similar to: Using lm coefficients in polyroot()

Dear useRs, Is there any more elegant way to convert dataframe to a vector of all its values than as.vector(as.matrix(x)) ? I did not have to do such conversion yet, so I am not sure... (of course as.vector() alone does not work). Regards, -- Lukasz Komsta Department of Medicinal Chemistry Medical University of Lublin Jaczewskiego 4, 20-090 Lublin, Poland Fax +48 81 7425165

Dear useRs, I would like to summarize results of several tests in groups of two columns - statistic, p-value, statistic, p-value etc. There would be nice to add significance stars, but printCoefmat allows to do it only to last column. Is there any way to do format such table without writing my own complicated function? Thank you in advance, -- Lukasz Komsta Department of Medicinal

Dear useRs, I have a curve which is a mixture of Gaussian curves (for example UV emission or absorption spectrum). Do you have any suggestions how to implement searching for optimal set of Gaussian peaks to fit the curve? I know that it is very complex problem, but maybe it is a possibility to do it? First supposement is to use a nls() with very large functions, and compare AIC value, but it is

Dear R Development Team, I have encountered the following difficulty in using the function polyroot under either NT4.0 (R version 1.2.1) or linux (R version 0.90.1). In the provided example, the non-zero root of c(0,0,0,1) depends on the results of the previous call of polyroot. R : Copyright 2001, The R Development Core Team Version 1.2.1 (2001-01-15) R is free software and comes with

In a bug report from Nov.28 2000, Li Dongfeng writes: ----- I have found that the polyroot() function in R-1.1.1(both solaris and Win32 version) gives totally incorrect result. Here is the offending code: # Polyroot bug report: # from R-1.1.1 > sort(abs(polyroot(c(1,-2,1,0,0,0,0,0,0,0,0,0,-2,5,-2,0,0,0,0,0,0,0,0,0,1,-2)))) [1] 0.8758259 0.9486499 0.9731015 1.5419189 1.7466214 1.7535362

I have found that the polyroot() function in R-1.1.1(both solaris and Win32 version) gives totally incorrect result. Here is the offending code: # Polyroot bug report: # from R-1.1.1 > sort(abs(polyroot(c(1, -2,1,0,0,0,0,0,0,0,0,0,-2,5,-2,0,0,0,0,0,0,0,0,0,1,-2,1)))) [1] 0.8758259 0.9486499 0.9731015 1.5419189 1.7466214 1.7535362 1.7589484 [8] 2.0216317 2.4421509 2.5098488 2.6615572

Hi, All: Is there a simple way to detect complex conjugates in the roots returned by 'polyroot'? The obvious comparison of each root with the complex conjugate of the next sometimes produces roundoff error, and I don't know how to bound its magnitude: (tst <- polyroot(c(1, -.6, .4))) tst[-1]-Conj(tst[-2]) [1] 3.108624e-15+2.22045e-16i

Dear useRs, I was searching CRAN for implementation of kurtosis and skewness tests, and found that there is some kind of lack on it. So, I have written three functions: 1. Anscombe-Glynn test for kurtosis 2. Bonett-Seier test based on Geary's kurtosis (which is not widely known, but I was inspired by original paper describing it, found coincidentally in Elsevier database) 3.

Hi, I'm using a fairly simple HP Compaq desktop PC running Windows 2K. When running a large process in R, the process "RGUI.exe" will never exceed 50% of the CPU usage. The program used to be able to use more of the computer, but does not now. I don't believe this is a multiple processor machine. Can anyone give any advice on how to solve the problem? Thanks, Michael

I'd just like to report a possible R bug--or rather, confirm an existing one (bug #751). I have had some difficulty using the polyroot() function. For example, in Win 98, R 1.1.1, > polyroot(c(2,1,1)) correctly (per the help index) gives the roots of 1 + (1*x) + (2*x^2) as [1] -0.5+1.322876i -0.5-1.322876i However, > polyroot(c(-100,0,1)) gives the roots of [1] 10+0i -10+0i

Dear All, I have small samples of data (between 6 and 15) for numerious time series points. I am assuming the data for each time point is normally distributed. The problem is that the data arrvies sporadically and I would like to detect the number of outliers after I have six data points for any time period. Essentially, I would like to detect the number of outliers when I have 6 data points then

Is there an easy way to compare complex numbers? Here is a small example: > (z1=polyroot(c(1,-.4,-.45))) [1] 1.111111-0i -2.000000+0i > (z2=polyroot(c(1,1,.25))) [1] -2+0i -2+0i > x=0 > if(any(identical(z1,z2))) x=99 > x [1] 0 # real and imaginary parts: > Re(z1); Im(z1) [1] 1.111111 -2.000000 [1] -8.4968e-21 8.4968e-21 > Re(z2); Im(z2) [1] -2

El problema del módulo es que pierde el signo. En tu caso sale igual porque has invertido el signo del coeficiente en el polinomio (en realidad se me pasó a a mí advertir que el término independiente debe ir con signo negativo): .> polyroot(z=c(0.5,0,0,0,0,1)) [1] 0.7042902+0.5116968i -0.2690149+0.8279428i -0.2690149-0.8279428i [4] 0.7042902-0.5116968i -0.8705506+0.0000000i .> .>

Dear R people: Is there a way to perform simple polynomial multiplication; that is, something like (x - 3) * (x + 3) = x^2 - 9, please? I looked in poly and polyroot and expression. There used to be a package that had this, maybe? thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: erinm.hodgess at

To R-helpers, R offers the polyroot function for solving mentioned equations iteratively. However, Dr Math and Mathworld (and other places) show in detail how to solve mentioned equations non-iteratively. Do implementations for R that are non-iterative and that solve mentioned equations exists? Regards, Mads Jeppe

Here is a relatively simple script (with comments as to the logic interspersed): # Some of these libraries are probably not needed here, but leaving them in place harms nothing: library(tseries) library(xts) library(quantmod) library(fGarch) library(fTrading) library(ggplot2) # Set the working directory, where the data file is located, and read the raw data

Full_Name: Kirk Hampel Version: 2.4.1 OS: Windows Submission from: (NULL) (144.53.251.2) Given an AR(p) model, the last p SE's are wrong. The source of the bug is that the C code (ver 2.4.0) assumes *npsi is the length of the psi vector (which is n+p), whilst the predict.ar function in R passes out as.integer(npsi), where npsi <- n-1. Some R code following reproduces the error. Let p=4,

hallo to all I've to calculate an arima model and I need only the first and 365 th parameter and also the sar1 and the intercept, so I'm traing with: arima(X,order=c(365,0,0),seasonal=list(order=c(1,0,0),..),fixed=c(NA,rep(0,363),NA,NA,NA),transform.pars=F) but the error answer is: Error in polyroot(z) : polynomial degree too high (49 max) also there are problems in allocating memory

R friends, I need to find the determinant of this matrix x 1 0 0 1 x 1 0 0 1 x 1 0 0 1 x det yields x^4-3x^2+1 I can then use polyroot to find the roots of the coefficients. The question is about the use of "x", which is what I'm solving for. thanks in advance, and this is a back-burner question. Apologies if I have posted this incorrectly/to the wrong place, I'm a newbie

Hi, I'm responding to a post about finding roots of a cubic or quartic equation non-iteratively. One obviously could create functions using the explicit algebraic solutions. One post on the subject noted that the square-roots in those solutions also require iteration, and one post claimed iterative solutions are more accurate than the explicit solutions. This post, however, is about