similar to: Calling variables dynamically [Broadcast]

I've looked in the online documentation for this, but have been unable to find an answer. I can name variables dynamically, but I cannot call them dynamically. Either a direct answer or a reference to an online resource that has the answer would be great. I'm trying to write a script that imports and compares results from various IR (information retrieval systems). Each imported file will

This patch introduces the concept of an "Advanced Storage Configuration", specifically for automated installs (although it could be hooked into manual install), while preserving the original style of specifying LVs. It accomplishes the following: * Add an optional AppVG * Offer new optional LVs - Data2,Swap2 - as part of the new AppVG * Allows the specification of multiple disks, per VG

Lately I was reading many articles about using boolean queries for search engine but I haven't seen any complete working demo. Therefore I put together very simple working demo of search engine using boolean query. Feel free to suggest any performance improvement or error while keeping it as simple as possible for understanding. Thanks, -Kevin Duraj http://myhealthcare.com

Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1)

Hi R helpers! I have a question. I'm trying to create a function for an exercise. Here are the arguments I should include: x and y are numeric z is a name ("plus","minus","multiply","divide") and swap is logical. Here is what the function should do: When z="plus", then x+y is performed and so on for the other z names. It should give a NA

Hi, I guess you meant this: dat2<- read.table(text=" patient_id????? t???????? scores 1????????????????????? 0??????????????? 1.6 1????????????????????? 1??????????????? 2.6 1????????????????????? 2???????????????? 2.2 1????????????????????? 3???????????????? 1.8 2????????????????????? 0????????????????? 2.3 2?????????????????????? 2???????????????? 2.5 2?????????????????????

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

Dear R users, I have a data structure as follows: id two res1 res2 c1 c2 inter 1 -0.786093166 1 0 1 2 6 3 -0.308495749 1 0 0 1 2 5 -0.738033048 1 0 0 0 1 7 -0.52176252 1 0

Hi, dat1<- read.table(text=" id??????????????? t???????????????????? scores 2???????????????? 0??????????????????????? 1.2 2???????????????? 2???????????????????????? 2.3 2???????????????? 3??????????????????????? 3.6 2???????????????? 4??????????????????????? 5.6 2???????????????? 6??????????????????????? 7.8 3???????????????? 0??????????????????????? 1.6 3????????????????

I noticed Doug only circulated the following on S-news, but it may be of interest to R users who don't follow S-news. His findings are certainly consistent with my own, where my problems often force me into the element by element type of situation. Paul Gilbert _______ At the risk of beating this example to death, I went back and compared the execution time for the element-by-element method

Dear R-help, I'd like ask for your opinion on choosing the "right" strategy for a particular dataset. We conducted 24-hour electric field measurements on 90 subjects. They are grouped by job (2 categories) and location (3 categories). There are four exposure metrics assigned to each subject. An excerpt from the data: n job location M OA UE all 0 job1 dist_200 0.297 0.072 0.171

Hi, I am new to clustering and was wondering why pvclust using "maximum" as distance measure nearly always results in p-values above 95%. I wrote an example programme which demonstrates this effect. I uploaded a PDF showing the results Here is the code which produces the PDF file: ------------------------------------------------------------------------------------- s <-

Dear R-users, I would like to use optim( ) to minimize a function which depends on 4 parameters: 2 vectors, a scalar, and a matrix. And I have a hard to define the parameters at the beginning of the function, and then to call optim. Indeed, all the examples I have seen dont treat cases where parameters are not all real. Here is my code, it doesnt work but its just to show you where is exactly my

HI, Using c11<- 0.01 c12<- 0.01 c1<- 0.10 c2<- 0.10 One possible problem is that: dim(res5) #[1] 513? 20 res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max) #Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) : ?# undefined columns selected A.K. ________________________________ From: Joanna Zhang <zjoanna2013 at gmail.com> To: arun <smartpink111 at

dear list, I stumbled over the follwing strange behaviour/error when using `nls' which I'm tempted (despite the implied "dangers") to call a bug: I've written a driver for `nls' which allows specifying the model and the data vectors using arbitrary symbols. these are internally mapped to consistent names, which poses a slight complication when using `deriv' to

Hello I'm trying to solve two sets of equations (each set has four equations and all of them share common parameters) with nlminb procedure. I minimize one set and use their parameters as initial values of other set, repeating this until their parameters become very close to each other. I have several parameters (say,param1, param2) and their constraints are given as inequality and depend

If you wanted to do a t.test res1<-do.call(cbind,lapply(seq_len(nrow(m)),function(i) do.call(rbind,lapply(split(rbind(m[i,-1],n),1:nrow(rbind(m[i,-1],n))), function(x) {x1<- rbind(x,m[i,-1]); t.test(x1[1,],x1[2,])$p.value})))) ?res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i) c(c(tail(res1[seq(1,i,1),i],-1),1),res1[-c(1:i),i]))) ?attr(res2,"dimnames")<-NULL ?res2

Hi, It would be better if you provided the output of dput(dataset).? I am not sure about the structure of your dataset. Just from reading the data as is shown. dat1<- read.table(text=" separator,tissID >,>,2 ,2,1 ,6,5 ,11,13 >,>,4 ,4,9 ,6,2 ,7,3 ,21,1 ,23,58 ,25,9 ,26,4 >,>,11 ,1,12 >,>,21 ,4,1 ,11,3

dear list members, my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as (I) expected. I'm posting this here since: (A) the problem might suggest a modification to the `m' component in the return argument of `nls' (making this post formally OK for this list) and (B) I got no

Dear R-users I have a problem choosing initial points for the function arms() in the package HI I intend to implement a Gibbs sampler and one of my conditional distributions is nonstandard and not logconcave. Therefore I'd like to use arms. But there seem to be a strong influence of the initial point y.start. To show the effect I constructed a demonstration example. It is reproducible