similar to: converting a row of a data.frame to a vector

I have a big contingency table, approximately of size 60*2*500*500, and I need to count the number of cells containing a count of 1 for each of the factors values defining the first dimension. Here is my attempt: tab1<-with(pisa1,table(CNT,GENDER,ISCOF,ISCOM)) tab2<-apply(tab1,1:4,function(x)ifelse(sum(x)==1,1,0)) tab3<-apply(tab2,1,sum) Computing tab2 is very slow. Is there a faster

I found writing the following default method the for the generic function "julian" causes R to crash. julian.default <- function(x, ...) { x <- as.Date(x) NextMethod("julian", x, ...) } Here is a test example > m <- as.Date("1972-09-27") + 0:10 > m [1] "1972-09-27" "1972-09-28" "1972-09-29"

I have a matrix of characters (actually numbers that have been read in as numbers), and I'd like to remove the NA. I'm familiar with na.omit, but is there an equivalent of na.omit when the NA are the actual characters "NA"? Thanks, Andrew [[alternative HTML version deleted]]

Hi R, I wanted to know how do we access the elements of a list. In particular, v=list(c(1,2,3,4,5),c(1,2,33,4,5),c(1,2,333,4,5),c(1,2,3333,4,5)) I want to access all the thirds items of the elements of the list. i.e., I want to access the elements, 3,33,333,3333. This can be done through sapply as: sapply(v,function(x) x[3]) But I need to access this without using

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While termplot is under discussion, here's another proposal. I'd like to change the default for partial.resid to TRUE, and for smooth to panel.smooth. I'd be surprised if those changes were to break existing code. John Maindonald email: john.maindonald at anu.edu.au phone : +61 2 (6125)3473 fax : +61 2(6125)5549 Centre for Mathematics & Its Applications, Room

The function "substitute" seems to fail to make a genuine substitution, although the printed verision seems fine. Here is an example. > m <- substitute(Y <- function(x) FUN(x+1), + list(Y = as.name("y"), FUN = as.name("sin"))) > m y <- function(x) sin(x + 1) > eval(m) > y function(x) FUN(x+1) However the story doesn't end there. The

I am converting some matlab code into R that use inverse of the complementary error function, erfcinv and did not find an equivalent in R, is there such a function in some contributed modules? Thanks.

I have just become painfully aware that objects of class "difftime", generated by the difference of two POSIXct objects, carry a "units" attribute, which flashes by when the object is printed, for example. The pain was occasioned when I tried to turn these objects into numberic objects for use elsewhere as a covariate. as.numeric(difftime object) simply turns off the units

I have a set data={A,B,C,D,E,F,G} I want to choose 2 letter from 8 letters, i.e. generate the combination set for choose 2 letters from 8 letters. I want to get the liking: combination set={AB,AC,AD,....} Does anyone konw how to do in R. thanks, Aimin

Hi, I have a matrix a = matrix (1:16, 4, 4) b = c (2,3) I want to find out which rows of a, where a[,1] equals any values of b? I know that if b is only one value, e.g, b=2, then what I want is a[a[,1] == 2,] But what about if it is not one value but a vector of values? Thanks much in advance. -- Waverley @ Palo Alto

Dear R-users, I am looking for a way to scan a directory and then make a string vector consisting of the file names in the directory. For example, under c:\temp\ there are 4 files a.txt, b.txt, c.txt, and d.txt I would like to have a string vector c("a.txt","b.txt","c.txt","d.txt") How do I do that? Thanks Taka,

This is not so much a bug as an infelicity in the code that can easily be fixed. The initialize expression in the quasi family function is, (uniformly for all links and all variance functions): initialize <- expression({ n <- rep.int(1, nobs) mustart <- y + 0.1 * (y == 0) }) This is inappropriate (and often fails) for variance function "mu(1-mu)".

Does anyone have, or has anyone ever considered making, a version of 'termplot' that allows the user to specify that all plots should have the same y-limits? This seems a natural thing to ask for, as the plots share a y-scale. If you don't have the same y-axes you can easily misread the comparative contributions of the different components. Notes: the current version of termplot

Does anyone have, or has anyone ever considered making, a version of 'termplot' that allows the user to specify that all plots should have the same y-limits? This seems a natural thing to ask for, as the plots share a y-scale. If you don't have the same y-axes you can easily misread the comparative contributions of the different components. Notes: the current version of termplot

I am looking for a solution to match 2 dataframes from pairs of values (x and y) as indicated thereafter : First dataframe : DATA1 x y a 1 30 40 0.2 2 21 130 0.3 Second dataframe DATA2 x y b 1 30 40 1 2 40 30 3 3 20 40 7 4 11 30 2 5 130 250 15 6 21 130 17 expected Results : DATA3 x y a b 1 30 40 0.2 1 2

Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): > x <- 1:3 > y <- c(1, 4, 9) performing a linear fit > f <- lm(y ~ poly(x, 2)) gives weird coefficients: > coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.6666667 5.6568542 0.8164966 However the fitted() result makes sense: >

hi, I am reading Modern Applied Statistics with S 4th ed$B!#(B page4 have these two lines: > library(MASS) > data(chem) # needed in R only but I find withou the line " data(chem)" I can still access chem, isn't it? is it unnecessary or something i missed here? thanks for the replay in advance.

Hi I run R in Windows. Is there a simple way of changing the prompt symbol ">" to, say, "R>" ? (Not just for a temporary session, but every time R command window is opened.) The documentation of doing this is rather "sparse". Much appreciated for your assistance. Jacob Jacob L van Wyk Department of Statistics University of Johannesburg, APK P O Box 524

Hi all, I used the function 'glm' to perform logistic regression, and then use 'summary' to display the fitting result. There are 'z value' and 'Pr(>|z|)' Do they simpily represent the result of 'z test' ? as I perform the 'z test' for a particular variable following the definition, it yields a different result. Stephen Lau