similar to: piecewise linear approximation

Hi all! I'm afraid I programmed something totally non-sensical and inefficient, but I can't figure out how to do it better. I have a list of ~ 40 000 characters. I want to take each element at a time, map it to a large data frame with hit=which(data.frame$column==elementFromList), then compute some statistic on data.frame[hit,] and return a result that consists of either 1) a list

Hello, I have been following the thread dated Monday, October 9, 2006 when Kamila Naxerova asked a question about plotting elliptical shapes. Can you explain the equations for X and Y. I believe they used the parametric form of x and y (x=r cos(theta), y=r sin(theta). I don't know what r is here ? Can you explain 1)the origin of these equations and 2) what is r? Sincerely, Mary A. Marion

Is there a way to plot elliptical shapes? symbols() only provides circles... Thanks! Kamila

Hi everyone, disturbingly, a PCA yields different results when I run the analysis under Windows and Linux. I open the same workspace (which is on an independent partition) and perform the exact same function (by using the history file). Interestingly, PC1 is the same, but PC2 comes out totally different. Thanks! Kamila Windows: platform i386-pc-mingw32 arch i386 os

Hi! I don't understand this: layout(matrix(c(1:10),5,2),heights=c(1,rep(2,4))) plot(1,1) error in plot.new() : plot region too large Why?????? Thanks! Kamila

Hi all, is the Matrix package no longer available for download via install.packages? When I try to install it (from any mirror), I get the following error message: install.packages("Matrix",lib="/home/kn52/R-2.5.0/library") Warning message: package 'Matrix' is not available in install.packages("Matrix",lib="/home/kn52/R-2.5.0/library") Other

I have a seasonal time series. I want to calculate the annual mean value of the time series at its peak (say the mean of the three values before the peak, the peak, and the three values after the peak). The peak of the time series might change cycle slightly from year to year. # E.g., nPts <- 254 foo <- sin((2 * pi * 1/24) * 1:nPts) foo <- foo + rnorm(nPts, 0, 0.05) bar <- ts(foo,

Hi, I need to fit a piecewise linear regression. x = c(6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50,50.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c(0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.186,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) there are two change points. so the fitted curve should look like \ \ /\

Hi, I'm using the R library(car) to draw confidence/prediction ellipses in a scatterplot. >From what i understood the ellipse() function return an ellipse based parameters: shape, center, radius . If i read dataEllipse() function i can see how these parameters are calculated for a confidence ellipse. ibrary(car) a=c(12,12,4,5,63,63,23) b=c(13,15,7,10,73,83,43) v <-

I need to find the area under a trapezoid for a research-related project. I was able to find the area under the trapezoid in MATLAB using the code: function [int] = myquadrature(f,a,b) % user-defined quadrature function % integrate data f from x=a to x=b assuming f is equally spaced over the interval % use type % determine number of data points npts = prod(size(f)); nint = npts -1; %number of

Im trying to ascertain whether or not the facilities of R are sufficient for solving an optimization problem I've come accross. Because of my limited experience with R, I would greatly appreciate some feedback from more frequent users. The problem can be delineated as such: A utility function, we shall call g is a function of x, n ... g(x,n). g has the properties: n > 0, x lies on the

Dear all, I coded a function called u.glm u.glm <- function (x,ahi,age,bmiz,gender) { library(nortest) lil.rslt <- lillie.test(x) if (lil.rslt$p.value >0.05) { cat("Logtrans=0, lillie=",lil.rslt$p.value,"\n") xmodel<-glm(x~ahi+age+bmiz+as.factor(gender)) summary(xmodel) confint(xmodel) } else { cat("Logtrans=1,

I have found some "issues" (bugs?) with nls confidence intervals ... some with the relatively new "port" algorithm, others more general (but possibly in the "well, don't do that" category). I have corresponded some with Prof. Ripley about them, but I thought I would just report how far I've gotten in case anyone else has thoughts. (I'm finding the code

Hi! I just started to use R recently, and would like to ask a help about automating the job. I need to use "kmeans" function with my own 300 data files, and wonder if it's possible to do it automatically. For example, > library (mva) > mydata <- read.table ("data1") > cl <- kmeans(mydata, 5, 20) and I just need to save "cl" info (i.e. the center

thanks to all for the 2d scatter plot. i have one more. how do i plot 'hist(y1, col="red") and hist(y2,col="blue") in the same window? thanks again.

1) FLAC supports only MSVS 2005 and newer, so (_MSC_VER >= 1400) is always true and can be removed. 2) The argument for FLAC__clz_uint32() is of FLAC__uint32 type, so FLAC__clz_soft_uint32() should have the same argument type. 3) The patch removes unnecessary parentheses around 'word' variable. 4) It also replaces "sizeof(FLAC__uint32) * CHAR_BIT - 1 -

A function I've been using for a while returned a surprising [to me, given the data] error recently: Error in plot.window(xlim, ylim, log, asp, ...) : Logarithmic axis must have positive limits After some digging I realized what was going on: x <- c(10460.97, 10808.67, 29499.98, 1, 35818.62, 48535.59, 1, 1, 42512.1, 1627.39, 1, 7571.06, 21479.69, 25, 1, 16143.85, 12736.96,

PR #7690 points out that if the confidence intervals (+/-1.58 IQR/sqrt(n)) in a boxplot with notch=TRUE are larger than the hinges -- which is most likely to happen for small n and asymmetric distributions -- the resulting plot is ugly, e.g.: set.seed(1001) npts <- 5 X <- rnorm(2*npts,rep(3:4,each=npts),sd=1) f <- factor(rep(1:2,each=npts)) boxplot(X~f) boxplot(X~f,notch=TRUE) I can

Dear All, I would like to plot text with a box around it. I used strwidth and strheight to compute the size of the box which is plotted with rect: z <- rnorm(10) # horizontal text works plot(rnorm(10)) x1 <- 5 y1 <- 0 label <- "Label" cha <- paste(" ", label, " ", sep = "") xh <- strwidth(cha, cex = par("cex")) yh <-

[posted to R-devel, no discussion: resubmitting it as a bug, just so it gets logged appropriately] Sorry to report further difficulties with nls and profiling and constraints ... the problem this time (which I didn't check for in my last round of testing) is that the nls profiler doesn't seem to respect constraints that have been set when using the port algorithm. See test code