similar to: Putting a loop in a function

Hello, What about an `invert` argument in grep, to return elements that are *not* matching a regular expression : R> grep("pink", colors(), invert = TRUE, value = TRUE) would essentially return the same as : R> colors() [ - grep("pink", colors()) ] I'm attaching the files that I modified (against today's tarball) for that purpose. Cheers, Romain --

Hello All, I'd like to read first words in lines into a new file. If I have a data file the following, how can I get the first words: apple, banana, strawberry? i1-apple 10$ New_York i2-banana 5$ London i3-strawberry 7$ Japan Is there any similar question already posted to the list? I am a bit new to R, having a few months of experience now. Cheers, John

Hi, I am trying to use sub, regexpr on expressions like log(D) ~ log(N)+I(log(N)^2)+log(t) being a model specification. The aim is to produce: "ln D ~ ln N + ln^2 N + ln t" The variable names N, t may change, the number of terms too. I succeded only partially, help on regular expressions is hard to understand for me, examples on my case are rare. The help page on R-help

hello everyone. Look at the following R idiom: a <- array(1:30,c(3,5,2)) M <- (matrix(1:15,c(3,5)) %% 4) < 2 a[M,] <- 0 Now, I think that "a[M,]" has an unambiguous meaning (to a human). However, the last line doesn't work as desired, but I expected it to...and it recently took me an indecent amount of time to debug an analogous case. Just to be explicit, I would

Hi, I have a correlation matrix of about 3000 items, i.e., a 3000*3000 matrix. For each of the 3000 items, I want to get the top 50 items that have the highest correlation with it (excluding itself) and generate a data frame with 3 columns like ("ID", "ID2", "cor"), where ID is those 3000 items each repeat 50 times, and ID2 is the top 50 correlated items with ID,

Hi Everyone, quick question before the end of the year. I have soem indices to select data from a bigger sample. I want to select n days before each index and n days after the index. Any clever way to do it. A for loop would do but I wanted to know if there is a moreR-friendly way to approach this Example # InitialIndices i2 = (90, 190, 290) # Indices I want to end up with i3 = c(85, 86, 87,

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Hi, I noticed that by() returns an object of class 'by', regardless of what its argument 'simplify' is. ?by says that it always returns a list if simplify=FALSE, yet by.data.frame shows: ---<--------------------cut here---------------start------------------->--- function (data, INDICES, FUN, ..., simplify = TRUE) { if (!is.list(INDICES)) { IND <-

Hi, I cannot go through the archives with which() as key-word... so common. Though I am sure to have seen something about this subject in the past could somebody put me on the track. I have a matrix (actually a data.frame) in which I would replace the non-null values by 1. I tried the following: indices<-which(myforetbin > 0,arr.ind=T) myforetbin[indices[,1],indices[,2]]<-1 and get

Dear R-users, I was wondering for the following: Let 'x' be a matrix and 'ind' and indicator matrix, i.e., x <- array(1:20, dim = c(4, 5)) ind <- array(c(1:3, 3:1), dim = c(3, 2)) I'd like to get (as a vector) the elements of 'x' which are not indexed by 'ind'. Since negative indices are not allowed in index matrices I thought of using something like:

summary: The function 'by' inconsistently strips class from the data to which it is applied. quick reason: tapply strips class when simplify is set to TRUE (the default) due to the class stripping behaviour of unlist. quick answer: This can be fixed by invoking tapply with simplify=FALSE, or changing tapply to use do.call(c instead of unlist executable example:

Dear list, As a minimal test of a more complex grid layout, I'm trying to find a clean and efficient way to arrange text grobs in a rectangular layout. The labels may be expressions, or text with a fontsize different of the default, which means that the cell sizes should probably be calculated using grobWidth() and grobHeight() as opposed to simpler stringWidth() and stringHeight().

I have a very large dataframe and wish to extract a subset of rows. I have a two column matrix listing the starting and ending indices of one subset on each row. My idea is to create a vector of indices that could be applied to the dataframe and I have a solution using a for loop (below). But surely there is some more elegant way to do this! I looked thorough the archives without

Consider the following little "benchmark" > require(Matrix) > tmp <- Matrix(c(rep(1,1000),rep(0,9000)),ncol=1) > ind <- sample(1:10000,10000) > system.time(tmp[ind,]) user system elapsed 0.004 0.001 0.005 > ind <- sample(1:1000,10000,replace=TRUE) > system.time(tmp[ind,]) user system elapsed 0.654 0.006 0.703 >

Hi, Jean-Marc Valin wrote: > Not sure I understand your question. Change the order of what within what? In cb_search, we iterate nb_subvect times over a codebook, finding nb_subvect codebook entries to quantize the excitation signal. After finding these nb_subvect codebook entries, they're written into the stream: /*save indices*/ for (i=0;i<nb_subvect;i++) {

by() fails for 1-column matrices and dataframes: X <- data.frame(a=1:10) g <- gl(2,5) by(X, g, colMeans) Suggested fix: --- by-old.R 2007-12-10 15:26:22.501086600 +0100 +++ by.R 2007-12-10 15:25:58.390477200 +0100 @@ -26,7 +26,7 @@ IND[[1]] <- INDICES names(IND) <- deparse(substitute(INDICES))[1] } else IND <- INDICES - FUNx <-

I have a matrix for which each row has 12 elements that represent the xyz coordinates of 4 points. So each row of M is (x1,y1,z1,x2,y2,z2,x3,y3,z3,x4,y4,z4). Some points have NA as z values. I want another matrix to be the same as M but with the coordinates of those points with z=NA placed last. For ezample if z1=NA then the new matrix row should be (x2,y2,z2,x3,y3,z3,x4,y4,z4,x1,y1,z1)

I need to split a data frame into 3 columns. The column I want to split contains indices of lag (prefix L1 or L2 and suffix 01, 03, 04), station name (shown in the sample data as capitalized G, P and S) and pollutant name. Names with no ?L? prefix or 01/04 suffix are lag 0. Lag 01 is average of lag 0 and 1, and 04 is average of 0 to 4 days. How can one do that in R? I will ignore the other

Hi, I would like to replicate a sort of Montecarlo experiment: I have to generate a random variable N(0,1) with 100 observations but I have to contaminate this values at certain point in order to obtain different vectors with different samples: tab<-function(N,n,E,L){ for(i in 1:100){ X1<-rnorm(N*(1-a),0,1) X2<-rnorm(N*(a),E,L) X_tab<-sample(c(X1,X2),n,replace=T) } return(X_tab)}

Rolf turner wrote: >I have been trying, unsuccessfully, to use identify() to (simply) >return a list of the indices of points clicked on and overplot (with >say a solid dot) each clicked-on point so that I can see where I've >been. I.e. I don't want to see the indices printed on the screen; I just want the points I've already selected to be highlighted. > >I