similar to: Need help on parsing dates

How do I parse a date "yyyymmdd"? I tried asking chron(s, "ymd") but that didn't work. Would the date parsing routines of the Date class of 1.9 grok this? -- Ajay Shah Consultant ajayshah at mayin.org Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi

I have a daily time series and have two questions to get some help with. Firs,t I have dates in simple numeric values. e.g. ymd [1] 20050104 20050105 20050106 20050107 20050110 20050111 20050113 20050114 [9] 20050118 20050120 20050121 20050124 20050125 20050126 20050127 20050128 [17] 20050201 20050202 20050203 20050204 Now, I'd like to compute statistics, e.g. acf, by business days. So, I

Currently I have the following indexscript: pid : unique=Q boolean=Q field=pid postdate : field=startdate author_name: unhtml boolean=XAUTHORNAME field=author author_id: boolean=XAUTHORID field=authorid url : field=url sample : weight=1 index field=sample How can I create the same indexing using PHP? With this, I can get an searchable index, but I have no idea how to set the fields, so that I

Dear all, I've been struggling to perform common operations on dates that I need to be able to correct draw date-time scales - in particular I need to be able to round/truncate/ceiling dates to arbitrary precision - e.g. to weeks, months or years (or multiples thereof). I haven't been able to find anything to do this in base R (trunc.Date only truncates to sub-day units), or in the date

I tried the code that Richard O'Keefe posted last week, to wit: library(chron) ymd.to.POSIXlt <- function (y, m, d) as.POSIXlt(chron(julian(y=y, x=m, d=d))) n <- 100000 y <- sample(1970:2004, n, replace=TRUE) m <- sample(1:12, n, replace=TRUE) d <- sample(1:28, n, replace=TRUE) system.time(ymd.to.POSIXlt(y, m, d)) [1] 8.78 0.10

Dear all, I have a problem with the function Julian, may be a bug in the function ? Here is a vector of character, which represents dates (May 18 to May 20 2000): > amj <- c("2000-05-18","2000-05-18","2000-05-18","2000-05-19","2000-05-19" > ,"2000-05-19", "2000-05-19", "2000-05-20", "2000-05-20",

I am having a problem with a gantt chart since moving to R2.4.0. from 2.3.1 I made some adaptations to the code from http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=74 and successful produced a simple gantt chart. However when I upgraded to 2.4.0 it no longer works as desired. See http://ca.geocities.com/jrkrideau/R/gantproblem.pdf for the two charts. The charts were produced

hi, Thanks for responding of gantt charts. but i have some problem regarding with gantt charts. i.e. Ymd.format <- "%Y/%m/%d" Ymd <- function(x){ as.POSIXct(strptime(x, format=Ymd.format))} gantt.info <- list( labels =c("First task","Second task","Third task","Fourth task","Fifth task"), starts

After v1.0 is released, I can finally get back to sane version numbers. But any comments on which one is better: a) Postfix-style: "1.1.UNSTABLE.YYYYMMDD" -> 1.1.0 (stable) b) Odd-even numbering: 1.1.x (unstable) -> 1.2.0 (stable) With a) style the releases could be done by simply copying a nightly snapshot to releases/ directory and announcing the changes since the last

Dear R-users, -I am new to R, and I am struggling with the following problem. -I am repeating the following operations hundreds of times, within a loop: I want to subset a data frame by columns. I am interested in the columns names that are given by the rows of another data frame that was built in parallel. The solution I have so far works well as long as the elements of the second data frame

Hi All, I am reading data file ( > 1B rows) and do some date formatting like dat=fread(mydatafile) dat$date1 <- as.Date(ymd(dat$date1)) However, I am getting an error message saying that Error: cons memory exhausted (limit reached?) The script was working when the number rows were around 650M. Is there another way to handle a big data set in R? Thank you.

Hola, Otra forma, utilizando la función de intervalos y la que comprueba si otro intervalo se solapa del paquete "lubridate": #---------------------- library(lubridate) fe.chas <- data.frame( entra=c('2001-01-01','2001-06-01','2003-01-01') ,sale=c('2002-01-01','2002-06-01','2004-01-01') ) ref <-

Hi, I don't know to solve this error that is returned, even though I understand it: library(plotrix) Ymd.format<-"%Y/%m/%d" gantt.info<-list(labels= c("First task","Second task (1st part)","Third task (1st part)","Second task (2nd part)","Third task (2nd part)", "Fourt task","Fifth task","Sixth

Hello, I have a set of data that has a Date column looks like this: 12/9/2007 12/16/2007 1/1/2008 1/3/2008 1/12/2008 etc. I'd like the date to look something like the follow (so that I could sort by date easily). 20071209 20071216 20080101 20080103 20080112 How to do it? Thank you very much Julia -- View this message in context:

Hi! Is not there an standard R function to retrieve the day of the year (since 1st Jan of the same year)? I know I can make my own using julian, but find it weird that having days(), months() etc doy() does not exist as an standard function. Also, is the following not a bit inconsistent? > a <- chron("20100506",format="ymd") > a [1] 100506 > years(a) [1] 2010

Hola. Tengo una duda con esta sintaxis. Tengo una variable con formato Date que por algún motivo (el data.table viene de una consulta con PostgreSQL): datos <- prov[, pprid, pprfecbaja] str(datos) Cuando intento quitar las fechas de bajas inválidas (0001-01-01) y convertirlas a NA, la variable resultante "pierde" su condición de Date. Probé con distintas formas, siempre con el mismo

How does one convert objects c("a","b","c") and "d" into "abcd"? > paste(c("a","b","c"), "d") of course yields [1] "a d" "b d" "c d" -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org

I know that this is correct: library(chron) x = dates("01-03-04", format="d-m-y", out.format="day mon year") print(x) It gives me the string "01 Mar 2004" which is correct. I also know that I can say: print(day.of.week(3,1,2004)) in which case he says 1, for today is monday. My question is: How do I combine these two!? :-) I have a

I have two data frames. Suppose the first has rows r1 r2 r3 and the second has rows R1 R2 R3 I'd like to generate the data frame: r1 R1 r1 R2 r1 R3 r2 R1 r2 R2 r2 R3 r3 R1 r3 R2 r3 R3 How would I go about doing this? I'm sure there's a clean way to do it but I find myself thinking in loops. -- Ajay Shah

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals