similar to: efficiency

i am sure that there is a simple solution that i am just overlooking, but there is no logical test for = there is one for != and >,<, >=, <=... what do i do if i need to check if something IS EQUAL or is there an easy way, in working with a matrix of size n x 3, to sort the matrix by one of the columns? specifically what i am trying to do is i have a matrix with individual and

i am probably just missing this one, but how can i check to see if a number is a part of a vector? i know i could do a loop checking for equivalence for each element, but i am thinking that there must be a simpler way to do it than that ( i think that i have even done it in the past, but can't remember how!) thanks what i am working with: (hope this makes sense) arclist [1] 2 2 6

me again, new to R, haven't programmed in a long time at all sorry to ask what is probably basic a while ago i asked how to remove an element from a list apparently what i am using is not a list but a vector, didn't realize there was a list data type could also use a matrix for what i am doing, is there a way to remove single elements from either of those? thanks for any help, if i was

i am having some problems with a program that i am writing and i think that knowing how the while command works will help me to figure out where i am going wrong *when do you get kicked out of a while loop?* (i.e., in the following example what would the final value of 'a' be?) i thought it would be 99 but running it in R i get 101 (and 101 for x too!) while (x <100) { for (i in

i have a function that returns a list containing a variety of variable types i am trying to run the function multiple times and return the output into a variable with a semi-consistent naming pattern i.e., for ten trials i want to return the list into variables trial1,trial2,...trial10 is there a generic way to get this to happen i have a similar process that does the same thing to an external

i am trying to use the sample command and have one question about it: i am getting the error Error in sample(length(x), size, replace, prob) : insufficient positive probabilities when i use something like: sample (1:4, prob=c(0,0,1,0))[1] i was expecting that to return a 3 every time while this is not exactly what i am using it for, i need the capabilities to deal with zeros (as the

i have a 2 x n matrix that is a paired list of connections, i am working with that looks something like: 1 2 1 3 1 5 2 1 2 3 3 1 3 2 4 5 5 1 5 4 for later operations it would be helpful if i could change this into the form of: 1 2 3 5 2 1 3 3 1 2 4 5 5 1 4 the initial list is randomly generated and the max number of times that any one number can appear is previoulsy defined by a limiting

Hello, I have a data frame, one element in that data frame is a LIST, with each element being a character string. I am trying to extract the first year listed in each of those character strings. The character elements are typically csv, but the position of the year can vary (think citations with varying citation standards). I.e., foo$a [[1]] [1] text, text, 2001, text [2] text, 2000, text [3]

As a novice to Wine I've been trying to run the old spreadsheet program Lotus 123 release 5 (1994) under Fedora Core3. I'm using the default wine config with one change under [Version] "Windows" = "win95". I have little knowledge of gdb. I get the following errors after I mount the first of seven 3.5in floppies. Can anyone give me some pointers? Please e-mail me

Dear All, I have a data set with the following structure: [A], [a], [B], [b] where [A] and [B] are measurements and [a] and [b] are the associated uncertainties. I produce [B]/[A] vs. [A] plots in R and would like to show uncertainties as error ellipses (rather than error bars). Would this be relatively easy to do in R? I would appreciate any help on this Thanks a lot Tibi

All! This piece of code: zzz1 <- as.POSIXct("1999-03-18", tz="CET") zzz2 <- as.POSIXlt("1999-03-18", tz="CET") zzz1 == zzz2 as.Date(zzz1) as.Date(zzz2) yields TRUE for "zzz1==zzz2", but the two dates returned by as.Date are different: > as.Date(zzz1) [1] "1999-03-17" > as.Date(zzz2) [1] "1999-03-18" I'm

Hello listers, Does anybody know how to add text to an xyplot without whipping out the existing curve? That's all. For instance, Lets say you generate a graph like this A <- data.frame(x = rnorm(100), y = rnorm(100)) xyplot(y ~ x, data = A) How would you add 'Hello world'? I tried 6.02E23 different partial solutions found on the web and failed. I just need one EXAMPLE that

Hello, I need to select a sample from a small population using simple random sampling without replacement. I've found two possibilities in R, the function sample() and the function S.SI() in the package TeachingSampling, but I don't know which one is better. Can someone help me? Thank you -- Lourdes Molera Peris M?todos Cuantitativos para la Econom?a y la Empresa Facultad de Econom?a

Hi all, In a data-frame, I have two columns of data that are categorical. How do I form some sort of measure of correlation between these two columns? For numerical data, I just need to regress one to the other, or do some pairs plot. But for categorical data, how do I find and/or visualize correlation between the two columns of data? Thanks!

How should I analysis it in R ???? all the resposes variables are ordinal from 0 to 10. and the explanatory variable is a factor ... -- View this message in context: http://old.nabble.com/Multivariate-problems-.-.-.-with-200-resposes-variables-and-1-explanatory-variable-tp26522912p26522912.html Sent from the R help mailing list archive at Nabble.com.

Hi dears while modeling an interaction random effect in lmer i receive the instantaneous error message > ldlM4<-lmer(ldl~rt*cd4+age+rf+pharmac+factor(hcv)+ + hivdur+(rt:cd4|id),na.action=na.omit,REML=F) *Warning message: In mer_finalize(ans) : false convergence (8) * I think the matter lies in syntax, 'cause i sistematically receive the same message even when changing response... PS:

Hello, I am learning R and I am fluent in Stata and I try to translate part of my Stata code to R to check the reliability of the data under R. I have a proportion variable as a dependent variable pQSfteHT . Independent variables are dummies for two categorical variables called dQSvacrateHTQuali3 and cluster_3. I am fitting a model with the Stata command below: glm pQSfteHT dQSvacrateHTQuali3_2

Hello, I plot box plot, and add a point to the box indicating mean. I also add some range to the box width. I want to add legend for both the range (line) and mean (point). However, I cannot add line legend and point legend together. The code looks like the following. (1) First, I tried the following code, however, it plot a line across the point legend("*"). boxplot(count ~ spray,

I'm still not fully understanding exactly how R is handling data frames, but am getting closer. Any help with this one will likely go a long way in getting me there. Let's say I have a data frame, let's call it "a". Within that data frame i have two variables, let's call them "b" and "c", where "b" is a single numeric value per observation,

Hi, x <- 0.2*5 is.integer(x) gives me FALSE because R stores it as a float number, right? Is there an elegant way to work around that problem? Right now I'm using x <- 0.2*5 round(x) == x which returns TRUE. But more strictly I should use all.equal(), right? I somehow just don't like the--pardon--ugliness of those pieces of code. Maybe there is a beautiful