similar to: vector replacement 1/0 to P/A

Displaying 20 results from an estimated 8000 matches similar to: "vector replacement 1/0 to P/A"

2008 Jun 27
2
converting numbers to characters
Hi, I need to convert numbers to characters in order to index an array when I encountered the number 100000 which turned into "1e+05". Does anyone know how to get around this problem? Lana Schaffer Biostatistics/Informatics The Scripps Research Institute DNA Array Core Facility La Jolla, CA 92037 (858) 784-2263 (858) 784-2994 schaffer at scripps.edu
2009 Feb 13
1
error with make
Hi, I am trying to compile the R-dev version on a unix Suse machine and got errors. Would someone be able to help me determine what to do to fix these errors: make[1]: Entering directory `/lustre/people/schaffer/R-devel/m4' make[1]: Nothing to be done for `R'. make[1]: Leaving directory `/lustre/people/schaffer/R-devel/m4' make[1]: Entering directory
2004 Aug 20
3
Partial Least Squares
Friends, Is there a Partial Least Squares package implemented in R? Thanks, Lana [[alternative HTML version deleted]]
2008 Jul 03
1
subset function within a function
Hi, I am using this subset statement and it works outside a function. LIS[[i]]<- lapply(LI, subset, select=cov[[i]]) However, wrapped inside a function this statement produces the same values for every LIS[[1]] which is only the first subset of LI. Does anyone know why is not working correctly inside a function? ff = factor(covariate) nLev <- nlevels(ff) cov <-
2004 Jul 23
2
graphing with error bars
I am not able to find information about doing line plots with error bars in the R- help information. Would someone like to tell me if this can be done in R and either get me started or lead me to information about how to do the error bars? Thanks, Lana
2004 Jul 07
1
command line interface
How can plots (histograms) be implemented with the command line interface to R? Lana Schaffer
2007 Jul 31
5
extract columns of a matrix/data frame
Hello all, I have a matrix whose column names look like a1 a2 b1 b2 b3 c1 c2 1 2 3 7 1 3 2 4 6 7 8 1 4 3 Now, I can have any number of a's. not just two as shown above and same goes for b's and c's. I need to extract all the a's columns and put them in another matrix, extract all b's columns and put them in some matrix
2013 Oct 11
3
matrix values linked to vector index
Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]
2012 Jul 11
4
Help with loop
Hi, I have two dataframes: The first, df1, contains some missing data: cola colb colc cold cole 1 NA 5 9 NA 17 2 NA 6 NA 14 NA 3 3 NA 11 15 19 4 4 8 12 NA 20 The second, df2, contains the following: cola colb colc cold cole 1 1.4 0.8 0.02 1.6 0.6 I'm wanting all missing data in df1$cola to be replaced by the value of df2$cola.
2013 Apr 26
2
speed of a vector operation question
Hello, I am dealing with numeric vectors 10^5 to 10^6 elements long. The values are sorted (with duplicates) in the vector (v). I am obtaining the length of vectors such as (v < c) or (v > c1 & v < c2), where c, c1, c2 are some scalar variables. What is the most efficient way to do this? I am using sum(v < c) since TRUE's are 1's and FALSE's are 0's. This
2010 Sep 16
1
Weibull simulation- number of items to replace is not a multiple of replacement length
Hi, I write below code for simulation for weibull- estimating parameters by weibullMLE function, Although I define metrix for the variables still I got this message: number of items to replace is not a multiple of replacement length Any suggestion > est=matrix (NA, 2,2) > se=matrix (NA, 2,2) > for ( p in 1:2) { + sampleSize <- 20 + shape.true <- 1.82 + scale.true <- 987 +
2010 Jul 15
2
replace negative numbers by smallest positive value in matrix
Hi Group, I have a matrix, and I would like to replace numbers less than 0 by the smallest minimum number. Below is an small matrix, and the loop I used. I would like to get suggestions on the "R way" to do this. Thanks, Juliet # example data set mymat <- structure(c(-0.503183609420937, 0.179063475173256, 0.130473004669938, -1.80825226960127, -0.794910626384209, 1.03857280868547,
2017 Jun 27
2
paste strings in C
Dear R-devs, Below is a small example of what I am trying to achieve, that is trivial in R and I would like to learn how to do in C, for very large matrices: > (mymat <- matrix(c(1,0,0,2,2,1), nrow = 2)) [,1] [,2] [,3] [1,] 1 0 2 [2,] 0 2 1 And I would like to produce: [1] "a*C" "B*c" Which can be trivially done in R via something like: foo
2008 Sep 21
1
How to put given values in lower triangle of splom-plot?
Dear R-experts, I have found a splom-modification online which is given below. This works perfectly, but I would like to have a matrix of given correlation values to be used in the lower triangular part (lower.panel) of the splom-plot instead of calculated correlation values. Here is the matrix I would like to use (it can be any other convenient data structure):
2007 Mar 12
1
How to avoid a for-loop?
Hi all, as I am trying to move slowly from just "working" to "good" code, I'd like to ask if there's a smarter way than using a for-loop in tasks like the example below. I need to obtain the extrema of the cumulated sum of a detrended time series. The following code is currently used, please have a look at the comments for my questions and remarks: system.time({ X
2009 Aug 26
3
changing equal values on matrix by same random number
Dear all, I have about 30,000 matrix (512x512), with values from 1 to N. Each value on a matrix represent a habitat patch on my matrix (i.e. my landscape). Non-habitat are stored as ZERO. No I need to change each 1-to-N values for the same random number. Just supose my matrix is: mymat<-matrix(c(1,1,1,0,0,0,0,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 0,0,0,0,2,2,2,0,0,0,0, 3,3,0,0,0,0,0,0,0,4,4,
2011 Oct 01
1
class definition
Hi everybody! I have a matrix of class "myClass", for example: myMat <- matrix(rnorm(30), nrow = 6) attr(myMat, "class") <- "myClass" class(myMat) When I extract part of ''myMat'', the corresponding class ''myClass'' unfortunately disappear: myMat.p <- myMat[,1:2] class(myMat.p) Please for any advice / suggestions, how
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 Jul 03
1
rownames, colnames, and date and time
Hi all I was wondering whether there has ever been an update on the rownames and colnames behaviour as described by Eric below? I still get the same behaviour, exactly as described by Eric, on my WinXP installation of R-2.3.0. I also posted a message to r-help on Friday but looking through the online archives it seems to have not made it to the list. I would agree with Eric that a consistent