similar to: Convert day of year back into a date format.

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

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

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, 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

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) +

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 --------------------

Hello, I've come across a problem in developing a set of custom functions to calculate the number of hours of daylight at a given latitude, and the number of days a date precedes or secedes the summer solstice. I discovered an inconsistency concerning leap years between my derived values and those from the US naval databases. It seems as far as I can figure that my inconsistency arises either

Hi, Is there a package for converting day-month-year type date to julian day number (JDN)? I looked around and I couldn't find any (I am pretty new to R...) Thanks and happy New Year to everybody! H. N?zhet Dalfes Professor, Istanbul Technical University Eurasia Institute of Earth Sciences +90 (532) 206-1308 [dalfes at itu.edu.tr]

Hello, I am trying to define a different interval for a "year". In hydrology, a "water year" is defined as the period between October 1st and September 30 of the following year. I was wondering how I might do this in R. Say I have a data.frame like the following and I want to extract a variable with the water year specs as defined above:

Hello, I am new to R and I would like to get the day of the year from numerous data in the following form: %a %d/%b/%Y %H:%M:%S (for example from Tu 10/Jul/2007 21:59:13 I would like to get 191) Whatever I try, I get NAs. Well... I need your help, please! Thanks in advance Yours sincerely, Solène Goy _________________________________________________________________ [[elided Hotmail spam]]

R-Help Please can you tell me if there is a function within R that will return the day of any given date? Using 19 March 2006 as an example: given 20060319 (or any other date format) will return the day Sunday. Many thanks Fiona.

I have a list that is generated from the resape package function 'cast'. It consists of three columns, Sku, DayOfYear, variable it is generated like: r2007 <- cast(m2008, DayOfYear ~ variable | Sku, sum) Now DayOfYear can range from 1:365 but there are not necessarily that many rows in the list. What I want to do is make every row in the list of lenght 365 and have the values

Hello! I've been trying to get this right for quite a while now and fear there is an easy solution I just don't see. I did not have this problem in Linux, and I searched r-help and Google but did not find a solution, but of course I am grateful for and resources I might not have found our not understood yet. I try to parse a time stamp with time zone. I essentially just want to parse the

Hello R-users, Does anyone know how obtain the week of a date? (in SPPS the instruction is "xdate.week") Thanks, Juan Ramon -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the

I have data in the following format: > DATE [1] "01/13/2004" In order to find the difference between two data points, I presently use brute force to calculate the day of the year: > strptime(DATE, format="%m/%d/%Y")$yday [1] 12 Although this works, it may not be robust over different years. I assume that R is sufficiently clever that a much simpler approach

> is.na(strptime("5/2/1992", format="%m/%d/%Y")) [1] FALSE > is.na(strptime("5/3/1992", format="%m/%d/%Y")) [1] TRUE Any idea what's going on with this? Running strptime against all dates from around 1946, only 5/3/1992 was converted as "NA". Even stranger, it still seems to have a value associated with it (even though is.na thinks

I have mange to use the library reshape to give me data structures that I want. Specifically: m2008 <- melt(t2008, id.var=c("DayOfYear","Category","SubCategory","Sku"), measure.var=c("Quantity")) m2007 <- melt(t2007, id.var=c("DayOfYear","Category","SubCategory","Sku"),

I have a set of data that is basically sales figures for a given year. It has columns for Yeaqr, Day Of Year, Sku, SubCatetory, and Category. The first few lines of data look like: Year DayOfYear Sku Quantity CatId Category SubCategory 1 2007 1 100091 1 10862 HOLIDAY Christmas 2 2007 1 100138 1 11160 PET COSTUMES Famous

Hello all, I have not been able to find an answer to this problem. I feel like it might be so simple though that it might not get a response. Suppose I have a dataframe like the one I have copied below (minus the 'calib' column). I wish to create a column like calib where I am subtracting the 'Count' when 'stain' is 'none' from all other 'Count' data for

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)) ##