similar to: explanation for left-side behaviour

libswfdec/swfdec_as_context.c | 20 + libswfdec/swfdec_as_interpret.c | 9 libswfdec/swfdec_as_string.c | 15 - libswfdec/swfdec_as_strings.c | 2 libswfdec/swfdec_as_types.c | 2 test/trace/Makefile.am | 28 + test/trace/isnan-5.swf |binary

Hi there, I'm looking for someone who can give me some hints how to make a nice levelplot. As an example, I have the following code: # create some example data # -------------------------------------- xl <- 4 yl <- 10 my.data <- sapply(1:xl, FUN = function(x) { rnorm( yl, mean = x) }) x_label <- rep(c("X Label 1", "X Label 2", "X Label 3", "X

Hi, I am searching for the equivalent of the function Index from SAS. In SAS: index("abcd", "bcd") will return 2 because bcd is located in the 2nd cell of the abcd string. The equivalent in R should do this: > myIndex <- foo("abcd", "bcd") #return 2. What is the function that I am looking for? I want to use the return value in substr, like I do

Hello, I also have problems to get to run the following lines. If I run the block instead of every single line, it simply does not wait for the input. Can anybody help me? ------------------------ pos_name <- readline("Please type: ") r <- substr(pos_name, 1,1) c <- substr(pos_name, 2,nchar(pos_name)) ------------------------ Thank you! Antje Peter Dalgaard schrieb: >

Hi there, I have a vector and would like to create a data frame, which contains all unique combination of two elements, regardless of order. myVec <- c(1,2,3) what expand.grid does: 1,1 1,2 1,3 2,1 2,2 2,3 3,1 3,2 3,3 what I would like to have 1,1 1,2 1,3 2,2 2,3 3,3 Can anybody help?

... when the replacement string is shorter than the portion of the string to be replaced? The documentation to substr (in R 1.3.1) gives me: If the portion to be replaced is longer than the replacement string, then only the portion the length of the string is replaced. And so I try: R> x <- "abcdef" R> substr(x,2,3) <- "xy" #ok R> x [1]

... when the replacement string is shorter than the portion of the string to be replaced? The documentation to substr (in R 1.3.1) gives me: If the portion to be replaced is longer than the replacement string, then only the portion the length of the string is replaced. And so I try: R> x <- "abcdef" R> substr(x,2,3) <- "xy" #ok R> x [1]

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hi All, I'm looking for a way to get many substrings from a longer string and then stitch them together. But, since the longer string is really, really long (like 250 MB long), I don't want to do this in a loop and load and re-load the longer string many times. Does anybody have an idea? Maybe I could pass in two vectors (the first would have the starting coordinates, and the second

Hello all Is there a way reducing the number of characters in a list so that just the left n numbers of characters is given? For example, If I have a list, listnames, which consists of 4 strings of 6 characters; >listnames [1] "item12" "item34" "item56" "item78" Is there a way to reduce this so only the 5 characters on the lefthand side are

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hi, In R < 3.0.0, we used to get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) A B A B A "bcdef" "2345" "bcd" "23" "b" But in R >= 3.0.0, we get: > substring(c(A="abcdefghij", B="123456789"), 2, 6:2) [1] "bcdef"

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

Hi there, I have a question concerning the behaviour of the colouring with levelplot. (I hope, I manage to explain) If I give the parameters "at" and "col.regions" like this: at <- c(1,2,3,4,5,6) col.regions <- c("blue","blue","blue","yellow","yellow","yellow") Which color would have the value 3.5? I would

Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice fit or not. So I have to apply them to observation data. But how does the method know about my observed

Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =