similar to: R process slow down after a amount of time

Hi R users, is there any possibilty that a while loop is working like that: z <- c(0,1,2,3,4,5,6,7,8,9) r <- 7 while(w == T) { for ( i in 1:10 ){ w <- r == z[i] print(w) } } The loop should stop if w == TRUE best regards -- View this message in context:

Dear members, I have a series of values in a vector and some value are missing and replaced with NA. For example: a <- c(27, 25, NA, NA, 24, 26, 27, NA, 26) I would like to replace the NAs with the value taken from the previous value that is non-NA. The output would be in this case: 27 25 25 25 24 26 27 27 26 Now I do that with a for loop, but I try to eliminate all the loops to gain in

hello R users, I didn't find a solution for a special problem. I have two dataframes. dataframe1: X value row col ID 1 8.973498062 5512625 3460000 1 2 11.656658570 5501625 3464000 2 3 11.121777570 5495625 3473000 3 4 9.310465964 5508625 3477000 4 5 8.883483845 5515625 3496000 5 dataframe2: X value

Dear all For a model I need to combine several netCDF files into one (which works fine). For better overview I'd like to delete/remove some of the attributes. Is there a simple way doing this? I'm using the package netcdf4, which creates an object of class(nc) = "ncdf4". It seems that for earlier versions of netcdf objects, there was the function att.delete.nc{RNetCDF}. But

Hello R-users, I have a little problem. I compare each row of a matrix with each row of another matrix. testmat1 <- matrix(c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16), nrow=4) testmat2 <- matrix(c(1,2,3,5,5,6,7,8,9,10,11,12,13,14,15,16), nrow=4) Both matrix differs in the last row. Now I create a loop: for (i in (1:4)){ for (j in (1:4)){ b <- (c(setequal(testmat1[j,],testmat2[i,])))

Hi R users, I want to colored points by their value for example: x <- c(1,2,3,4) y <- c(1,2,3,4) z <- c(2,3,4,9) y and x are coordinates z is the value of the coordinates points(x,y,col= rainbow(z)) something like that But haven't found any solution at the moment. Thanks. Chris -- View this message in context:

Hello, I don't know if the problem originates from R-devel 3.2 or Rcurl itself. I post this message to the R-devel list and to the author of RCurl (duncan at r-project.org). > library("RCurl") Le chargement a n?cessit? le package : bitops > print(sessionInfo()) R Under development (unstable) (2015-02-03 r67717) Platform: x86_64-apple-darwin13.4.0 (64-bit) Running under:

I define priors in jags within r using a gamma distribution. I would like to control the shape but I have problems. Any help will be usefull. From help of dgamma ___________________ The Gamma distribution with parameters shape = a and scale = s has density f(x)= 1/(s^a Gamma(a)) x^(a-1) e^-(x/s) and rate=1/scale From jags user manual ____________________ dgamma(r, mu) has a density of

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Hi, I'm new to R. I have some daily soil moisture data for the year 1979 in netcdf format such as these sm19790101.1.nc sm19790102.1.nc . . . sm19791231.1.nc I need to average a variable called "sm" to monthly resolution. I've done these days = formatC(1:31, width=2, flag="0") ncfiles = lapply(days, function(d){ filename = paste("sm197901", d,

Hi, all I have a medium sized data, 6 years. Each observation is a case with a date variable, such as '2004-08-02'. Some of the months didn't occur a case. I want to plot the 6 years data by month, and the Y_axis is the freqency of cases for each month, meaning 12*6=72 bars or points in the figure. I though of a method, 1st, using the months function, then ploting. But I need to

Hi, I try to make my first package? The HelloWorld.R file is: #### HelloWorld.R #### #' showHello est une fonction R permettant d'afficher le message #' "Hello World!" sur la console. #' @title la fonction showHello() showHello <-function(){ cat("Hello World!\n") } I use the following procedure to get the tar: # set the working directory where the file is

When ploting a timeseries, the months are shown with abbreviation in the current language of the system. For example, x <- seq(from=as.Date("2000-04-01"), length.out=100, by=1) y <- rnorm(length(x), 5,2) plot(x, y) Show for me "avi mai jui jul" as I use R with French language localization. I see in the Windows FAQ how to completely change the language of R, but I

Dear all, I was wondering how to read DNA sequences in R, is there a specific function and/or a specific package for that? Thank you very much in advance, Gian ** [[alternative HTML version deleted]]

Hello all. I am using method "bag()" to develop a bagging with my dataset. When I do bag( dataP, dataP$Score, B=10) I obtain this error: error en bag.default(Score ~ ., data = dataP, B = 10) : entrada en evaluacion: recursivo por defecto o problemas anteriores? ( in english, evaluating input: default recursive or previous problems?) Any idea? Thanks! [[alternative HTML

HI, I have the pattern like "`Year_Month)201103`" I want to delete single quotes within double quetes. I want to change it into "Year_Month)201103" , how to do it in r? Thanks a lot. Tammy [[alternative HTML version deleted]]

how can i draw the graph in R?

I want to fit a multinomial model with logit link. For example let this matrix to be analyzed: male female aborted factor 10 12 1 1.2 14 14 4 1.3 15 12 3 1.4 (this is an example, not the true data which are far more complex...) I suppose the correct function to analyze these data is polr from MASS library. The data have been

I use R 2.15.2 on MacosX 10.8.2 I have a problem checking for a package that I want to submit in CRAN. When I check like that, no problem or warning are detected. system("R CMD check '[path]/phenology'") I create the .tar.gz package with that. The file phenology_3.43.tar.gz is generated. I can install the package and it works. system("R CMD build

Hello R users, I want to readout the row and column postion from a certain matrix value into a csv file. I have only found this syntax "which(a == b, arr.ind = T)" so I get a = matrix [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1 1 2 2 2 1 1 2 [2,] 1 2 3 3 3 4 4 3 [3,] 3 3 3 5 5 6 5 4 [4,] 4 4 4 3