similar to: write.csv feature/bug with pipe

I don't understand why I can not close 'f'. This may be very simple, but I don't see why. Could somebody let me know? $ cat gzfile.csv "","V1","V2","V3","V4","V5" "1",1,5,9,13,17 "2",2,6,10,14,18 "3",3,7,11,15,19 "4",4,8,12,16,20 $ Rscript gzfile.R > f =

I am experiencing some problems at working with lists at high level. In the following "coef" contains the original DWT coefficients organized in a list. Thorugh applying the following two commands: coef.abs <- lapply(unlist(coef,recursive=FALSE,use.names =TRUE),abs) coef.abs.sorted <- sort(unlist(coef.abs),decreasing=TRUE) I get vector "coef.abs.sorted" containing

Dear R wizards: is there a clean way to assign to elements in a list? what I would like to do, in pseudo R+perl notation is f <- function(a,b) list(a+b,a-b) (c,d) <- f(1,2) and have c be assigned 1+2 and d be assigned 1-2. right now, I use the clunky x <- f(1,2) c <- x[[1]] d <- x[[2]] rm(x) which seems awful. is there a nicer syntax? regards, /iaw ---- Ivo Welch

I would like to insert a few modest size data frames directly into my R code. a short illustration example of what I want is d <- read.csv( _END_, row.names=1 ) , "col1", "col2" "row1",1,2 "row2",3,4 __END__ right now, the data sits in external files. I could put each column into its own vector and then combine into a data frame, but this seems

Dear R-list members, I've had a hard time trying to solve a non-linear system (nls) of equations which structure for the equation i, i=1,...,4, is as follows: f_i(d_1,d_2,d_3,d_4)-k_i(l,m,s) = 0 (1) In the expression above, both f_i and k_i are known functions and l, m and s are known constants. I would like to estimate the vector d=(d_1,d_2,d_3,d_4) which is solution

dear r-experts: there is probably a very easy way to do it, but it eludes me right now. I have a large data frame with, say, 26 columns named "a" through "z". I would like to define "sets of regressors" from this data frame. something like myregressors=c("b", "j", "x") lm( l ~ myregressors, data=... ) is the best way to create new

dear R experts---I am using the coef() function to pick off the coefficients from an lm() object. alas, I also need the standard errors and I need them fast. I know I can do a "summary()" on the object and pick them off this way, but this computes other stuff I do not need. Or, I can compute (X' X)^(-1) s^2 myself. Has someone written a fast se() function? incidentally, I think

this is not an important question, but I wonder why lm returns an error, and whether this can be shut off. it would seem to me that returning NA's would make more sense in some cases---after all, the problem is clearly that coefficients cannot be computed. I know that I can trap the lm.fit() error---although I have always found this to be quite inconvenient---and this is easy if I have only

dear R experts---now I have a case where I want to estimate very large regression models with many fixed effects---not just the mean type, but cross-fixed effects---years, months, locations, firms. Many millions of observations, a few thousand variables (most of these variables are interaction fixed effects). could someone please point me to packages, if any, that would help me estimate such

Dear Sir, Please excuse my akwardness as I a new to R and computers, but would kindly appreciate help. { a <- sample (1:10,100,replace=T ) b <-sample(10:20,100,replace=T) c <- sample(20:30,100,replace=T) d <- sample(30:40,100,replace=T) e <- sample(40:50,100,replace=T) } d1 <- a d2 <- b d3 <-c d4 <- d d5 <- e data.frame(d1,d2,d3,d4,d5) dd <-

Hi Berend, I see you are one of the contributors to the rbecnhmark package. I am sorry that I am bothering you again. I have tried to run your code (slightly tweaked) involving the benchmark function, and I am getting the following error message. What am I doing wrong? Error in benchmark(d1 <- s1(df), d2 <- s2(df), d3 <- s3(df), d4 <- s4(df), : could not find function

Hi All I am trying to use the round function on some columns of a dataframe while leaving others unchanged. I wish to specify those columns to leave unchanged. My attempt is below - here, I would like the column d3 to be left but columns d1, d2 and d4 to be rounded to 0 decimal places. I would welcome any suggestions for a nicer way of doing this. d1= rnorm(10,10) d2= rnorm(10,6) d3=

Hello, The obvious way would be to preallocate the resulting data.frame, to expand an empty one on each iteration being a time expensive operation. n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D4 <- data.frame(distance=integer(n),difference=integer(n)) k <- 0 for (i in 1:nrow(D1)){ for (j in 1:nrow(D2)) { k <- k + 1 D4[k, ] <-

Hi R help, Hi R help, Which is the easiest (most elegant) way to force "aov" to treat numerical variables as categorical ? Sincerely, Andrea Bernasconi DG PROBLEM EXAMPLE I consider the latin squares example described at page 157 of the book: Statistics for Experimenters: Design, Innovation, and Discovery by George E. P. Box, J. Stuart Hunter, William G. Hunter. This example use

I am new to R and have a problem that I haven't been able to find the answer to in the guides or online. I have multiple datasets, D1, D2, D3, D4, D5, D6, D7 and D8, and I would like to produce two plots side by side using mfrow.  The first plot should contain data from D1-D4, the second should contain D5-D8. I can plot these separately using the code, par(mfrow=c(1,1))

Hi, I am trying to aggregate the sum of my test data.frame as follow: testDF <- data.frame(v1 = c("a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "c", "c", "c", "c", "c", "d", "d", "d", "d",

Hello, Another way would be n <- nrow(expand.grid(1:nrow(D1), 1:nrow(D2))) D5 <- data.frame(distance=integer(n),difference=integer(n)) D5[] <- do.call(rbind, lapply(seq_len(nrow(D1)), function(i) t(sapply(seq_len(nrow(D2)), function(j){ c(distance=sqrt(sum((D1[i,1:2]-D2[j,1:2])^2)),difference=(D1[i,3]-D2[j,3])^2) } )))) identical(D3, D5) In my first answer I forgot to say that

For certain reason, the content was not visible in the last mail, so posting it again. Dear Members, I have two different dataframes with a different number of rows. I need to apply a set of functions to each possible combination of rows with one row coming from 1st dataframe and other from 2nd dataframe. Though I am able to perform this task using for loops, I feel that there must be a more

Hello, I have a longitudinal dataset where each individual has a different number of entries. Thus, it is of the following structure: x <- runif(12) id.var <- factor(c(rep("D1",4),rep("D2",2),rep("D3",3),rep("D4",3))) dat <- as.data.frame(x) dat$id.var <- id.var dat > dat x id.var 1 0.9611269 D1 2 0.6738606 D1 3

Dear R experts--- after many years, I am planning to give in and write my first R package. I want to combine my collection of collected useful utility routines. as my guide, I am planning to use Friedrich Leisch's "Creating R Packages: A Tutorial" from Sep 2009. Is there a newer or better tutorial? this one is 4 years old. I also plan on one change---given that the