similar to: Turning strings into expressions

Hello, I'm not sure how to state my question in a technically accurate manner, so I'll use a short example to clarify my problem: Say I have a vector, vec<-c(1,2,3,2) I would like to be able to "reference" the vector by using the string containing the name of the vector - that is, I would like to know of some function which is able to convert the string "vec"

Hello, I am using R 1.9.1 on Windows 2000 SP4. I have the following problem: Say I have a matrix, >my.matrix [,1] [,2] [,3] [1,] "A" "B" "C" [2,] "D" "E" "F" [3,] "G" "H" "I" I would like to apply an operation to this matrix which returns a list my.list containing the

Hello, I have the following very simple problem: Say I have two vectors, a<-c(1,7,4,5,9,11) b<-c(7,4,9) I would like to create a vector containing the elements in a which are not in b. Obviously, this is possible by writing a[a!=b[1] & a!=b[2] & a!=b[3]] But I would like a solution which is applicable to the situation where the number of elements in b is unknown. I have

Hi there fellow R-users, I'm stuck on this seemingly trivial problem. All I want to coerce a character string into a command. For example: x<-rnorm(20) y<-rnorm(20) str<-"lm(y~x)" I want to evaluate the "str" command. I have tried eval(as.expression(str)) But it doesn't seem to work. I am aware of the call command, but for reasons I won't go

Dear R-helpers: thank your for your attention. i am a newer to R and i am doing some protein category classification based on the amino acid sequence.while i have some question urgent. 1. any packages for analysis amino acid sequence 2. given two sequences "AAA" and "BBB",how can i combine them into "AAABBB" 3. based on "AAABBB",how can i get some

Hi R, Again struck with regular expressions... Suppose, S=c("World_is_beautiful", "one_two_three_four","My_book") I need to extract the last but one element of the strings. So, my output should look like: Ans=c("is","three","My") gsub() can do this...but wondering how do I give the regular expression....

I'm having a Thursday morning mental block, any suggestions on the following would be most appreciated... I have (as an example) surgery = c("d48", "d67", "dnc37", "a75", "d10", "a78", "d31", "d55", "d1") before each number part the possibilities are c("a", "d",

Hi If string <- "xyz" f <- function(x){1 + sin(cos(x)) + exp(x^2)} How do I manipulate "string" and f() to give the string "1 + sin(cos(xyz)) + exp(xyz^2)" ? -- Robin Hankin Uncertainty Analyst National Oceanography Centre, Southampton European Way, Southampton SO14 3ZH, UK tel 023-8059-7743

Hi, all. I'm trying to automate some regression operations in R but am confused about how to evaluate expressoins that are expressed as character strings. For example: y <- ifelse (rnorm(10)>0, 1, 0) sex <- rnorm(10) age <- rnorm(10) test <- as.data.frame (cbind (y, sex, age)) # this works fine: glm (y ~ sex + I(age^2), data=test, family=binomial(link="logit"),

Dear R users, I was wondering if someone could suggest a few lines of code for my problem. I want to count the number and the length of strings of NA in a vector. For example: vec <- c(1, 2, 1, NA, NA, 1, 2, NA, NA, NA, 3, 4, NA, NA) has 2 strings of NA's of length 2 and 1 string of NA' of length 3. I can easily count the number of NA's per vector, but I am having a hard time

Say you have a vector named x and a function which returns the character string "x" . How would I take "x" as an input and return the vector x? DISCLAIMER: This e-mail message and any attachments are inte...{{dropped}}

Hi all, My question is not really urgent. I can write a loop and solve the problem. But I know that I'll be in a similar situation many more times so it would be useful to find out the answer Is there a fast way to perform linear fit to all the columns of a matrix? (or in the one dimension of a multi-dimensional array.) I'm talking about many single linear fits, not about a multiple fit.

Thanks Dimitris!!! That's much clearer now. Still have a lot of work to do this weekend to understand every bit but your code will prove very useful. Cheers, Aziz -----Original Message----- From: Dimitrios Rizopoulos [mailto:Dimitris.Rizopoulos at med.kuleuven.be] Sent: May 12, 2006 4:35 PM To: Chaouch, Aziz Subject: RE: [R] Maximum likelihood estimate of bivariate

Hi everyone, I have a vector of strings, each string made up by different number of words. I want to get a new vector which has only the first word of each string in the first vector. I came up with this: str <- c('aaa bbb', 'cc', 'd eee aa', 'mmm o n') str1 <- rep(1, length(str)) for (i in 1:length(str)) { str1[i] <- strsplit(str, "

Dear R-devels, I'd like to create a plot method for a class of objects that passes the '...' argument to both plot() and legend(), e.g., x <- list(data = rnorm(1000)) class(x) <- "foo" plot.foo <- function(x, legend = FALSE, cx = "topright", cy = NULL, ...){ dx <- sort(x$data) plot(dx, dnorm(dx), type = "l", ...) if (legend)

I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order does not matter. paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'. Any idea? -- Regards, Anh Tran [[alternative HTML version deleted]]

hi all: xlim and ylim are used to define the interval limits of a plot. I'm interested in the scale of values between this limits. suppose xlim=c(0,10) we can have e.g. 0 5 10 0 2 4 6 8 10 0 1 2 3 4 5 6 7 8 9 10 which is the parameter that allows me to modify this? thanks in advance alexandre

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

I have a vector that has 1,974 elements and each element is one of the following (B, F, N, Y). How do I recreate that vector accept in the place of N put 0 and in the place of B, F or Y put a 1? Thanks, Jacob [[alternative HTML version deleted]]

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323