similar to: Plotting linear fit

Is there an R function to generate a radar or spider graph from a table - e.g. radar(table(x)) or some such? ================================================== Isaac T. Van Patten, Ph.D. Professor Department of Criminal Justice Box 6934, Radford University Radford, VA 24142 540-831-6148 ivanpatt@radford.edu <mailto:ivanpatt@radford.edu> http://ivanpatt.asp.radford.edu

I have a factor vector of subject races (Asian, Black, Hispanic, White; n=30) that I want to plot with a Cleveland dotplot or dotchart. I tried the following in R2.12.1 : > dotchart(table(school$Race)) Error in plot.xy(xy.coords(x, y), type = type, ...) : invalid plot type Using the same data set in R2.11.1 the operation succeeded (I tried several variations to be sure): >

R 2.7 Windows XP I have two model that have been run using exactly the same data, both fit using glm(). One model is a linear regression (gaussian(link = "identity")) the other a quasipoisson(link = "log"). I have log likelihoods from each model. Is there any way I can determine which model is a better fit to the data? anova() does not appear to work as the models have the

Dear list members I have a doubt on how p-values for t-statistics are calculated in the summary of Linear Models. Here goes an example: x <- rnorm(100,50,10) y <- rnorm(100,0,5) fit1<-lm(y~x) summary(fit1) summary(fit1)$coef[2] # b summary(fit1)$coef[4] # Std. Error summary(fit1)$coef[6] # t-statistic summary(fit1)$coef[8] # Pr(>|t| summary(fit1)$df [2] # degrees of freedom #

Dear R users, I d like to assess the effect of "treatment" covariate on a disease relapse risk with the package cmprsk. However, the effect of this covariate on survival is time-dependent (assessed with cox.zph): no significant effect during the first year of follow-up, then after 1 year a favorable effect is observed on survival (step function might be the correct way to say that ?).

Dear list, I am getting weired with fitting data with a 1/x-polynomial. Suggest I have the following data: x <- c(1,2,3,4,5,6,7) y <- c(100,20,4,2,1,.3,.1) I may fit this with a linear model fit1 = lm(y ~ I(x)) Getting plot out of this model I applied library(polynom) pol1 = polynomial(fit1$coefficients) f1 = as.function(pol1) plot(x,y) lines(x, f1(x), col = 2) Clearly, this model

I am trying to extract coefficients from lars fit and can't find how to get intercept. E.g. y = rnorm(10) x = matrix(runif(50),nrow=10) X = data.frame(y,x) fit1 = lars(as.matrix(X[,2:6]),as.matrix(X[,1])) fit2 = lm(y~.,data=X) Then, if I do: > predict(fit1,s=1,mode='fraction',type='coefficients')$coef X1 X2 X3 X4 X5 0.3447570

Dear all, I have 3 models (from simple to complex) and I want to compare them in order to see if they fit equally well or not. From the R prompt I am not able to see where I can get this information. Let´s do an example: fit1<- lm(response ~ stimulus + condition + stimulus:condition, data=scrd) #EQUIVALE A lm(response ~ stimulus*condition, data=scrd) fit2<- lm(response ~ stimulus +

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

I am merging two dataframes using a relational key (incident number and incident year), but not all the records match up. I want to be able to review only the records that cannot be merged for each individual dataframe (essentially trying to select records from one dataframe using a multi-value relational key from the other dataframe). The following code shows what I am trying to do. The final

Hi, There is something I don't get with object of class "mle" returned by a function I wrote. More precisely it's about the behaviour of method "confint" and "profile" applied to these object. I've written a short function (see below) whose arguments are: 1) A univariate sample (arising from a gamma, log-normal or whatever). 2) A character string

here is the data: y<-c(5,2,3,7,9,0,1,4,5) id<-c(1,1,6,6,7,8,15,15,19) t<-c(50,56,50,56,50,50,50,60,50) table1<-data.frame(y,id,t)//longitudinal data what I want to do is to use the linear model for each id ,then get the estimate value,like: fit1<-lm(y~t,data=table1,subset=(id==1)) but ,you can see the variable "id" is quite irregular,they are not arranaged in order

Hi! I have fitted two glms assuming a poisson distribution which are: fit1 <- glm(Aids ~ Year, data=aids, family=poisson()) fit2 <- glm(Aids ~ Year+I(Year^2), data=aids, family=poisson()) I am trying to work out how to represent the fitted regression curves of fit1 and fit2 on the one graph. I have tried: graphics.off() plot(Aids ~ Year, data = aids) line(glm(Aids ~ Year,

Dear all, I have been trying to fit my data (only right censored) with gumbel distribution using fitdistrplus. I am getting very high standard error. I have been wondering why. The followings are the outputs: fit1=fitdistcens(dr0, "gumbel", start=list(a=99, b=0.6), optim.method= "L-BFGS-B", lower = 0.0, upper = Inf) > summary(fit1) FITTING OF THE DISTRIBUTION ' gumbel

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

Hi, I'm really struggling with barplot I have a data.frame with 3 columns. The first column represents an "incident" type The second column represents a "month" The third column represents a "time" Code for a sample data.frame incidents <- rep(c('a','b','d','e'), each =25) months <- rep(c(1,2), each =10) times

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Hi, I just encountered what I thought was strange behavior in MDS. However, it turned out that the mistake was mine. The lesson learned from my mistake is that one should plot on a square pane when plotting results of an MDS. Not doing so can be very misleading. Follow the example of an equilateral triangle below to see what I mean. I hope this helps others to avoid this kind of headache.

Dear List, I used rlm to calculate two regression models for two data sets (rlm due to two outlying values in one of the data sets). Now I want to compare the two regression slopes. I came across some R-code of Spencer Graves in reply to a similar problem: http://www.mail-archive.com/r-help at stat.math.ethz.ch/msg06666.html The code was: > df1 <- data.frame(x=1:10, y=1:10+rnorm(10))

I'm attempting to do model selection with AIC, using a glm and a lognormal distribution, but: fit1<-glm(BA~Year,data=pdat.sp1.65.04, family=gaussian(link="log")) ## gives the same result as either of the following: fit1<-glm(BA~Year,data=pdat.sp1.65.04, family=gaussian) fit1<-lm(BA~Year,data=pdat.sp1.65.04) fit1 #Coefficients: #(Intercept) Year2004 # -1.6341