similar to: extract week from date

Hi, I have a function which takes some time to execute (20 sec.) and I want to show the progress of the execution with simple messages such as "10% done", "20% done", ... (with a simple "cat" function). The problem is that all the messages appear together when the execution of the function is over. How can I show each message at their corresponding time? Thanks,

Hello listers, I have installed Tcl/Tk v8.4 on my system (Win XP, R 1.5.1). When I try to load the Tcl/Tk package I get this error: > library(tcltk) Error in firstlib(which.lib.loc, package) : TCL_LIBRARY is not set Error in library(tcltk) : .First.lib failed Can someone help me, please? Thanks, Juan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

Hi, Does anyone know if exists a similar funcion to %in% from Hmisc S-plus library in R? Thank you -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- 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 subject !) To: r-help-request at

Hello R-users, I need to create a vector inserting an 1 after each value of another vector. For example: vec1<-c(2,3,4) I need to create a vector with the values 2,1,3,1,4 Does anyone know how create this vector without loops (vec1 could have 1000 elements) Thank you, Juan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear R-listers, Does anybody know what is the correct source of "rats" dataset in survival package? The help gives the following information: Rat data from survival5 Description: 48 rats were injected with a carcinogen, and then randomized to either drug or placebo. The number of tumors ranges from 0 to 13; all rats were censored at 6 months after randomization.

Dear R-listers, I have created a "dll" and I call it from a R function. However R stops and shows the message: "Unhandled exception in Rgui.exe (R.DLL): 0xC0000005: Access Violation" I get this error in other situations and I solved it verifying that all parameters are called fine. (e.g., double precision -> as.double, integer -> as.integer,...). I have verify this.

Hello listers, I would like to know how I can get the resulting graphic of the function plot.hclust (from the package cluster) in landscape orientation. Is it possible? Thanks, Juan

Hello R-users, I'm using the boot's library. I have two questions: 1. I execute for example data<-rnorm(100) boot(data,mean,500) and I obtain 500 equal bootstrap samples of data. Why? 2. I need to calculate the bootstrap confidence intervals for a statistic, but when I use the function boot.ci it shows the error message: Error in if

Hello R-listers, I am developing a function which needs to create jpeg files from some plots. To create the jpeg files I use the instruction dev.print(jpeg, file=c://graph.jpg",width=400,height=400). Until now everything has been fine, but now I have a special plot which is formed by 3 separate plots joined with the instruction "layout": nf <-

Hello, I have made a dll with mingw32 fortran compiler. I load my dll with the dyn.load function, but when I call one of my Fortran procedures with the .Fortran instruction, an error message appears saying that my function is not in the load table. However, if I call my Fortran routines with ".C" it works fine. I would like to know what is happening here. Is it usual to call Fortran

# hi all, # I'm trying to plot temperatures by date in a trellis plot by their stations # I'm plotting the following data.frame library(lattice) h <- structure(list(station_name = structure(c(3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L, 5L, 8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L, 5L, 8L, 9L, 3L, 4L, 2L, 10L, 11L, 12L, 6L, 7L), .Label = c("Ashqelon", "Beer Sheva",

Dear R People: I'm trying to convert a daily zoo object to a weekly zoo object: xdate <- seq(as.Date("2002-01-01"),as.Date("2010-07-10"),by="day") library(zoo) length(xdate) xt <- zoo(rnorm(3113),order=xdate) xdat2 <- seq(index(xt)[1],index(xt)[3113],by="week") xt.w <- aggregate(xt,by=xdat2,mean) Error: length(time(x)) ==

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

Hi, I have the following type of data: myData <- data.frame(x = 1:5, y = letters[1:5], xDate = seq(as.Date("2001/2/1"), as.Date("2005/2/1"), by="year") ) > myData x y xDate 1 a 2001-02-01 2 b 2002-02-01 3 c 2003-02-01 4 d 2004-02-01 5 e 2005-02-01 What I need is a new column, say xDate2, that for each xDate (or for each unique combination of

Dear R People: Now I have a function with 3 graphics windows. I'd like for them all to be visible. However, the first window goes blank. Here is the basic setup. The first plot is the entire series. The user selects two points, which in turns generates a subset plot and an EWMA subset plot. Finally, if things worked right, the user should be able to click on "Tile Vertically"

Hi, I have some confusion in applying a function over a column. Here's my function. I just need to shift non-March month-ends to March month-ends. Initially I tried seq.dates, but one cannot give a negative increment (decrement) here. return(as.Date(seq.dates(format(xdate,"%m/%d/%Y"),by="months",len=4)[4]) ) Hence this simple function: > mydate <-

Javier, no s? si es esto lo que buscas. Te copio una l?nea de uno de mis scripts. En el eje X, van los a?os desde 1990 a 2022, pero solo salen los a?os 90, 95, ... y el resto solo la marca secundaria sin el a?o scale_x_continuous(breaks = seq(1990, 2022, by = 1), labels = c(1990, rep("", 4), 1995, rep("", 4), 2000, rep("", 4),

Hola, Como te comentan, shiny es la mejor opción que tienes, solo añadir que no necesitas Rstudio connect. Se puede hacer el volcado de datos con conexiones ODBC perfectamente, hay varios paquetes de R que te sirven para ello. Lo que si es cierto es que necesitarás dedicarle un tiempo a cómo utilizarlos de manera razonable, probableente tendrás que pelear un poco para configurarlos correctamente.

Hola: Por si no os ha llegado por otras vías, os informo que se están gestando las próximas (¡las 4as!) Jornadas de Usuarios de R en Castellano. Ved el mensaje más abajo de Juan Ramón González. En breve, los que queráis echar una mano, hacer sugerencias, etc. , podéis hacerlo en el foro que se ha creado en la web de la comunidad (hay gente que no está - aún - en esta lista, etc):