search for: h_a_patience

Hi there, I have a spreadsheet in excel which consists of first column of dates and then subsequent columns that refer to prices of different securities on those dates. (the first row contains each series name) I saved the excel file as type csv and then imported to excel using prices=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Option prices.csv",header = TRUE, sep

Hi there, What is the best way to get a time series of daily stock price observations into R (from excel). The time series are daily but there are spaces for w/e's and holidays etc. So I am not sure a ts object will be suitable but I am not sure what I should use? What ever package you recemmned i need to be able to run a corrolelagram. Thanks Hugh -- View this message in context:

Hi there, I have a ts object that I would like wo split into arbirary sizes. Can'tfind how to do this? I realise its probaly very simple buy I can't sem to find the right function? -- View this message in context: http://r.789695.n4.nabble.com/Split-an-array-into-arbirtary-sizes-tp3948938p3948938.html Sent from the R help mailing list archive at Nabble.com.

Hi there, I have a largish optimisation problem (10 years of daily observations). I want to optimise between 4 and 6 parameters. I'd like to utilise parallel computing if I can as I will have to run it with different starting values etc. I have a quad core PC with 16GB ram running windows 7. Having done a little reading it seems the two best options for me are: 1.) Use the academic

m0<-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x) params<-all.vars(m0) this reads all the params from m0 so theta1,2 and 3 correct? params<-params[-which(params=="x")] checks which params are multiplied by x? np<-length(params) for(i in 1:6){ esp<-get(sprintf("m%d",i-1))

Hi there, I ran the following code: vols=read.csv(file="C:/Documents and Settings/Hugh/My Documents/PhD/Swaption vols.csv" , header=TRUE, sep=",") X<-ts(vols[,2]) #X dcOU<-function(x,t,x0,theta,log=FALSE){ Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t) Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2]) dnorm(x,mean=Ex,sd=sqrt(Vx),log=log) }

Hi there, I am reletively new user I only dowloaded is about a week ago. I was getting along fine but last night I tried to selected save workspace. Since then R will not work!!!! And I really really need it. There are two eror massages: The first if a pop up box is "Fatal error: unable to resolve data in .RData." The second is a message in the GUI: "Error in

Hi there, I am trying to run the following code: > dcOU<-function(x,t,x0,theta,log=FALSE){ + Ex<-theta[1]/theta[2]+(x0-theta[1]/theta[2])*exp(-theta[2]*t) + Vx<-theta[3]^2*(1-exp(-2*theta[2]*t))/(2*theta[2]) + dnorm(x,mean=Ex,sd=sqrt(Vx),log=log) + } > OU.lik<-function(theta1,theta2,theta3){ + n<-length(X) + dt<-deltat(X) +

I get the following error: Error in backSpline.npolySpline(sp) : spline must be monotone Has anyone had this error before? any ideas on a workaround? > > vols=read.csv(file="C:/Documents and Settings/Hugh/My > Documents/PhD/Swaption vols.csv" + , header=TRUE, sep=",") > X<-ts(vols[,2]) > #X > > > dcOU<-function(x,t,x0,theta,log=FALSE){ +

Hi there, following the instructions shown here: http://www.r-bloggers.com/ec2-micro-instance-of-rstudio/ I used sudo apt-get install r-baseto insall R on an Amazon EC2 server, but when I do that I get an old cersion of R > R.version.string [1] "R version 2.12.1 (2010-12-16)" Do you know how I can change this command so that it installs the latest version?Also are the ubuntu

Hi there, When I run the code below I get the error Error in xy.coords(x, NULL, log = log) :(list) object cannot be coerced to type 'double' Any tips how I can resolve this? > > library("waveslim") > > vols=read.csv(file="C:/Users/ocuk/My Documents/Abs Vol.csv", header=TRUE, > sep=",") > x<-c(vols[,1]) > #x > #data(ibm) >

x<-matrix(c(1,0,0,0,1,0,0,0,1),nrow=3) > y<-matrix(c(0,0,0,1,0,0,1,1,0),nrow=3) > z<-matrix(c(0,1,0,0,1,0,1,0,0),nrow=3) > x[z]<-y[z] The resultant matrix x is all zeros except for the last two diagonal cells which are 1's. While y is lower triangualr 0's with the remaining cells all ones. I really don't understand how this deceptively simple looking piece of