similar to: Calculating quarterly statistics for time series object

Hi all, Suppose I have following dataset : library(zoo) SD = 1 date1 = seq(as.Date("01/01/90", format = "%m/%d/%y"), as.Date("12/31/08", format = "%m/%d/%y"), by = 1) len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow = len1), date1) Now I want to extract monthly observation. obs =

I have a data set with quarterly time series for several variables. The time index is recorded in column 1 of the dataframe as a character vector "Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want to produce line plots with ggplot2, but it seems I need to convert the time index from character to date class. Is that right? If so, how

On Sun, 28 Jan 2018, phil at philipsmith.ca wrote: > I have a data set with quarterly time series for several variables. The > time index is recorded in column 1 of the dataframe as a character > vector "Q1 1961", "Q2 1961","Q3 1961", "Q4 1961", "Q1 1962", etc. I want > to produce line plots with ggplot2, but it seems I need to

Using Achim's d this also works to generate z where FUN is a function used to transform the index column and format is also passed to FUN. z <- read.zoo(d, index = "time", FUN = as.yearqtr, format = "Q%q %Y") On Sun, Jan 28, 2018 at 4:53 PM, Achim Zeileis <Achim.Zeileis at uibk.ac.at> wrote: > On Sun, 28 Jan 2018, phil at philipsmith.ca wrote: > >> I

I have financial data on a a set of firms, with a quarterly period (fundamental data). The data spans 10 years, and four quarters per year. The present file (.csv) reads the Date columns as "200706" for the second quarter of 2007; "199809" for the third quarter of 1997. Is there a way I can convert it to something like "2007 Q2", "1998 Q3"? I am aware of

Hi all, I'm having a little bit of trouble with some date conversions and am hoping someone can help me out. Thanks in advance. OK, I have two sources of data that provide date info in a csv file differently. I've attached a small zipped file with two text files that illustrate both. (Is it ok to send attachments to this list? Not sure. It's very small.) I need to be able to

Hi all, I am wondering how to write a 'list' object to a file. I already gone through some threads like http://mail.python.org/pipermail/python-list/2001-April/080639.html, however could not trace out any reliable solution. I tried following : > write.table(calc, file="c:/data.csv") Error in data.frame("200501" = c(-0.000387071806652095, -0.000387221689252648, :

Hi everybody, I am beginer in R and I need your precious help. I want to create a small function in R as in sas to retrieve date. I have a file with data that import in R. DATE PAYS nb_pays.ILI. 1 24/04/2009 usa 0 2 24/04/2009 usa 0 3 24/04/2009 Mexique 0 4 24/04/2009

Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev

I am trying to calculate the median of each row of a data frame where the data frame consist of columns of class Date. 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,

Hi All, I have an script in R which accepts user inputs for certain parameters, particularly dates, which the user inputs as character strings. eg: > date1 <- "2009-01-21" The script later parses the input via the as.Date function: > as.Date(date1) However, as.Date encounters an error when the string does not represent an actual date. eg: > date1 <-

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)

I am currently puzzled by a seach path behavior. I have a library of a dozen routines getlabs(), getssn(), getecg(), ... that interface to local repositories and pull back patient information. All have a the first 6 arguments in common, and immediately call a second routine to do initial processing of these 6. The functions "joe" and "fred" below capture the relevant

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

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

I Have been trying to convert a vector of dates into julian dates using the following commands: as.date & as.numeric. I can convert a date no problem by doing the following: as.numeric (as.date ("9/21/2004")) but as soon as I try to do an entire vector I am given the following: as.numeric (as.date(date1)) Error in as.date(date1) : Cannot coerce to date format I have tried

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

Hi Elisa, Try this: date1<-format(seq.Date(as.Date("1991.1.1",format="%Y.%m.%d"),as.Date("1996.12.31",format="%Y.%m.%d"),by="day"),"%Y.%m.%d") ?length(date1) #[1] 2192 mat1<-matrix(c(.314,.314,.273,.273,.236,.236,.236,.236,.273,.314,.403,.314),ncol=1) res1<-

Hello all, Can somebody point 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

Hello all, I'm having a problem with data handling. My input data is (dput in the after the signature): Date Time Fraction 06/19/13 22:15:39 0.3205 06/19/13 22:15:44 0.3205 06/19/13 22:15:49 0.3205 06/19/13 22:15:54 0.3205 06/19/13 22:15:59 0.3205 06/19/13 22:16:09 0.3205 Date in format month/day/year, Time in HH:MM:SS and fraction represents the fractions of