similar to: Rounding

Hi, Is it possible to draw barplot with x-axis being shown? I looked up the help file and I couldn't seem to find it. For example, I tried to do x <- 1:9 p <- log10(1 + 1/x) barplot(p, xlab = "Digit d", ylab = "Probability", ylim = c(0, 0.35), axes = F, main = "Benford's Law Probability") axis(1, 1:9) axis(2, seq(0, 0.35, by = 0.05),

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, Is it possible to add a subtitle that appears directly below the main title? I tried the "sub" parameter, but it adds sub-title to the bottom of the plot. Cheers, Kevin ------------------------------------------------------------------------------ Ko-Kang Kevin Wang Postgraduate PGDipSci Student Department of Statistics University of Auckland New Zealand Homepage:

Hi, Suppose I have created something like this in R: foo <- "myfoo" and I want to find out the number of character in foo (in other words, R should return 5 since "myfoo" has 5 charactors. How can I do it? I tried: length(foo) but it returned 1. Cheers, Kevin ------------------------------------------------------------------------------ Ko-Kang Kevin Wang

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 did something like: par(mfrow = c(1, 2)) plot(1:10) plot(1:10) and I'd like to have an overall title, i.e. a title that would appear, centered, on the top of both plots, rather than the top of the last plot. I tried: title("FOO", outer = T) but it doesn't appear right. Half of the FOO is outside the figure region and can't be seen... Is there another

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

Hi, Suppose I have a function: myfunc <- function(x, y) { ... } And within the function I want to print out the name of the x, y vectors. For example, if I do: > myfunc(foo, goo) [1] "foo" "goo" It shall return "foo", "goo" (with or without quotes is fine), where foo and goo are two vectors with numbers. I know this sounds strange, but I'd

Hi, Suppose I have the following lines at the end of a function: answer <- c(2, 1, 0, 4, 5) # In fact, answer will be generate in my # function print(answer) # Print the answer # Now, find the best fitted n degree polynomial print(paste("The best fit is with", which.min(answer) - 1, "-degree polynomial")) this will return:

Hi, Does anyone know how to manually configure the number of digits printed out from xtable()? For example, I'm exporting a data frame through xtable() into a LaTeX table, I only have two columns in the data frame so by default I only get two decimal places. But I'd like at least 5 decimal places. I had a look at ?xtable() but can't seem to find an example. Cheers, Kevin

Dear all, I would like to (i) produce boxplot graphs with axis in logarithm in base 10 and (ii) showing the values on the axis in 10^exponent format rather than 10E+exponent. To illustrate with an example, I have some widely spread data that I chart plot using boxplot() [figure on the left]; the log="y" option of boxplot() I obtained the natural logarithm conversion of the data and

Hi, I'm learning Fortran and trying to load a Fortran subroutine into R. I've done: R SHLIB Fibonacci.f and it compiled fine. Then I went into R and done: > dyn.load("Fibonacci.so") > Fib <- function(n) { + .Fortran("Fibonacci", + as.integer(n))[[1]] + } > Fib(5) Error in .Fortran("Fibonacci", as.integer(n)) :

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)

Hi, I was trying to read in the following data set by using: gas <- read.table("gas.tab", header = T) when I got the error message: Error in "row.names<-.data.frame"(*tmp*, value = row.names) : duplicate row.names are not allowed Any help are appreciated. Thanks, Ko-Kang Wang

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, I'm also cc'ing it to r-help. On Wed, 6 Nov 2002, Kenneth Cabrera wrote: > Date: Wed, 06 Nov 2002 10:09:21 -0500 > From: Kenneth Cabrera <krcabrer at perseus.unalmed.edu.co> > To: kwan022 at stat.auckland.ac.nz > Subject: some questions! > > Hello Dear Ko-Kang Wang: > > I am trying to compile R v 1.6.1 . > > I am following the

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, 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"

Dear all, I am trying to apply the logistic regression to determine the limit of detection (LOD) of a molecular biology assay, the polymerase chain reaction (PCR). The aim of the procedure is to identify the value (variable "dilution") that determine a 95% probability of success, that is "positive"/"total"=0.95. The procedure I have implemented seemed to work looking

Hi, Is it possible to create a ts() object, whose data is daily based BUT measured only on working days? In other words, suppose I have a data set with 255 observations, measured from 29 June 1959 to 30 June 1960. How would I create such a data? I tried something like: ts(c(...), start(1959, 180)) but I'm not sure what to use for frequency. In other words I don't know how to