similar to: why the dim gave me different results

HI, Dear R community, I have one data set like this, What I want to do is to calculate the cumulative coverage. The following codes works for small data set (#rows = 100), but when feed the whole data set, it still running after 24 hours. Can someone give some suggestions for long vector? id reads Contig79:1 4 Contig79:2 8 Contig79:3 13 Contig79:4 14 Contig79:5 17

Hi, Dear R- community, I am use the intToChar function to convert the integers to letters. But the output is mess. Can you guys give some suggestions? Thanks! > outcome.predict [1] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 7 4 4 4 4 4 4 4 4 [26] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 7 4 4 4 4 4 7 4 4 4 4 [51] 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

HI, Dear R community, I am writing the following function to create one data set(*tree.pred*) and one vector(*valid.out*) from loops. Later, I want to use the data set from this loop to plot curves. I have tried return, list, but I can not use the *tree.pred* data and *valid.out* vector. auc.tree<- function(msplit,mbucket) { * tree.pred<-data.frame()

HI, Dear R community, I want to split a data frame by using two variables: let and g > x = data.frame(num = c(10,11,12,43,23,14,52,52,12,23,21,23,32,31,24,45,56,56,76,45), let = letters[1:5], g = 1:2) > x num let g 1 10 a 1 2 11 b 2 3 12 c 1 4 43 d 2 5 23 e 1 6 14 a 2 7 52 b 1 8 52 c 2 9 12 d 1 10 23 e 2 11 21 a 1 12 23 b 2 13 32 c 1 14

HI, GUYS, I used the following codes to run SVM and get prediction on new data set hh. dim(all_h) [1] 2034 24 dim(hh) # it contains all the variables besides the variables in all_h data set. [1] 640 415 require(e1071) svm.tune<-tune(svm, as.factor(out) ~ ., data=all_h, ranges=list(gamma=2^(-5:5), cost=2^(-5:5)))# find the best parameters. bestg<-svm.tune$best.parameters[[1]]

HI, Dear R community, My original file has 1932 lines, but when I read into R, it changed to 1068 lines, how comes? cdu@nuuk:~/operon$ wc -l id_name_gh5.txt 1932 id_name_gh5.txt > gene_name<-read.table("/home/cdu/operon/id_name_gh5.txt", sep="\t", skip=0, header=F, fill=T) > dim(gene_name) [1] 1068 3 -- Sincerely, Changbin -- Changbin Du DOE Joint Genome

Hi I'm using kronecker() with a matrix and a vector. I'm interested in the column names that kronecker() returns: > a <- matrix(1:9,3,3) > rownames(a) <- letters[1:3] > colnames(a) <- LETTERS[1:3] > b <- c(x=1,y=2) > kronecker(a,b,make.dimnames=TRUE) A: B: C: a:x 1 4 7 a:y 2 8 14 b:x 2 5 8 b:y 4 10 16 c:x 3 6 9 c:y 6 12 18 > The

Dear R gang, Can anyone help me sort out how to append one data frame to another while adding a factor to distinguish which was the original frame? For example, I have two frames, x and y > x exp size 1 a 10 2 b 9 3 c 10 4 d 12 5 e 11 > y exp size 1 a 13 2 b 15 3 c 12 4 d 20 5 e 15 and I'd like to create a new frame that looks like

HI, Dear R community, How to extract the variables actually used in tree construction? I want to extract these variables and combine other variable as my features in next step model building. > printcp(fit.dimer) Classification tree: rpart(formula = outcome ~ ., data = p_df, method = "class") Variables actually used in tree construction: [1] CT DP DY FC NE NW QT SK TA WC WD WG WW

Or just dim(x) <- NULL. (as matrices in base R are just vectors with a dim attribute stored in column major order) ergo: > x [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 > x<- 1:20 ## a vector > is.matrix(x) [1] FALSE > dim(x) <- c(5,4) > is.matrix(x) [1] TRUE > attributes(x) $dim [1] 5 4 > ## in painful and unnecessary detail as dim() should

Hi All, How do i calculate KMO for a dataset? *Dataset:---------------------* m1 m2 m3 m4 m5 m6 m7 m8 1 2 20 20 2 1 4 14 12 2 9 16 3 5 2 5 5 15 3 18 18 18 13 17 9 2 4 4 7 7 2 12 2 11 11 11 5 7 8 5 19 5 2 20 18 6 7 4 7 4 7 9 3 3 7 5 5 5 12 5 13 13 12 8 6 6 4 3 5 17 17 16 9 12 12 4 2 4 4 14 14 10 5 14

Hi list, I am looking at the data yarn in package, I don't understand what is dimension of this data set. I did the following: > library(pls) > data(yarn) > dim(yarn) [1] 28 3 > head(yarn) NIR.1 NIR.2 NIR.3 NIR.4 NIR.5 NIR.6 NIR.7 NIR.8 NIR.9 NIR.10 NIR.11 1 3.06630 3.08610 3.10790 3.09720 2.99790 2.82730 2.62330 2.40390 2.19310 2.00580 1.83790 2

I am writing a function in fortran 95, but the intrinsic function MATMUL is not working properly. Here's an example. SUBROUTINE mymult(x,y,res,m,n) IMPLICIT NONE INTEGER :: m,n REAL :: x, y, res DIMENSION :: x(m,n), y(n,m), res(m,m) res = MATMUL(x,y) END SUBROUTINE mymult R CMD SHLIB mat.f95 In R: dyn.load("mat.so") x <-

You could also do dim(x) <- c(length(x), 1) On Sat, Aug 5, 2023, 20:12 Steven Yen <styen at ntu.edu.tw> wrote: > I wish to stack columns of a matrix into one column. The following > matrix command does it. Any other ways? Thanks. > > > x<-matrix(1:20,5,4) > > x > [,1] [,2] [,3] [,4] > [1,] 1 6 11 16 > [2,] 2 7 12 17 > [3,]

Hello R-users, I'd like to load data from a CSV-file to a 3-dimensional array. However the default seems to be 2-dimensional. It would also be ok to load the data and then convert it to a 3-dimensional structure. I've been trying: dat = read.csv(filename) myArr = as.array(dat, dim = c(12,100,3)) However, I get an error message: Error in `dimnames<-.data.frame`(`*tmp*`,

Stacking columns of a matrix is a standard operation in multilinear algebra, usually written as the operator vec(). I checked to see if there is an R package that deals with multilinear algebra. I found rTensor, which has a function vec(). So, yet another way to accomplish what you want would be: > library(rTensor) > vec(as.tensor(x)) Eric On Sun, Aug 6, 2023 at 5:05?AM Bert Gunter

Hello, I can't get Kerberos authentication works with my Linux clients. Server : samba 4.1.4 (compiled from source) Client : Debian Wheezy with sernet-samba 4.0.17-8 Without Kerberos authentication, everything works : -> the domain users can log with pam_winbind (with ssh, gdm ....). -> "kinit myuser at MYREALM" works fine. -> "wbinfo -K MYDOM\\myuser" works.

HI, Dear R community, My data set has 2409 variables, the last one is response variable. I have used the nnet after feature selection and works. But this time, I am using nnet to fit a model without feature selection. I got the following error information: > dim(train) [1] 1827 2409 nnet.fit<-nnet(as.factor(out) ~ ., data=train, size=3, rang=0.3, decay=5e-4, maxit=500) # model

Dear all, I have a problem that may be someone of you can help. I am a newbie and do not find how to do it in manuals. I have two arrays, for example: ar1 <- array(data=c(1:16),dim=c(4,4)) ar2 <- array(data=c(1:16),dim=c(4,4)) colnames(ar1)<-c("A","B","D","E") colnames(ar2)<-c("C","A","E","B") > ar1

OK guys another problem. I have a 3D array "x" with dim(x)=c(a,a,b^2) and I want to rearrange the elements of x to make a matrix "y" with dimensions c(a*b,a*b). Neither a nor b is known in advance. I want the "n-th" a*a submatrix of y to be x[,,n] (where 1 <= n <= b^2). Needless to say, this has gotta be vectorized! Toy example with a=2, b=3 follows: