similar to: Sampling from a Data Frame

Hi, I'm not sure if this can be done but.. I know that with ifelse() I can do something like: ifelse(x <= 3, 1, 2) to go through each element in my vector x, and if x_i <= 3 substitute the number with 1 else with 2. Essentially I'll get a vector with 2 levels. Can I tweak it so I can get 3-levels? For example: if(x <= 3) then 1 elseif(3 < x <= 4) then 2

Hi, I hope I haven't got this wrong, but I don't think one needs Perl installed if they install the "precompiled" packages for Windows right? I'm just wondering if the first sentence in the "R Installation and Administration" can be re-worded a bit so newbies won't get confused.... ---------- Forwarded message ---------- Date: Mon, 2 Jun 2003 08:56:37 +1200

Hi, I'm doing a presentation on Neural Networks and Tree-Based Models in two weeks, at the moment I'm looking for a data set to use in the presentation. What I would like to use is a good old data, like the Iris data, that is already known by every statisticians. MASS4 uses the cpus data in Chapter 8.10 and the Cushing's syndrome in Chapter 12.4. These two data sets plus the

Hi, If I have: foo <- survfit(y ~ x) where y is a survival object and x is a n-level factor. The documentation says when I plot(foo), the confidence intervals will not be plotted (which I guess is understandable as otherwise the plot will get really messy). I tried to plot with confidence intervals by using: plot(foo, conf.int = TRUE) and indeed the resulting plot is messy. However

Hi, I have a column in a dataframe in the form of: > as.vector(SLDATX[1:20]) [1] "1/6/1986" "1/17/1986" "2/2/1986" "2/4/1986" "2/4/1986" [6] "2/21/1986" "3/6/1986" "3/25/1986" "4/6/1986" "4/10/1986" [11] "4/23/1986" "4/30/1986" "5/8/1986"

Hi, Suppose I have: # Fit a base model d1.ph <- coxph(Surv(start, stop, event)~ ejec + diavol + score + smoking + beta + surg.done, data = data.frame(foo)) summary(update(d1.ph, . ~ . + td1)) summary(update(d1.ph, . ~ . + td2)) As I have many columns in my data frame, foo, called td's. e.g. td1, td2, td3, .... And I'd like to

Hi, If I do something like: biplot(x, cex.axis = .7) then it will only change the font size for axis 1 and 2, but not 3 and 4. Is there a way to change the fonts on axis 3 and 4? -- Cheers, Kevin ------------------------------------------------------------------------------ /* Time is the greatest teacher, unfortunately it kills its students */ -- Ko-Kang Kevin Wang Master of Science

Hi, Say I have a factor with 20-levels: 1, 2, 3, ..., 20, called foo. If I do plot(foo) it will draw a barplot. But the x-axis is sorted alphanumerically, i.e. 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 2, 3, ..., 9, which is not what I want. I'd like to x-axis to be in the order of 1 ~ 20 numerically. So, the question is, how do I change the order on the x-axis in this case? --

Hi, What does this mean when I have something like: > qda.boot <- boot(train, qda.bootstrap, R = 500) Error in qda.default(structure(data.matrix(x), class = "matrix"), ...) : Rank deficiency in group M with my qda.bootstrap() looks something like: > qda.bootstrap <- function(data, index) { + boot.qda <- qda(x = data[index, 2:9], group = data[index, 1]) + qda.pred

Hi, Suppose I have: > summary(manova(plank.man)) Df Pillai approx F num Df den Df Pr(>F) plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 *** Residuals 58 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 My understanding is the MANOVA summary returns a list.

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Hi, Suppose I use the following codes to read in a data set. ############################################### > rating <- scan("../Data/Rating.csv", + what = list( + usage = "", + mileage = 0, + sex = "", + excess = "", + ncd = "", +

Hi, This may actually be a theoretical question. When I tried to do the following: ########################################################## > colnames(rating.adclms) [1] "usage" "mileage" "sex" "excess" "ncd" [6] "primage" "minage" "drivers" "district" "cargroup" [11]

Hi, I'm a bit confused how julian() works. If I understand right, it returns the number of days since the origin. I have a vector: > SLDATX[1:10] [1] "1986-01-06" "1986-01-17" "1986-02-02" "1986-02-04" [5] "1986-02-04" "1986-02-21" "1986-03-06" "1986-03-25" [9] "1986-04-06"

Hi, I think I found the problem. It lies in my Fortran program. Is there a way, after a DO loop, to make sure it does NOT return anything? Cheers, Kevin On Mon, 15 Apr 2002, Ko-Kang Kevin Wang wrote: > Date: Mon, 15 Apr 2002 17:27:20 +1200 (NZST) > From: Ko-Kang Kevin Wang <kwan022 at stat1.stat.auckland.ac.nz> > To: R Help <r-help at stat.math.ethz.ch> > Subject:

Hi, I have a function foo(x,y) and a dataframe, DF, comprised of two vectors, x & w, as follows : x w 1 1 1 2 2 1 3 3 1 4 4 1 etc I would like to apply the function foo to each 'pair' within DF e.g foo(1,1), foo(2,1), foo(3,1) etc I have tried >apply(DF,foo) >apply(DF[,],foo) >apply(DF[DF$x,DF$w],foo) However, none of the above worked. Can anyone help ?

Hi, Now suppose I have just one list called FOO, which has 25 objects, e.g.: [[1]] 1 2 3 4 5 [[2]] 6 7 8 9 10 . . . And I want to do something like: FRED <- data.frame(cbind(unlist(FOO[[1]]), unlist(FOO[[2]]), # ... for all 25 subsets )) Is it possible to do this, without doing unlist(FOO[[i]]) 25

Hi, Say I want to do something like fitting 10 different sized trees with rpart() function. The only modification I need to do is to set 10 different cp's, which I have in a vector called foo. Can I do something like: for(i in 1:10) { rpart(y ~ ., cp = foo[i], data = mydata) } My problem is, I wish to save the 10 rpart objects into 10 different names, my.rpart1 ~ my.rpart10, for

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

Hi, Is there a function in R that does Cleveland's Cut-and-Stack plot (Page 190 -- 191, The Elements of Graphing Data, William S. Cleveland)? Or do I need to do it the hard way, i.e. set par(mfrow = c(m, n)) then do it one-by-one? (I have a time series data set that is almost identical to the description in Cleveland's book, hence I'm interested in trying the Cut-and-Stack plot)