates are in the mmddyy format, but when I attempt to format these in R with the function, date.mmddyy(Date1), I get very odd results. Any help on...

...est of the search list. In code sourced to the global environment, only the third of these is searched. Since base is in the second one, it is found first in the package version. Duncan Murdoch > > Terry Therneau > > > code: > joe <- function(id, data, subset, na.action, date1, date2, other.args) { > Call <- match.call() > if (!missing(data)) temp <- fred(Call) > > temp > } > > fred <- function(Call) { > # get a first copy of the id and date variables > index <- match(c("id", "date1"...

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x) data.frame(date1=as.Date(paste(x[,2],x[,3],x[,4],sep="-"),format="%Y-%m-%d"),discharge=x[,...

...eful with envr and enclos arguments -- how does base end up earlier in the search path? Perhaps this is clearly stated in the docs and just not clear to me? A working solution to the dilemma is of course more than welcome. Terry Therneau code: joe <- function(id, data, subset, na.action, date1, date2, other.args) { Call <- match.call() if (!missing(data)) temp <- fred(Call) temp } fred <- function(Call) { # get a first copy of the id and date variables index <- match(c("id", "date1", "date2"), names(Call), nomatch=0)...

I have time series observation on daily frequencies : library(zoo) SD=1 date1 = seq(as.Date("01/01/01", format = "%m/%d/%y"), as.Date("12/31/02", format = "%m/%d/%y"), by = 1) len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow = len1),  date1) plot(data1) Now I want to calculate 1. Quarterly statistics like...

Full_Name: Alexander L. Belikoff Version: 2.8.1 OS: Ubuntu 9.04 (x86_64) Submission from: (NULL) (67.244.71.200) I'm trying to reshape the following data frame: ID DATE1 DATE2 VALUE_TYPE VALUE 'abcd1233' 2009-11-12 2009-12-23 'TYPE1' 123.45 ... VALUE_TYPE is a string and is a factor with only 2 values (say TYPE1 and TYPE2). I need to transform it into the following data frame ("wide" tr...

Hi, I have a query about sqldf, and dates in general. I couldnt find much on the net or on the forums, hence I am here. Here is the issue: I want to write a function that accepts 3 arguments: date1, date2 and a dataframe, say 'df'. Within the function, I want to populate a temp dataframe which essentially contains the output of the query "select * from df where DATE between date1 and date2". DATE is a column (of class Date) which will be present in the input dataframe. This...

...to convert a vector of dates into julian dates using the following commands: as.date & as.numeric. I can convert a date no problem by doing the following: as.numeric (as.date ("9/21/2004")) but as soon as I try to do an entire vector I am given the following: as.numeric (as.date(date1)) Error in as.date(date1) : Cannot coerce to date format I have tried starting with a number of different date formats including dd/mm/yy, dd/mm/yyyy, month, day, year.. Following is a sample of the vector of dates that I am using: > date1 [1] 7/1/2005 4/27/2005 7/15/2004 11/30/2004 1/8...

i have a data frame with 2 columns of dates. with str(dataframe) i have ensured myself that they were indeed formatted as dates. one column has NA's in it. the aim is now to make a third column that chooses date1 if it is a date, and choose date2 if it is a NA. i am trying df$date3=ifelse(is.na(df$date1), df$date2, df$date1). this leads to unexpected behaviour: the resulting column is numeric, and shows numbers like 16000. i have no idea what this is and how to solve it? henk [[alternative HTML versio...

...ave a vector in a format year-month-day hh:mm:ss.0000. Or, in format: format = "%F %H:%M:%OS4", as POSIXct class. I made the the conversion step-by-step to have sure that nothing is missed in the way: > options (digits.sec = 4) > getOption ("digits.sec") [1] 4 > teste$Date1 = as.Date (teste$Date, format = "%m/%d/%y") > class (teste$Date1) [1] "Date" > teste$Fraction = sub ("0.", "", teste$Fraction) > teste$TimeC = paste (teste$Time, teste$Fraction, sep = ".") > teste$TimeCC = paste (teste$Date1, teste$TimeC...

Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ??? 2,A,9-Feb-01,24-May-01 ??? ??? 2,B,24-May-01,25-May-01 ??? ??? ??? ??? ??? 2,...

Folks: With the help of David L. Reiner, I've developed a function that computes the number of months between 2 dates, x and y. num.months <- function ( x , y ) { x <- as.Date( x ) y <- as.Date( y ) seeq <- seq(from=x , to=y , by="months") ans <- length( seeq ) - 1 if ( max( seeq) > y ) ans <- ans -1 ans } To ease your reading this function, I've

...alf Of Ana Patricia Martins Sent: Wednesday, April 19, 2006 8:45 AM To: Smith, Phil Cc: r-help at stat.math.ethz.ch Subject: Re: [R] Function for computing the difference between 2 dates inmonths Hi, I tried to apply the function you developed to my data file, which had to be created as follows: date1<-sub(' - - ','',paste(substring(ie[,],257,260),substring(ie[,],255,256),substring(ie[,] ,253,254),sep="-")) Your functions works well only in example #1 but when I applied a "cycle" (if clause) doesn't work (example #2 and #3). Example #1: a35641<...

Hi all, following sample code illustrates the problem : Date1 <- Date2 <- as.POSIXlt(seq.Date(as.Date("2010-04-01"),as.Date("2011-04-01"),by='day')) identical(Date1,Date2) all.equal(Date1,Date2) identical() gives the correct answer. As there is no all.equal method for POSIXlt objects, all.equal.list is used instead. Subs...