similar to: Data with different time lines for one analysis

OK, .Internal is not necessary to reproduce oddity in this area. I also see things like (notice 1980) > strptime(paste0(sample(1900:1999,80,replace=TRUE),"/01/01"), "%Y/%m/%d", tz="CET") [1] "1942-01-01 CEST" "1902-01-01 CET" "1956-01-01 CET" "1972-01-01 CET" [5] "1962-01-01 CET" "1900-01-01 CET"

I am trying to extract only the winters (defined to be 01-Dec through 28-Feb) of daily data from 1948-2002. There are 90 days in each winter season. I wrote the following code to gather the winter dates into a single vector: DJF <- NULL for(year in 1949:1999) { temp.begin <- strptime(paste("01/12", year-1, sep="/"), "%d/%m/%Y") temp.end <-

On 3/12/16 12:33 AM, peter dalgaard wrote: >> On 12 Mar 2016, at 00:05 , Mick Jordan <mick.jordan at oracle.com> wrote: >> >> This is definitely obscure but we had a unit test that called .Internal(strptime, "1942/01/01", %Y/%m/%d") with timezone (TZ) set to CET. > Umm, that doesn't even parse. And fixing the typo, it doesn't run: > >>

I am new to R and new to time series modeling. I have a set of variables (var1, var2, var3, var4, var5) for which I have historical yearly data. I am trying to use this data to produce a prediction of var1, 3 years into the future. I have a few basic questions: 1) I am able to read in my data, and convert it to a time series format using 'ts.' data_ts <- ts(data, start = 1988, end =

Hi, again i am stuck in my presentation, and i have never learn R before in my life but need this to be done, so please help me out for a favour: http://www.nabble.com/file/p20333155/kew.dat kew.dat run this in R and these comes up: Month Year Rain 1 Jan 1900 74.400000 2 Feb 1900 80.500000 3 Mar 1900 23.600000 4 Apr 1900 23.600000 5 May 1900 25.100000 6

I have a problem to plot label (Year) only for significant values (in this case spoz and sneg). I use this code, but don't work with labels. library(ggplot2) ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+ geom_bar(stat="identity",position="identity")+ scale_y_continuous(breaks = round(seq(-100, 100, by = 10),10))+ theme_bw() Thank you! the data used is:

Hello everybody, I am sorry if my questions are too simple or not easily understandable. I’m not a native English speaker and this is my first analysis using this function. I have a problem with a cross correlation function and I would like to understand how I have to perform it in R. I have yearly data of an independent variable (x) from 1982 to 2010, and I also have yearly data of a variable

>>>>> peter dalgaard <pdalgd at gmail.com> >>>>> on Sat, 12 Mar 2016 19:11:40 +0100 writes: > OK, .Internal is not necessary to reproduce oddity in this area. I also see things like (notice 1980) >> strptime(paste0(sample(1900:1999,80,replace=TRUE),"/01/01"), "%Y/%m/%d", tz="CET") ............... >

How does one convert to a date format when survey respondents have used two different date formats whilst entering their data. There were clearly told to use mm/dd/yyyy but humans being humans some entered mm/dd/yy. There was even validity checks on the forms but I allowed them to be overridden since the data is more holy than the format. The data was downloaded as a csv and read.csv was used to

R-devel: Some very inconsistent behavior, that I can't seem to find documented. Sys.setenv(TZ="GMT") str(unclass(strptime("1969-12-31 23:59:59","%Y-%m-%d %H:%M:%S"))) List of 9 $ sec : num 59 $ min : int 59 $ hour : int 23 $ mday : int 31 $ mon : int 11 $ year : int 69 $ wday : int 3 $ yday : int 364 $ isdst: int 0 - attr(*, "tzone")= chr

Hi all, I am getting different results from ccf and cor, Here is a simple example: set.seed(100) N <- 100 x1 <- sample(N) x2 <- x1 + rnorm(N,0,5) ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1] cor(x1[-N], x2[-1]) Results: > ccf(x1,x2)$acf[ccf(x1,x2)$lag == -1] [1] -0.128027 > cor(x1[-N], x2[-1]) [1] -0.1301427 Thanks, Tal ----------------Contact

Dear All, I wonder how to compute the age from the date of birth and the date of examination. Suppose that have the following data: df <- as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"), c(1,"10/08/1950","20/07/1998"), c(1,"10/08/1950","23/10/1998"))) names(df) <- c("ID", "date_birth",

Dear all, Can anybody tell me how to export a 3d figure made with the plot3d function? I'm careless about whether it's still interactive or not in another format, as long I can get it out of R. Thanks! Alejandro Gonz?lez Departamento de Biodiversidad y Conservaci?n Real Jard?n Bot?nico Consejo Superior de Investigaciones Cient?ficas Claudio Moyano, 1 28014 Madrid, Spain Tel +0034

I'm installing R on an Irix 6.5 machine and R fails to understand dates prior to 1 Jan 1970. This first cropped up early in the test scripts (Examples/base-Ex.R): > > seq(as.Date("1910/1/1"), as.Date("1999/1/1"), "years") > Error in fromchar(x) : character string is not in a standard unambiguous format The root there is the

Uwe, You said you used testdata <- data.frame(date = strptime(c("31121991", "31121991"), "%d%m%Y"), nothing = c(NA, NA)) but that's not the same object, and that one does work for me. > dput(testdata) structure(list(date = structure(c(694137600, 694137600), class = c("POSIXt", "POSIXct")), nothing = c(NA, NA)),

Hi all, I have been chipping away at a problem I encountered in calculating rates per year from a moderately large data file (46412 rows). When I ran the following command, I got obviously wrong output: interval<- c(NA,as.numeric(diff( strptime(mkdf$MEAS_DATE,"%d/%m/%Y")))/365.25) The values in MEAS_DATE looked like this: mkdf$MEAS_DATE[1:10] [1] 1/5/1962 1/5/1963

Hi. It's my understanding that a cross-correlation function of vectors x and y at lag zero is equivalent to their correlation (or covariance, depending on how the ccf is defined). If this is true, could somebody please explain why I get an inconsistent result between cov() and ccf(type = "covariance"), but a consistent result between cor() and ccf(type = "correlation")? Or

hello, i have been using ccf() to look at the correlation between lightning and electrogamnetic data. for the most part it has worked exactly as expected. however, i have come across something that puzzles me a bit: > x <- c(1, 0, 1, 0, 1, 0) > y <- c(0, 0, 0, 0, 0, 0) > ccf(x, x, plot = FALSE) Autocorrelations of series 'X', by lag -4 -3 -2 -1 0

Dear all, I am trying to avoid a for loop here and wonder if the following is possible: I have a data.frame with 6 columns and i want to get a cross-correlogram (by using ccf) . Obivously ccf only accepts two columns at once and then returms a list. In fact, with a for loop i?d do the following for (i in 1:6) { x[[i]]=ccf(mydf[,i],mydf[,6]) } Is there any chance to the same with

Hi all, I have a problem concerning my understanding of the cross-correlation (ccf) function in R. assume a time serie as: > t<-seq(0,6.28,by=0.01); > my_serie<-ts(sin(t),start=0,end=6.28,deltat=0.01) then I generate an other one shifted by 12 time points: > my_shifted_serie<-ts(sin(t),start=0+0.12,end=6.28+0.12,deltat=0.01) if I do the cross-correlation I get that the two