similar to: Need help with labeling a plot correctly

Hi Hadley et all, I am struggling with reshape2 and melt works and melt_check (filtered151) seems fine My cast command was acast (filtered151, Time ~ Species ~ Number) > melt_check (filtered151) Using time, Species as id variables $id [1] "time" "Species" $measure [1] "Number" When I execute cast the data matrix is in the correct order but the data

Two different ways: library(ggplot2) x=5 size=50 A=data.frame(X=sample(x, size, replace=T), Y=sample(x, size, replace=T),a=rep(1:2,each=25));A # Facetting qplot(X,Y,data=A) + geom_jitter(position=position_jitter(width=.03)) + facet_grid(.~a) # Or with vp p=qplot(X, Y, data=A) + geom_jitter(position=position_jitter(width=.03)) ggsave(p, file='main.png') p1=qplot(X, Y, data=A) +

H all, I have relative abundance data from >100 sites. This is from acoustic monitoring and usually the data is for 2-3 nights but in some cases my be longer like months or years for each location.. The data output from my management data base is proved by species by night for each location so data frame would look like this below. What I need to do is sum the Survey_time by Spec_Code for

Hi there, I set up my company's back up server using rsync. And I've got a strange problem. I searched in the archives of this list, but none of them seems not giving me an idea to solve the problem. If anyone can help, it would be grateful. I'm using cron by a user (non wheel/admin) to rsync everyday during the night. The cron is set in the server to transfer the backing-up

Full_Name: John Brzustowski Version: R-devel-trunk OS: linux (problem under Windows too) Submission from: (NULL) (74.101.124.238) (This bug was discovered by Phil Taylor, Acadia University.) I'm not sure from reading the documentation whether strptime(x, "%j") is meant to be supported, but if so, there is a bug which prevents it from working on the first day of months after

Hi all: I am having trouble dropping levels, got a few hints online?without success. Please consider the dataset below: ?I was under the inpression that subset(......drop=TRUE) would work but it doesn't library(ggplot2) ??? library(hmisc) x <- structure(list(first = c(38.2086, 43.1768, 43.146, 41.8044, 42.4232, 46.3646, 38.0813, 40.0745, 40.4889, 38.6246, 40.2826, 41.6056, 34.5353,

Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

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, Two years back, I posted a small function for getting the ISO 8601 defined week number of a date (such as the week number used in all Swedish calendars), in a os-independent manner. I've since discovered an inaccuracy in that code, and so I thought I'd repost the corrected version. Hopefully this will come in handy for someone searching the mailing list archives in the future.

x = structure(list(date = structure(list(sec = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), min = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), hour = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), mday = c(1L, 2L, 3L, 4L, 5L, 8L, 9L, 10L, 11L, 12L, 15L, 16L, 17L, 18L, 19L, 22L, 23L, 24L,

Hi all, Is there a simple function already implemented for getting the ISO weeks of a Date object? I couldn't find one, and so wrote my own function to do it, but would appreciate a pointer to the "default" way. If a function is not yet implemented, could the code below be of interest to submit to CRAN? Best Regards, Gustaf --------------------

Dear useRs, I'm trying to use ggplot2 in Sweave (R 2.5.1). The plots use the alpha channel, so I need to use pdf version 1.4. Search the mailinglist archive I found two solutions: \SweaveOpts{echo = FALSE, pdf.version=1.4} and explicit writing to a pdf 1.4 file. The latter works but the first doesn't. Probably I'm doing something wrong (see the Rnw file below). The file tmp_tmp.pdf is

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

Hi everyone,I'm trying to figure out how to use gPath and the documentation is not very helpful :( I have the following plot object: plot-surrounds:: background plot.gTree.378:: background guide.gTree.355:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) guide.gTree.356::

Can someone, please, explain the difference is results below (notice the isdst value) > unlist(as.POSIXlt('2005-7-1')) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 1 > unlist(as.POSIXlt(as.Date('2005-7-1'))) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 0

Hi all, I'm usually comfortable using the *apply functions for vectorizing loops in R. However, my particular problem now is using it in a sequential operation, which uses values evaluated in an offset of the loop vector. Here is my example using a for loop approach: dat <- data.frame(year=rep(1970:1980,each=365),yday=1:365) dat$value <-

Full_Name: Michael Toews Version: R-devel and 2.4.1 OS: Debian etch and WindowsXP Submission from: (NULL) (142.58.206.114) This is a quick demonstration of the present time limitation of 'sprintf' on long vectors with a suggestion for significant optimization. First, consider a data.frame with numeric (double) values: dat <- data.frame(year=as.numeric(rep(1970:2000,each=365)),

Have you noticed any problems with big dates (>=1/1/2040) in R? Here is the bit of code that I'm having trouble with: > test.date <- strptime("1/1/2040",format="%m/%d/%Y") > > unlist(test.date) sec min hour mday mon year wday yday isdst 0 0 0 1 0 140 0 0 0 > > date.plus.one <- as.POSIXct(test.date) +

Dear R Users, I have data for more than 3 years. For each year I want to find the day corresponding to Jaunary 1 of that year. For example: > x <- c('5/5/2007','12/31/2007','1/2/2008') > #Convert to day of year (julian date) - > strptime(x,"%m/%d/%Y")$yday+1 [1] 125 365 2 I want to know how to do the same thing but with time added. But I still

?Hi again, I take advantage of my previous mail to ask you a question for which I was looking for an answer when detected the behaviour I previously told. In the help of DataTimeClasses one can read: "POSIXlt" objects will often have an attribute "tzone", a character vector of length 3 giving the time zone name from the TZ environment variable and the names of the base time