similar to: Help with segmented package

Hi everyone. I'm trying to use the segmented function with the following data: For instance, I use segmented package as follow: myreg2 = lm(xy$y ~ xy$x) mysegmented = segmented(myreg2, seg.Z=~x, psi=c(245000), control = seg.control(display=FALSE)) Which get me to the following error : As a break point, a starting guess of 245000 seems fair. Anyone has an idea why I'm getting such

I am trying to fit a very simple broken stick model using the package "segmented" but I have hit a roadblock. > str(data) 'data.frame': 18 obs. of 2 variables: $ Bin : num 0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 ... $ LnFREQ: num 5.06 4.23 3.50 3.47 2.83 ... I fit the lm easily: > fit.lm<-lm(LnFREQ~Bin, data=id07) But I keep getting an error

Hi everyone. I'm trying to use small letter phi in a graph produced in R. However, the small letter phi does not look as it should. In fact, it looks like this: http://r.789695.n4.nabble.com/file/n3963311/Untitled.png instead of what is here http://en.wikipedia.org/wiki/Phi Here's the code I use: expression(phi [1]) Anyone has an idea? With regards, Phil -- View this message

Hi everyone. I have these data : myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type = c('x','x','y','z')) which gives me: Name length type 1 a 1 x 2 a 2 x 3 b 3 y 4 b 4 z I would group (mean) this DF using 'Name' as grouping factor. However, I have a

Dear R users. I'm currently making a report with knitr (RStudio) where I would like to plot a googleVis map. However, the map generated is an HTML file which I don't know how to integrate it in my report. So my question is how to include a map generated with googleVis in a PDF created with knitr/sweave. Regards, Phil -- View this message in context:

Hi everyone. Is it possible in R to create a matrix or a list (vector) or R object. For instance, I have f1 <- function(x) sqrt(x%*%x); f2 <- function(x) (2x+1); I would like to do something like L <- List(); L[1] = f1; L[2] = f2; So, is there a way to create matrix or vector that can contains R object. With regards, Phil -- View this message in context:

Hi everyone, while trying to use 'segmented' (R i386 2.15.0 for Windows 32bit OS) to determine the breakpoint I got stuck with an error message and I can't find solution. It is connected with psi value, and the error says: Error in seg.glm.fit(y, XREG, Z, PSI, weights, offs, opz) : (Some) estimated psi out of its range This is the code I am using:

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Hi everyone. I performing a simple PCA using the princomp function. Then, I use the biplot function to show it. However, the function use line number to represent samples. I would like to know if there's a way to use a dot (point) instead of the line number when using the biplot function. With regards, Phil -- View this message in context:

Hi everyone. I'm using matrix product such as : #Generate some data NCols = 5 NRows = 5 A = matrix(runif(NCols*NRows), ncol=NCols) B = matrix(runif(NCols*NRows), ncol=NCols) #First calculation R = A%*%B for(i in 1:100) { R = R%*%B } I would like to know if it was possible to avoid the loop by using something like mapply or anything else. Tx in advance, Phil -- View this message

Hi every one. I have a exponential function (3 fitting parameters) that I would like to use to produce data (6 series) without having to use a loop. Here wl = seq(300,500,1) k1 = c(1.2e-6, 4.9e-6, 9.6e-6, 2.7e-10, 6.7e-8, 7.44e-6) k2 = c(726, 352, 128, 5232, 1538, 128) k3 = c(-176, -224, -257, 88.7, -111, -256) stations = c('R5d', 'R5a', 'R9', '108',

Dear R users, I'm currently trying to re-project a geotiff in another coordinate system. For instance, I have a tif image in UTM 19 zone which I would like to reproject into UTM 18. I was wondering if it was possible in R. Furthermore, I looked into 'rgdal' package, but I can't really find out if I'm doing the right thing. So far, here is what I'm doing: library(rgdal)

Hi all. I'm ploting a raster and I can't find the proper way to move the legend. For example, r = raster(system.file("external/test.grd", package="raster"))plot(r) How can I put the legend at the desired position? Thank in advance,Phil [[alternative HTML version deleted]]

Hi everyone. I have have a dataset with daily measurement from January 1st of 1966 up to December 31th of 2011. Here's the first part of the data: Date SLEV 1/1/1966 1.086 1/2/1966 1.079 1/3/1966 1.133 1/4/1966 1.261 1/5/1966 1.391 1/6/1966 1.571 1/7/1966 1.728 1/8/1966 1.823 1/9/1966 1.97 1/10/1966 1.804 1/11/1966 2.02 1/12/1966 2.017 1/13/1966 1.86 1/14/1966 1.96 1/15/1966 1.813 1/16/1966

Hello, I am interested in fitting a segmented model with unknown joint points in nls and perhaps eventually in nlme. I can fit this model in sas (see below, joint points to be estimated are a41 and a41), but am unsure how to specify this in the nlm function. I would really appreciate any suggestions or example code. Thanks a lot. -andy proc nlin data=Stems.Trees; params b41=-3 b42=1.5

Hi everyone. I have a quick question regarding the look of my legend in my plot. As you can see in the next figure, I have 3 series. http://r.789695.n4.nabble.com/file/n3067466/legend.png However, I find rather difficult to differentiate the series 1 and 3 according to their line type (lty). I would like to know if it was possible to make the line type in the legend to appear more clearly.

Most of the data sets I'm dealing with exhibit a time trend. We would like to get rid of the time trend. The plot shows in some cases a monotonic increase of the dependent variable with time. This is the easiest case. In some other cases the plot shows a time trend where the dependent variable changes slope 4-5 times along the observations measurement period. I've attempted a segmented

Hi, I appreciate your help with the segmented function. I am relatively new to R. I followed the introduction of the 'segmented'-package by Vito Muggeo, but still it does not work. Here are the lines I wrote: data_test<-data.frame(x=c(1:10),y=c(1,1,1,1,1,2,3,4,5,6)) lr_test<-lm(y~x,data_test) seg_test<-segmented(lr_test,seg.Z~x,psi=1) /error in segmented.lm(lr_test, seg.Z ~ x,

I tried to install "segmented" from three different repositories but I keep getting the following message on R window: ... Please, advice where to find this package. Thank you in advance. Maura > utils:::menuInstallPkgs() --- Please select a CRAN mirror for use in this session --- Warning: package 'segmented' is in use and will not be installed > utils:::menuInstallPkgs()

Hi everyone. I have a long label that I would like to split. I found that I could use "strwrap" for simple text. However, this is not working with this label: str = expression(paste("< 20 ?m phytoplankton ","(cells ? ",mL^-1,")")) plot(...., ylab = strwrap(str,20),...) I suspect this is because I'm using "expression" for form my label.