similar to: ROWNAMES error message

Displaying 20 results from an estimated 20000 matches similar to: "ROWNAMES error message"

2008 Mar 25
2
help with rowsum/aggregate type functions
Hi-- This is a question with a trivial and obvious answer, I'm sure, but I can't seem to find it in the help files and books that I have handy. I have a dataframe consisting of two columns, "Gene_Name," a list of gene symbols, and "Number," a numeric measure of how frequently a tag representing that gene showed up in a SAGE library. Several of the genes are
2004 Aug 20
3
Loss of rownames and colnames
Hi, I am working on some microarray data, and have some problems with writing iterations. In essence, the problem is that objects with three dimensions don't have rownames and colnames. These colnames and rownames would otherwise still be there in 2 dimensional objects. I need to generate multiple iterations of a 2 means-clustering algorithm, and these objects thus probably need 3
2004 May 13
2
xtable without rownames
When I tried to read all the entries (after searching the FAQ) for "row names xtable", I get START-INFO-DIR-ENTRY * R FAQ: (R-FAQ). The R statistical system ... ... xtable* Export data to LaTeX and HTML tables. ... For dropping the row names of a matrix `x', it may be easier to use `rownames(x) <- NULL', similarly for column ...
2010 Feb 22
1
rownames cannot allocate vector of size
Hi, On R 2.10.1 for Windows, when I do the following to duplicate the structure of a large numeric matrix called matrix1: matrix2 <- matrix(0,nrow=nrow(matrix1),ncol=ncol(matrix1)) and then rownames(matrix2) <- rownames(matrix1) I get a "cannot allocate vector of size xxMb" error but if I instead do: rnames <- list() rnames <- rownames(matrix1) matrix2 <-
2005 Jul 07
1
Plotting Character Variable
Any ideas about the following problem: I have a matrix (A) that looks like this: gene_names values hsa-mir-124 0.3 hsa-mir-234 0.1 hsa-mir-344 0.4 hsa-mir-333 0.7 ..... ....... (This is a 2 by 22283 matrix: quite large) I would like to plot the values, but output the gene_names as the plotting symbol. I have tried
2011 Apr 12
1
extract element from list by rownames
Hi, I've a list of list. I want to extract an element by the rownames. I can extract it by: data[[1]][[1]][[4]][1] But I want to exctract it by a command like this: data[[1]][["B0"]][["smac"]][["cont"]][1] It's possible? Thanks, Alfredo > str(data) List of 1 $ :List of 4 ..$ :List of 4 .. ..$ : num [1, 1:3] 0.4 0.458 0.5 .. ..$ : num
2006 Mar 21
1
rownames, colnames, and date and time
I noticed something surprising (in R 2.2.1 on WinXP) According to the documentation, rownames and colnames are character vectors. Assigning a vector of class POSIXct or POSIXlt as rownames or colnames therefore is not strictly according to the rules. In some cases, R performs a reasonable typecast, but in some other cases where the same typecast also would be possible, it does not. Assigning a
2006 Mar 21
1
rownames, colnames, and date and time
I noticed something surprising (in R 2.2.1 on WinXP) According to the documentation, rownames and colnames are character vectors. Assigning a vector of class POSIXct or POSIXlt as rownames or colnames therefore is not strictly according to the rules. In some cases, R performs a reasonable typecast, but in some other cases where the same typecast also would be possible, it does not. Assigning a
2006 Sep 14
1
R-devel: rownames of a data.frame
Hello! Data.frames have new rownames funcionality, however in use of colnames<- in R-devel "changes" this. Here is the example: > df1 <- data.frame(letters[1:5]) > attributes(df1) $names [1] "letters.1.5." $row.names [1] 1 2 3 4 5 $class [1] "data.frame" > colnames(df1) <- "bla" > attributes(df1) $names [1] "bla"
2009 Mar 02
3
Error setting rowname if rowname currently NULL
Hi, My first post here and new to R so please bear with me (long time programmer though, helping a friend with some scripts). I've noticed a behaviour when using rownames() that I think is odd, wondering if I'm doing something wrong. To illustrate, say I create a very simple matrix (called fred): fred<-matrix(,4,2) It looks like this: [,1] [,2] [1,] NA NA [2,] NA NA
2008 Jan 04
1
predict.lm removes rownames for a single row. Why?
predict.lm keeps row names when working from several rows in newdata, but always removes rowname from a single row. The rownames are removed by the line in predict.lm predictor <- drop(X[, piv, drop = FALSE] %*% beta[piv]) What is the reason for that decision? I usually want to retain the row names. tmp <- data.frame(x=1:4, y=c(1,3,2,5)) tmp.lm <- lm(y ~ x, data=tmp) tmp.new <-
2002 Aug 13
2
Misalignment of <NA> in rownames (PR#1905)
An NA in the rownames of a matrix (or dataframe) causes misalignment when the matrix is printed: R> x <- matrix(1:12, 3,4, dimnames=list(letters[1:3], LETTERS[1:4])) R> rownames(x)[2] <- NA R> x A B C D a 1 4 7 10 <NA> 2 5 8 11 c 3 6 9 12 The bug is in function Rstrlen, in src/main/printutils.c. MatrixRowLabel and MatrixColumnLabel (same file) rely on Rstrlen
1998 Nov 09
1
Proposal for discussion: COLNAMES & ROWNAMES
Looking into several different parts of R../src/library/base/R/*.R has led me to the conclusion that quite a bit of code doubling could be saved by using the following two functions whose naming philosophy is derived from that of NROW() & NCOL(): COLNAMES <- function(x) if(is.null(n <- colnames(x))) paste(seq(length=NCOL(x))) else n ROWNAMES <- function(x)
2010 Jun 03
2
Matrix subsetting with rownames
Hi, I have a large matrix containing 10000 rows and 60 columns. I would like to subset this matrix by using a list of several hundred row names. I tried to find how to do that, eg with %in% or subset commands, but I never succeed and I do not find the proper method on internet. How to proceed ? Thanks Muppy -- View this message in context:
2011 Mar 01
2
Entering table with multiple columns & rows
Hi, I'm having difficulty with getting a table to show with multiple rows and columns. Below is the commands that I've typed in and errors that I am getting. Thank you. Laura Table trying to enter: Diet: Binger-yes: Binger-No: Total: None 24 134 158 Healthy 9 52 61 Unhealthy 23 72 95 Dangerous 12 15 27 >
2006 Mar 14
1
MEMORY limit in R
I sent this to Bioconductor, but I think it is more of an R listserv question, because of memory allocation trouble I am having trouble getting the function heatmap() to work on the following gene expression > dim(SAMPLES_log) [1] 12626 20 sample1 sample2...................sample20 gen1 gen2 gen3 .... gen12626 I have converted SAMPLES_log to a numeric matrix using:
2009 May 17
1
[wishlist, patch] make row() and col() preserve dimnames (PR#13705)
Full_Name: Ben Goodrich Version: 2.9.0 OS: Linux (Debian unstable) Submission from: (NULL) (128.103.220.16) row(x), col(x), and functions that call them like lower.tri(x) and upper.tri(x) do not retain the rownames or colnames of x in the matrix that is returned. Example from R version 2.9.0 : x <- matrix(1:9, nrow = 3, ncol = 3) rownames(x) <- LETTERS[1:3] colnames(x) <- letters[1:3]
2006 Dec 01
3
Make many barplot into one plot
Dear all, ## I have 4 tables like this: satu <- array(c(5,15,20,68,29,54,84,119), dim=c(2,4), dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", "Blond"))) dua <- array(c(50,105,30,8,29,25,84,9), dim=c(2,4), dimnames=list(c("Negative", "Positive"),
2009 Jul 15
4
Extract pairs (rowname, columname) from a matrix where value is 0
Dear sir, I have a matrix like a<-matrix(c(0,2,0,4,0,6,5,8,0),nrow=3) colnames(a)<-c("F1","F2","F3") rownames(a)<-c("A1","A2","A3") a F1 F2 F3 A1 0 4 5 A2 2 0 8 A3 0 6 0 I want to extract all pairs (rownames, columnames) from which the value in the matrix is 0 The result should be something like this A1, F1 A2,
2013 Feb 23
1
how to calculate left kronecker product?
For an application, I have formulas defined in terms of a left Kronecker product of matrices, A,B, meaning A \otimes_L B = {A * B[i,j]} -- matrix on the left multiplies each element on the right. The standard kronecker() function is the right Kronecker product, A \otimes_R B = {A[i,j] * B} -- matrix on the right multiplies each element on the left. The example below shows the result of