search for: nmof

Dear all, version 1.2-2 of package NMOF is on CRAN now. NMOF stands for 'Numerical Methods and Optimization in Finance'. The package provides R code and datasets for the book with the same name, written by Manfred Gilli, Dietmar Maringer and Enrico Schumann, published by Elsevier/Academic Press in 2011. The package has finally...

Dear all, version 0.20-0 of package NMOF is now on CRAN. 'NMOF' stands for 'Numerical Methods and Optimization in Finance'. The package accompanies the book with the same name, written by Manfred Gilli, Dietmar Maringer and Enrico Schumann, published by Elsevier/Academic Press in 2011. The package contains, in particul...

Dear all, version 0.20-0 of package NMOF is now on CRAN. 'NMOF' stands for 'Numerical Methods and Optimization in Finance'. The package accompanies the book with the same name, written by Manfred Gilli, Dietmar Maringer and Enrico Schumann, published by Elsevier/Academic Press in 2011. The package contains, in particul...

Dear All, I am using package fPortfolio to run minimum variance portfolio optimizations in R. I already know how to set portfolioSpecs, portfolio objects and constraints. Unfortunately I am not able to set the following type of constraints. I have a timeSeries object with returns data for roughly 1.5k assets for 261 subperiods (workingdays) and want to compute the global minimum variance

dear R wizards: `optimize()` requires the user to provide the brackets. I can write a bracketing routine, given a function and a starting point, but I was wondering whether there was already a "standard" user-exposed implementation. (Presumably, this is used in nlm, too; alas, nlm is in C, not native R.) regards, /iaw

Hi, I'm an R newbie and I've been struggling with a optimization problem for the past couple of days now. Here's the problem - I have a matrix of expected payouts from different stock option strategies. Each column in my matrix represents a different stock and each row represents the return to the strategy given a certain market move. So the rows are not a time series of percentage

Dear R-helpers, www.statistik.lmu.de/~*leisch*/*Sweave*/*Sweave*-Rnews-2003-2.pdf says that R CMD build creates the pdf from a .Rnw file. But when I do R CMD build followed by R CMD check it gives me a warning that there is a vignette without a PDF. My query is : when R CMD build works it creates a pdf in the doc subdirectory of the directory returned by system.file(package = mypackage) and not

I'd like to "step" through 24 histograms by using the return or click button option, as shown in the demo(graphics) demonstration. I've searched for "interactive graphics", and "waiting for page change" in R documentation but with no result. I'm sure that this is a relatively straightforward procedure. Can anyone point me to the correct solution?   Jabez

Hello, With which package can I build a yield curve using swap data with the bootstrapping method? Thanks for the reaction, André [[alternative HTML version deleted]]

Dear R users, I want to subset a daily zoo series according to its month, find % of "NA" in each month. I am finding it difficult to subset the daily dataset into monthly for the given operation.I am planning to do this for a huge dataset. Thanks in advance. Regards Vikram [[alternative HTML version deleted]]

Hi one col in my Excel file contains many numbers. But on line 3000 and some other lines are strings like "FG 1". "RODBS" seems to omit this lines. "gdata" works, but is much slower. Is this a bug of RODBC or do I apply it wrong? Example with the same "file.xlsx" library(RODBC); excel <- odbcConnectExcel2007("file.xlsx") tab <-

I don't quite know how to word what I want, but if I have (1, 2, 3); (a, b, c); (x, y) I want: 1 a x 1 b x 1 c x 1 a y 1 b y 1 c y 2 a ... and so forth What is the appropriate command? Best, Don -- View this message in context: http://r.789695.n4.nabble.com/Create-combinations-of-rows-tp3872641p3872641.html Sent from the R help mailing list archive at Nabble.com.

Hi, I need some neat ways of determing a subset of binary strings. For example, x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a subset of y, but x is not a subset of z. I tried to search R functions and packages but no hits. Any ideas? Best, Jing -- Jing Tang, PhD Senior Researcher Finnish Institute of Molecular Medicine (FIMM) FI-00014 University of

#using "rollapply" to calculate a moving sum or running sum? #I am tryign to use rollapply to calcualte a moving sum? #I tried rollapply and get the error message #"Error in seq.default(start.at, NROW(data), by = by) : # wrong sign in 'by' argument" #example: mymatrix <- ( matrix(data=1:100, nrow=5, ncol=20) ) mymatrix_cumsum <- ( matrix(data=NA, nrow=5,

Would someone be so kind as to provide example code where they use the suggestions argument in the rgba function In genalg? I can't get it to work. The following code works just fine: GenFit <-rbga(Lower, Upper, evalFunc = evaluate) Lower and Upper are each numeric vectors with 7 elements. Evaluate is an objective function. However, when I want to use a suggested chromosome, I get an

Dear All I have a constrained optimisation problem, I want to maximise the following function t(weights) %*% CovarianceMatrix %*% weights for the weights, subject to constraints on each element within the weights & the weights vector summing to 1. i.e. weights = (x1, x2, x3), where x1 is within some given range (a +b, a - b). I have tried to do this using the optim function in R,

Folks, I have a data frame with 4861469 rows that contains an ip address xxx.xxx.xxx.xxx as one of the columns. I want to assign a site to each row based on IP ranges. To do this I have a function to split the ip address as character into class A,B,C and D components. It works but is horribly inefficient in terms of speed. I can't quite see how one of the l/s/m/t/apply functions could be

How do I sum 15 years to my dataset? original dataset dates val 1 2001-01-12 1.2 2 2001-02-12 1.2 3 2001-03-12 1.2 result dates val 1 2016-01-12 1.2 2 2016-02-12 1.2 3 2016-03-12 1.2

I would like to run a for loop with an index going from 0 to 499 but the following seems to miss out the first value: C <- 499 for (i in 0:C) The alternative is: C <- 500 for (i in 1:C) { #Then every time I use i, I replace it with i-1 } Is this a good way to do it or is tere a better way? Thank you, ThomasThis message and any attachment are intended solely for the addressee and may

Dear R-help, I have been trying really hard to generate the following vector given the data (x) and parameter (alpha) efficiently. Let y be the output list, the aim is to produce the the following vector(y) with at least half the time used by the loop example below. y[1] = alpha * x[1] y[2] = alpha^2 * x[1] + alpha * x[2] y[3] = alpha^3 * x[1] + alpha^2 * x[2] + alpha * x[3] ..... below are