similar to: expand.grid applied to a matrix

<<insert bug report here>> > apply(array(1:20,c(2,2,5)),2:3,function(x) x) Error: length of dimnames must match that of dims > Changing: dimnames = if (is.null(dn.ans)) list(ans.names, NULL) else c(list(ans.names), dn.ans) To: dimnames = if (length(dn)==0) NULL else if (is.null(dn.ans)) list(ans.names, NULL) else c(list(ans.names), dn.ans) seems to fix this. Chuck

kronecker, with make.dimnames=TRUE uses a hardwired sep=":" in the line tmp <- outer(dnx[[i]], dny[[i]], FUN = "paste", sep = ":") For an application in which dimnames arise from an n-way array, where different dimensions have different roles, and I would like to be able to use kronecker in the form kronecker(A, B, make.dimnames=TRUE,

Dear R-users, I am looking for a way to assign to slices of arrays where dimensionality of the array is not a-priory known. Specifically, I would like to be able to generalize the following example of dimensionality 2 to an arbitrary diminsionality: In this example we create an array x, a smaller array y and then assign y to a slice of x. > dimnmx <- list(c('a','b'),

hello, i have been trying to convert my data frames to matrices in the hopes of speeding up some of my more complicated scripts. to assist with this, i am trying to create a "matrix column operator" like $: "%$%" = function(data,field) { as.numeric(data[,grep(field,unlist(dimnames(data)[2]))]) } the idea here is that you can use a matrix like a dataframe:

Hi, when I pick out one element from a matrix, the attribute name is kept, but when more than one elements are extracted, the attribute name lost;   >a<-matrix(c(1,2,3,11,12,13,45,56,76),ncol=3,dimnames=list(c(),c("c1","c2","c3"))) > k<-a[a[,"c3"]>50,"c3"] > kk<-a[a[,"c3"]>60,"c3"] > attributes(k) NULL

Ok, so I'm slowly figuring out what a factor is, and was able to follow the related thread about finding a mode by using constructs like my_mode = as.numeric(names(table(x))[which.max(table(x))]) Now, suppose I want to keep looking for other modes? For example, Rgames> sample(seq(1,10),50,replace=TRUE)->bag Rgames> bag [1] 2 8 8 10 7 3 2 9 8 3 8 9 6 6 10 10 7 1

How can I see more of the structure than displayed by 'str'? Consider the following: tstDF <- data.frame(a=1, row.names='b') > str(tstDF) 'data.frame': 1 obs. of 1 variable: $ a: num 1 The object 'tstDF' has row.names, but I have to suspect they are there -- AND know a function like 'row.names' or 'dimnames' -- to see

Hello, All, I have a dataset that looks like this: x <- matrix(c( 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1), ncol = 5, byrow = T, dimnames = list(1:10, c("gender", "race", "disease"))) I want to write a function to produce several matrices including only ?TRUE? and ?FALSE? for the different levels of the

Hi Im pretty new to R and I have run in to a problem. How do I name all the levels in this list. Lev1 <- c("A1","A2") Lev2 <- c("B1","B2") Lev3 <- c("C1","C2") MyList <- lapply(Lev1,function(x){ lapply(Lev2,function(y){ lapply(Lev3,function(z){ paste(unlist(x),unlist(y),unlist(z)) })})}) I would like to name the different

This is a follow-up on my note of Saturday. Let me start with two important clarifications - I think this would be a nice addition, but I've had exactly one use for it in the 15+ years of developing the survival package. - I have a work around for the current case. Prioritize accordingly. The ideal would be to change survexp as follows: fit <- survexp( ~ gender,

I have been through the help file archives a number of times, and still cannot figure out what is wrong. I have a tab-delimited text file. 76Mb, so while it's large.. it's not -that- large. I'm running Win7 x64 w/4G RAM and R 2.10.1 When I open this data in Excel, i have 27 rows and 450932 rows, excluding the first row containing variable names. I am trying to get this into R as a

Dear R gurus, to compute the correlation matrix of "n" variables with "n_obs" observations each, possibly including NA, I use cor(M, use="pairwise.obs") where m is a "n" x "nobs" matrix. Now I want to know the number of observations actually used in this computation, namely for each pair of columns in M, say pair (i,j), I want to compute sum(

Dear community, I had been looking for an easy way to produce latex tables from R output. xtable() and the package apsrtable produce good outputs but they are not exactly what I was looking for. I wrote this code that generates regression tables from multiple R linear models. I want to share it because it might be useful for someone else, and because I would appreciate comments on how to

I wanted to print the first and last rows of some dataframes in Sweave using dots in columns to separate the two parts. Head and tail almost work, but I have problems with factors and row names. z<-data.frame(id=letters[1:26], x=sample(1:26,26)) rbind(head(z,3), ".", tail(z,1)) id x 1 a 18 2 b 8 3 c 14 4 <NA> . 26 z 10 Warning message: invalid

Dear Happy R-users & experts, I am in need of advice, While working with spatial data (x & y coordinates of seed locations) I have come accross the problem that I need to convert my point data into a matrix or grid system. I then need to count how often a point falls into a certain position in the matrix or grid. I have searched all day online, asked collegeas but nothing works. Sadly

Hi all, I was wondering whether it is possible to use the lapply() function to alter the value of the input, something in the spirit of : a1<-runif(100) a2<-function(i){ a1[i]<-a1[i-1]*a1[i];a1[i] } a3<-lapply(2:100,a2) Something akin to a for() loop, but using the lapply() infrastructure. I haven't been able to get rapply() to do this. The reason is that the "real"

Hi there, I have a binary matrix (dim 100x100) filled with values 0 and 1. I need select a record "n" positions of that matrix when values are 1. How can I do that? Thanks for all, Miltinho Brazil --------------------------------- [[alternative HTML version deleted]]

Is this a bug? > matrix(1:4,nc=2,dimnames=list(1:2,1:2))[c(1,3)] 1 NA 1 3 It has some annoying consequences, e.g. > median( matrix(1:9,nc=3,dimnames=list(1:3,1:3))[1,,drop=F] ) NA 4 > The above was for Version: platform = sparc-sun-solaris2.7 arch = sparc os = solaris2.7 system = sparc, solaris2.7 status = major = 1 minor = 3.0 year = 2001 month = 06 day = 22

Hi there, I run two regressions: y = a1 + b1 * x + e1 y = a2 + b2 * z + e2 I want to test against the null hypothesis: b1 = b2. How do I design the test? I think I can add two equations together and divide both sides by 2: y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2). or just y = a3 + 0.5*b1 * x + 0.5*b2 * z + e3 If I run this new regression, I can test against

Hello, I am trying to figure out how to use optimize() with array variables as inputs. I have a for loop in the function definition: SS <- function(int,slo,x,y){ for(i in 1:length(x)) ((int+slo*x[i])-y[i])^2->squares[i] sum(squares)->>sum_squares output_txt = c ("The sum of squares is", sum_squares) print(output_txt, quote=FALSE)} Even assuming I make x and y