similar to: histogramming dates

Hello R-experts! I am using R Version 1.7.1 (2003-06-16) on a Debian Linux box and I have discovered an odd result when plotting data involving dates. Please try this minimal example: a = seq(ISOdate(2000,1,1), ISOdate(2001,1,1), "months") b = 1:13 plot(a,b, col="red") What I get is a plot that looks as expected except the x-axis is mostly red. Can anyone reproduce this

Although the documentation is somewhat sketchy, I() can be used to create objects of class AsIs: > I("a") [1] "a" attr(,"class") [1] "AsIs" "character" > I(4) [1] 4 attr(,"class") [1] "AsIs" "numeric" > I(4 + 0i) [1] 4+0i attr(,"class") [1] "AsIs" "complex" > This

I'd like to use stripplot for some plots because I want to use the jitter parameter. On the other hand, I'd like to use dotplot because I'd like to have the horizontal lines that it includes. dotplot doesn't have a jitter option and I'm not having any success with getting panel.grid(h=-1) with stripplot. Can anyone show me how to make dotplot-like lines on a stripplot? Or

Hi I'm fairly new to R and the list, so please take what I say accordingly! Far as I can see, strptime gives you a string in some specified format. In order to do any kind of numerically-based modelling with that, you need to obtain a number to work with. One way to do this is by getting the time with Sys.time() instead and coercing it to a number using as.integer(): >

Dear R-help, I have a data set consisting of measurements made on multiple subjects. Measurement sessions are repeated for each subject on multiple dates. Not all subjects have the same number of sessions. To create a factor that represents the session, I do the following: data <- read.csv('test-data.csv') # data appended below data$date <- as.Date(data$date,

HiI'm fairly new to R and the list, so please take what I say accordingly!Far as I can see, strptime gives you a string in some specified format. In order to do any kind of numerically-based modelling with that, you need to obtain a number to work with.One way to do this is by getting the time with Sys.time() instead and coercing it to a number using as.integer():> as.integer(Sys.time())I

Dear R-gang, I have a question about handling underscores in names in R 1.8.1 and 1.9.0. I recently installed 1.9.0 on a machine and found that many codes no longer work as a result of the changed behavior in make.names. I have numerous data files that have dashes, periods and underscores in the header row. I've got numerous R codes that read those files with read.table and read.csv and

I am trying to plot a image where the x axis has the units of time. When I issue the image(x,y,z) command with x as a POSIXct object, it fails to put a time stamp on the x axis. Instead I get a warning "Incompatible methods" warning and no dates on my x axis. This example shows my problem: Rmat=t(matrix(data=rnorm(1:500),ncol=10,nrow=50)) tax=seq(ISOdate(2010,4,14,12,0,0),

Hello, For basic date time arithmetics operations, AFAK, there're actually the function difftime() and the (dt + num) operations. I'm wondering if other basic operations exist, like add(dt, num, unit) where unit would be "y", "q", "m", etc. Also for the function seq.dates (or seq.POSIXt), the case for by="months" would be more useful if it

Hello, For basic date time arithmetics operations, AFAK, there're actually the function difftime() and the (dt + num) operations. I'm wondering if other basic operations exist, like add(dt, num, unit) where unit would be "y", "q", "m", etc. Also for the function seq.dates (or seq.POSIXt), the case for by="months" would be more useful if it

Dear R Gang, I'm interested in using R and the nls package for fitting kinetic models. I'm having some difficulty getting a model specified for nls though. The math for the model that I want to fit is dg(t)/dt = K1 f(t) - k2 g(t) where g(t) and f(t) are measured data at a sequence of times t. K1 and k2 are the parameters of the model. If I solve this, the solution is g(t) = K1

hi, I've troubles with some difftime objects. e.g. ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) - 2 works, telling me "Time difference of 57 days". But when I'd like to add days, such as ISOdate(2001, 4, 26) - ISOdate(2001, 2, 26) + 2 the function gives me an error. Function "as.COMDate.chron" of the Rbloomberg package doesn't work for that reason. I'm

Dear all I try to make hourly average by cut() function, which almost works as *I* expected. What puzled me is that if there is only one item at the end of your data it results in NA. Example will explain what I mean datum<-seq(ISOdate(2004,8,31), ISOdate(2004,9,1), "min") cut(datum[1370:1381],"hour", labels=F) [1] 1 1 1 1 1 1 1 1 1 1 1 NA

Dear list: working with date/times I have come across a problem that ISOdate and ISOdatetime are too slow on large vectors of data. I was surprised just until I looked at the implementation and the man page: "ISOdatetime and ISOdate are convenience wrappers for strptime". In other terms, they convert data to character representation first in order to create a POSIXlt object that is then

Dear R-people! I am using R 1.8.0, under Windows XP. While using ISOdate() and strptime(), I noticed the following behaviour when "wrong" arguments (e.g., months>12) are given to these functions: > ISOdate(year=2003,month=2,day=20) #ok [1] "2003-02-20 13:00:00 Westeurop?ische Normalzeit" > ISOdate(year=2003,month=2,day=30) #wrong day, but returns a value [1]

I am a beginner in R with a background in SAS. Are there built-in R methods of reading dates for calculating elapsed days between two calendar dates? If so, are there any examples I can browse? Thanks in anticipation. John Byrne. Lecturer in Information Systems. Australian Catholic University.

Dear all, I have found the following (for me) incomprehensible behaviour of ISOdate (POSIXct): > ISOdate(1900,6,16) [1] "1900-06-15 14:00:00 Westeurop?ische Sommerzeit" > ISOdate(1950,6,16) [1] "1950-06-16 14:00:00 Westeurop?ische Sommerzeit" Note that in the first case I get the 15th of June back, not the 16th as I would have expected! This happened under R-1.7.1 on

Hi, Given a date, how do I get the last date of that month? I have data in the form YYYYMM, that I've read as a date using > x$Date <- as.Date(ISOdate(substr(x$YearEnd,1,4),substr(x$YearEnd,5,6),1)) But this gives the first day of the month. To get the last day of the month, I tried > as.Date(as.yearmon(x$Date,frac=0)) But I don't get the last day of the month here. (Tried

Full_Name: andrea capodicasa Version: 1.7.0 OS: w2k sp3 Submission from: (NULL) (212.17.194.154) Hi all, the problem is when you try to round.Posix an empty vector of dates Please give a look to this code: > data=seq(ISOdate(2001,1,1),by="day",length=3) > data [1] "2001-01-01 13:00:00 ora solare Europa occidentale" [2] "2001-01-02 13:00:00 ora solare Europa

The interface for dates in R is a little confusing to me. I want to create a vector of Date objects from vectors of years, months, and days. One solution I found is: years <- c(1991, 1992) months <- c(1, 10) days <- c(1, 2) dates <- as.Date(ISOdate(years, months, days)) But, in this solution the ISOdate function converts the vectors into characters, which can cause serious