search for: date2

...etimes 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(Date2) Date2 Date2 = strptime(Date2 + 19e6L, "%Y%m%d&quo...

...009 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(date1,date...

Hello, probably a newbie question, but i didnt find any information on plotting/regressing w.r.t. a date data type. My trials were unfruitful. Can anyone help ? Thanks in advance! Here is my interaction with R: > tabelle date number date2 1 2009-01-1 1673 2009-01-01 2 2009-12-1 2111 2009-12-01 3 2010-7-1 2487 2010-07-01 4 2013-2-1 4301 2013-02-01 > regression.punkte<-lm(tabelle$number ~ tabelle$date2) Fehler in model.frame.default(formula = tabelle$number ~ tabelle$date2, : ung?ltiger Typ (list) f?r die Variable 'tabelle$d...

...d dataset based on the measurement date of my observed data. Here, is an example similar to what I have library(chron);library(zoo) DATE<- seq(as.Date("2009-01-01"), as.Date("2009-05-01"), by = 1) mydat<- rnorm(length(DATE), 20,5) myzoo<- zoo(mydat, order.by = DATE) DATE2<- seq(as.Date("2009-01-01"), as.Date("2009-01-30"), by = 7) Now I need to create a new zoo object with index as DATE2 and corresponding data values. Any kind of help will be highly appreciated. Thanks in advance you very much -- Acharya, Subodh [[alternative HTML versi...

...gets one color and the treatment phase another? I want to be able to show how long we have been in contact with our patients, how much of the contact time that was assessment and how much that was actual treatment. Below is an example (I call it the not-working example) df2 <- data.frame( ?date2 = seq(Sys.Date(), len= 156, by="2 day")[sample(156, 78)], ?patient2 = factor(rep(1:26, 3), labels = LETTERS) ) df2 <- df2[order(df2$date2), ] dt2 <- qplot(date2, patient2, data=df2, geom="line") dt2 + scale_x_date(major="months", minor="weeks") d...

...ate. 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(unlist(x)) med <- median(obj, na.rm...

...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", &quot...

...th 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) temp...

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" transpose) based on...

...nt 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 this matte...

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 ??? ???...

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 is how...

I have following object : > date2 [,1] [,2] [1,] "apr" "1992" [2,] "aug" "1992" [3,] "dec" "1992" [4,] "feb" "1992" [5,] "jan" "1992" [6,] "jul" "1992" [7,] "jun" "1992"...

...date1 = timeDate(charvec = Sys.Date(), format = "%Y-%m-%d") date1 dow = 3; for (i in 1:length(V4) ) { x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings=""); y = x[,1]; year = V2[[i]]; week = V3[[i]]; dtstr = sprintf("%i-%i-%i",year,week,dow); date2 = timeDate(dtstr, format = "%Y-%U-%w"); resultsdataframe$dt[[i]] <- difftimeDate(date1,date2,units = "weeks"); fp = fitdistr(y,"exponential"); print(c(V1[[i]],V2[[i]],V3[[i]],fp$estimate,fp$sd)); print(c(year,week,date2,resultsdataframe$dt[[i]])); resultsd...

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. Subsetting using...

..."20.4"), PS = c("1014.8", "912.8", "1003", "1014.4", "967.8", "NULL"), FF_10M = c("2.91", "9.91", "1.94", "4.08", "0", "6.02" ), DD_10M = c(220, 180, 140, 180, 0, 320), date2 = c("2011051312", "2011051312", "2011051312", "2011051312", "2011051312", "2011051312" )), .Names = c("Station_NO", "TMAX_2M", "TMIN_2M", "TD_2M", "PS", "FF_10M", "DD_10M&quo...

...om dropdowns, pick a category and a report is generated.) <%= start_form_tag :action => ''display_report'' %> From: <%= date_select ''report'', ''date1'' %> <br /> To: <%= date_select ''report'', ''date2'' %> <br /> Category: <%= select(:report, :category_id, @categories) %><br /> <%= submit_tag ''Show Report'' %> <%= end_form_tag %> When I submit this form (back to the same action) all of the dropdowns are reset to default values, inst...

Hi Ahmed, Your problem appears trivial as you have already specified the form of the calculation. Learn how to "extract" specified elements from a data structure: # first value sum(dataset1$NPP[dataset1$date >= date1 & dataset1$date <= date2]) # second value dataset2$biomass[dataset2$date == date2] - dataset2$biomass[dataset2$date == date1] # third value dataset3$littfall[dataset3$date == date2] Note that you may have to convert character strings to dates to do the above - see a function like "as.Date". Obviously I do not k...

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 version deleted]]

Hello everyone, I'm having a problem performing reshape() on a large data frame. The operation is fairly trivial but it makes R run out of memory. The data frame has the following structure: ID DATE1 DATE2 VALTYPE VALUE 'abcd1233' 2009-11-12 2009-12-23 'TYPE1' 123.45 ... VALTYPE 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" transpose) based on common ID and DA...