similar to: Turning Data Frame Columns into Vectors

Hi, I am trying to create a script that will evaluate each column of a data frame, regardless of # columns, using some function and sorting the results by an index vector: #upload data (112 rows x 73 columns) SD <- read.csv("/Users/johnjacob/Desktop/StudentsData_RInput.csv", header=TRUE) #assign index vector ID <- SD[ ,2] #write indexed mean function meanfun <- function(x) {

Hello, I have a question regarding subsetting of vectors. Here's an example of what I'm trying to do: vect.1 <- c(76,195, 290, 380) vect.2 <- c(63, 95, 133, 170, 215, 253, 285, 299, 325, 375) I would like to subset vect.2 so that it has the same length as vect.1, and its numbers are the first corresponging higher value compared to vect.1. The output should be: final.output =

As a newer R practicioner, it seems I stump myself weekly (at least) with issues that have spinning my wheels.  Here is yet another... I'm trying to turn a list of numeric vectors (of uneual length) inot a dataframe.  Each vector held in the list represents a row, and there are some rows of unequal length.  I would like NAs as placeholders for "missing" data in the shorter vectors. 

Hi ? This is giving me a headache. I?m trying to do a relatively simple optimization ? actually trying to approximate the output from the Excel Solver function but at roughly 1000x the speed. ? The optimization parameters look like this. The only trouble is that I want to add a constraint that sum(wgt.vect)=1, and I can?t figure out how to do that in optim. Mo.vect <-

Using the proposed patch for enabling the i386 multilib under the x86_64-apple-darwin build... http://lists.cs.uiuc.edu/pipermail/llvmdev/2009-August/025040.html the following gfortran testsuite results are obtained... Native configuration is x86_64-apple-darwin10 === gfortran tests === Running target unix/-m32 FAIL: gfortran.dg/aint_anint_1.f90 -O (internal compiler error) FAIL:

Hi, I got caught out by this behaviour in 1.8.0 and I wondered why this happens: I have a list of vectors and was using lapply and grep to remove matched elements that occur in only a subset of the elements of the list: locs <- lapply(locs, function(x){x[- grep("^x", x)]}) The problem is that where the grep finds no matches and hence returns a vector of length 0, all the

Folks; I am having a problem with the cv.glm and would appreciate someone shedding some light here. It seems obvious but I cannot get it. I did read the manual, but I could not get more insight. This is a database containing 3363 records and I am trying a cross-validation to understand the process. When using the cv.glm, code below, I get mean of perr1 of 0.2336 and SD of 0.000139. When using a

Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > On May 3, 2018, at 3:01 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > You can't -- at least as I read the docs for ?optim (but I'm pretty > ignorant

I'm trying to use Rprof() to identify bottlenecks and speed up a particullary slow section of code which reads in a portion of a tif file and compares each of the values to values of predictors used for model fitting. I've written up an example that anyone can run. Generally temp would be a section of a tif read into a data.frame and used later for other processing. The first portion

Hi Michael, A few comments 1. To add the constraint sum(wgt.vect=1) you would use the method of Lagrange multipliers. What this means is that in addition to the w_i (the components of the weight variables) you would add an additional variable, call it lambda. Then you would modify your optim.fun() function to add the term lambda * (sum(wgt.vect - 1) 2. Are you sure that you have defined

You can't -- at least as I read the docs for ?optim (but I'm pretty ignorant about this, so maybe there's a way to tweak it so you can). See here: https://cran.r-project.org/web/views/Optimization.html for other R optimization capabilities. Also, given your credentials, the r-sig-finance list might be a better place for you to post your query. Cheers, Bert Bert Gunter

Hello, I was used to count how many times every factor is found the following way : (assuming 'vect' is a vector of factors) tapply(vect,vect,length) but recently I experienced weird results with this In fact, my command line was tapply(ORGMORE[[1]][vect],ORGMORE[[1]][vect],length) where - 'vect' is a vector of indices - 'ORGMORE[[1]]' is a slightly long vector of

On Thu, May 3, 2018 at 2:03 PM, Michael Ashton <m.ashton at enduringinvestments.com> wrote: > Thanks Bert. But everyone on that forum wants to use finance tools rather than general optimization stuff! And I am not optimizing a traditional Markowitz mean-variance problem. Plus, smarter people here. :-) > I'm very confused by these statements. Most of the "finance tools"

The current llvm/llvm-gcc4.2 svn is now fixed with respect to the extra warnings that were being emitted by the gfortran compiler. The gfortran testsuite results under Intel Darwin9 are appended below. Jack Native configuration is i686-apple-darwin9 === gfortran tests === Running target unix/-m32 FAIL: gfortran.dg/aint_anint_1.f90 -O (internal compiler error) FAIL:

Hi, I was trying to get error bars in my matplot. I looked at an earlier thread, and the sample code that I made is: #------------------ library(plotrix) mat1 <- matrix(sample(1:30,10),nrow=5,ncol=2) ses <- matrix(sample(1:3,10,replace=T),nrow=5,ncol=2) vect <- seq(20,100,20) rownames(mat1) <- rownames(ses) <- vect colnames(mat1) <- colnames(ses) <- letters[1:2]

Hi there, I have a numeric vector let say: Vect <- c(12.234, 234.5675, 1.5) Now I want a string vector like: changedVec <- c("012.234", "234.568", "001.500") Would be grateful if somebody help me how can I do that. Thanks and regards, [[alternative HTML version deleted]]

Building the prerelease of llvm-gcc 2.4 on Intel darwin9 with the following patch... --- llvm-gcc-4.2-2.3.999-20081024.source/gcc/stub-c.c.org 2008-10-30 18:55:45.000000000 -0400 +++ llvm-gcc-4.2-2.3.999-20081024.source/gcc/stub-c.c 2008-10-30 18:57:29.000000000 -0400 @@ -157,3 +157,27 @@ { gcc_assert(0); } + + +bool cvt_utf8_utf16 (const unsigned char *, size_t, unsigned char **, +

Hi, I search a solution to record data in dynamic structures in R. I have an algorithm that will be executed each step and whose output is an array of doubles with unknown size. The solution I found is to use lists 1) I initialise my list l <- list() 2) and at a step numbered i I conacatain the new tab to the list vect <- algorithm() l <<--c( l , list(stepi=vect)) The problem is

Hi, Try this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun1<- function(dat){??? ? ??? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),]) ???

Hi R-users, I have to define a noise level function L and its energy in the various moment of the day by: if time is between 18:00:00 and 23:59:59 then L[j] <- L[j]+5 and W <- 10^((L+5)/10) if time is between 22:00:00 and 05:59:59 ==> L <- L+10 and W <- 10^((L+10)/10) else L=L and W = W Could someone help me to realize this function please? You will find my following