similar to: how to convert string to object?

If I have a model line = lm(y~x1) and I want to use a for loop to change the number of explanatory variables, how would I do this? So for example I want to store the model objects in a list. model1 = lm(y~x1) model2 = lm(y~x1+x2) model3 = lm(y~x1+x2+x3) model4 = lm(y~x1+x2+x3+x4) model5 = lm(y~x1+x2+x3+x4+x5)... model10. model_function = function(x){ for(i in 1:x) { } If x =1, then the list

s= "Hey" a = "Hello" table = rbind(s,a) write.table(table,paste("blah",".PROPERTIES",sep = ""),row.names = FALSE,col.names = FALSE) In my table, how do I output only the words and not the words with the quotations? -- View this message in context: http://r.789695.n4.nabble.com/How-to-unquote-string-in-R-tp3025654p3025654.html Sent from the R

If I have a list: list = c(~garch(1,1), ~arma(1,1)) and I run typeof(list[1]), the output is a list object. But I want each element in the list to be a language object. How do I transform these list objects to language objects? -- View this message in context: http://r.789695.n4.nabble.com/how-to-convert-list-to-language-object-tp2966813p2966813.html Sent from the R help mailing list archive at

R-help - I have this set of aggregated tables (sample data below via dput()). And I would like to have delayValue as the column variables with the "temp" (temp1, temp2, temp3) values as the row variables. However I would like to have the temp variables *aggregated into single rows* so that I have the frequency ("Freq" | counts) of each time each "delayValue" occurs

Hi, Could somebody say if it is possible to compute APARCH-models with garchFit commands. I have earlier used aaa (garchOxFit) and now I try to use bbb (look below) aaa <- garchOxFit(formula.mean=~arma(1,0),formula.var=~aparch(1,1),series=nyk,cond.dist=c('gaussian')) bbb <- garchFit(formula=~arma(1,0)+aparch(1,1),data=nyk) aaa works well, but I need other characteristics of

i have a time Series of IBM closing px from 1/1/2000 to today I want to graph the time serie by dividing the graph by year and month all the monthly graphs with the same year will go to one page. so from 1/1/2000 to 11/19/2010. i will have 11 pages, and each page will have 12 graphs (jan to dec) except for 2010. I am able to do it in R, but when i use sweave, I can only print the last page.

Hi all, I have a C code in Linux, it has 7 pointers and compile e run OK, but when I run in R happens problems with calloc function, it returns NULL. ############################################### > int *temp1,*temp2,*temp3,*temp4; temp1 = (int *)calloc(col,sizeof(int)); if(temp1 == NULL){ printf("\n\n No Memory1!"); exit(1); } temp2 = (int *)calloc(col,sizeof(int));

Hi there, I'm trying to fit a GJR-GARCH Model using fGarch. I wanted to try that by fitting an APARCH model with a fixed delta of 2 and a non-fixed gamma. So I was simply trying to use: spec <- garchFit(~aparch(1,1),data=garchSim(),delta=2) coef(spec) And sometimes, it's working like a charm and delta is indeed exactly 2 in the resulting coefficient vector. Frequently, though, the

I am using R 2.1.1 in an windows XP environment. I have 2 dataframes, temp1 and temp2. Each dataframe has 20 variables (“cocolumns") and 525 observations (“rows”). All variables are numeric. I want to create a new dataframe that also has 20 columns and 525 rows. The values in this dataframe should be the sum of the 2 other dataframe. (i.e. temp1$column

Hi, temp3<- read.table(text=" ID CTIME WEIGHT HM001 1223 24.0 HM001 1224 25.2 HM001 1225 23.1 HM001 1226 NA HM001 1227 32.1 HM001 1228 32.4 HM001 1229 1323.2 HM001 1230 27.4 HM001 1231 22.4236 #changed here to test the previous solution ",sep="",header=TRUE,stringsAsFactors=FALSE) ?tempnew<- na.omit(temp3) ?grep("\\d{4}",temp3$WEIGHT) #[1] 7 9 #not correct

HI Irucka, Please check this: temp<- structure(list(`:Bostoncitydata` = structure(list(Month = c(1L, 2L, 3L, NA), Data1 = c(1.5, 12.3, 11.4, NA), Data2 = c(9.1342, 12.31, 3.5, NA)), .Names = c("Month", "Data1", "Data2"), class = "data.frame", row.names = c(NA, -4L)), `:Chicagocitydata` = structure(list(Month = c(1L, 2L, 3L, 4L, 5L, NA), Data1 = c(1.52,

I stumbled onto this working on an update to coxph. The last 6 lines below are the question, the rest create a test data set. tmt585% R R version 2.12.2 (2011-02-25) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: x86_64-unknown-linux-gnu (64-bit) # Lines of code from survival/tests/singtest.R > library(survival) Loading required package: splines

Hi, I am writing a function to plot a pdf of a distribution, GNL.pdf.fn = function(x,mu,sigma,alpha,beta,rho) { y = x-rho*mu cf.fn = function(s){ cplex = complex(1,0,1) temp1 = alpha*beta*exp(-sigma*s^2/2) temp2 = (alpha-cplex*s)*(beta+cplex*s) out = (temp1/temp2)^rho out } temp.fn = function(s){ (Mod(cf.fn(s)))*cos(Arg(cf.fn(s))-s*y) } int.fn =

On Mon, September 14, 2015 21:28, Always Learning wrote: > > On Mon, 2015-09-14 at 14:51 -0400, James B. Byrne wrote: > >> The Logwatch imapd service script distributed with CentOS-6 does not >> generate anything when I run logwatch --service all on a cyrus-imapd >> host. Is this expected behaviour? Is there a separate script for >> cyrus-imapd or are their

Dear all, I am working on a data set in which I have sequentially measured egg temperatures ("eggtemp") in birds incubating in different ambient temperatures ("treat", sample data set below), "id" is not replicated within treatment. id treat eggtemp 1 79 3 30.90166 2 42 3 34.94044 3 10 3 32.69945 4 206 3 36.64127 5 23 3 31.80055 6

Dear All, I am very new to R - trying to teach myself it for some MSc coursework. I am plotting temperature data for two different sites over the same time period which I have downloaded from a university weather station data archive. I am using the following code to create the plot plot ( x = TEMP3[,"TIME"], y = TEMP3[,"TEMP"], type = "p", col =

Hi Irucka, I tried it and was able to plot it without any errors.? Here, your code indicates you need two lines. temper[[i]][1] ?temper[[1]][1] # which is the column 1. ? Month 1???? 1 2???? 2 3???? 3 ?temper[[1]][2] #? Data1 #1?? 1.5 #2? 12.3 #3? 11.4 Suppose I use names(temper) instead of seq_along(temper) pdf("irucka.pdf") ?lapply(names(temper),function(i)

Full_Name: Jim Breaux Version: 1.9.0 OS: WinXP Submission from: (NULL) (209.78.110.135) dir.create() is reporting a warning that a directory already exists when in fact the directory does NOT already exist. However, the erroneous behavior does not manifest itself until dir.create() has been called at least once to create a directory that DOES already exist. For example: # Create a directory

What is the syntax for this? If you have: vector = c(1,2,3,4), how would you output this to column A of an excel spreadsheet? -- View this message in context: http://r.789695.n4.nabble.com/how-to-you-output-a-vector-to-a-column-in-excel-tp2530470p2530470.html Sent from the R help mailing list archive at Nabble.com.

If you have: Name Value A 2 A 3 A 4 B 5 B 6 B 7 In R how do you assign one value to the name: A:9 B:18 -- View this message in context: http://r.789695.n4.nabble.com/for-loop-help-tp2539621p2539621.html Sent from the R help mailing list archive at Nabble.com.