similar to: Problem with by statement for spaghetti plots

Hi all First post on the Wine forum and a newbie to Linux but I'm loving it. I was surprised how well Sugarsync ran in Wine until, when a window opened in Sugarsync and prompted me to update I accepted. Now it won't synchronise and has corrupted the cloud storage. I tried uninstalling then reinstalling but it was still the same. The version of Sugarsync I have upgraded to is 1.9.35 and

Greetings, I am new to R. Right now I'm most interested in the spaghetti plots of achievement over time by student ID associated with longitudinal analysis. How can I do a spaghetti plot of all students, but color the lines by group membership such as gender or race, and indicate this color scheme in the legend? Any advice would be much appreciated. Jack Jack B. Monpas-Huber, Ph.D.

I would like to plat some spaghetti plots from my data , ma data is as follows ak[1:3,] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [1,] 0.3211745 0.4132568 0.5649930 0.6920562 0.7760113 0.8118568 0.8609301 0.9088819 0.9326736 [2,] 0.3159234 0.4071270 0.5579212 0.6844584 0.7684690 0.8243702 0.8677043 0.8931288 0.9261926 [3,] 0.3075260 0.3993699

Hello, I am trying to create individual concentration-time spaghetti plots sorted by dose, and within each dose to show two different colors for a categorical variable (gender). I can’t find a way to add the color for gender. Groups=ID enables individual lines for each subject. If I use groups=gender, there will be two colors, but the lines between subjects are connected. This is what I have

Dear Rxperts, Is there a simpler way to generate multipanel grouped individual profile plots? All individuals of a group within a panel have the same color. As of now I am using lattice::xyplot to get the desired effect. Please feel free to suggest other ideas. Also, I am trying to create a generalized function which goes on similar lines like this.. grpPlot <- function(dat, mpgrp=quote(G),

Hi Rosa, You pass a vector to ggplot, which expects a data.frame. I am sure you meant to do this: point7$y_point7 <- point7$beta0_7 + point7$beta1_7*point7$time + point7 $epsilon_7 ggplot(point7, aes(time, y_point7)) + geom_line() HTH Ulrik On Wed, 19 Jul 2017 at 20:37 Rosa Oliveira <rosita21 at gmail.com> wrote: > Hi everyone, > > I?m trying to do a spaghetti plot and I

Hi everyone, I?m trying to do a spaghetti plot and I know I?m doing all wrong, It must be. What I need: 15 subjects, each with measurements over 5 different times (t1, ..., t5), and the variable that I need to represent in the spaguetti plot is given by: PCR = b0 + b1 * ti + epsilon B0, - baseline of each subject B1 - trajectory of each subject over time (so multiply by t) Epsilon - error

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 #

Hello - I am using glmnet to generate a model for multiple cohorts i. For each i, I run 5 separate models, each with a different x variable. I want to compare the fit statistic for each i and x combination. When I use auc, the output is in some cases is < .5 (.49). In addition, if I compare mean MSE (with upper and lower bounds) ... there is no difference across my various x variables, but

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

Hi everyone, I have a data frame (d), wich has the results of mosquitoes trapping in three different places. I suspect that one of these places (Local=='Palm') is biased by low numbers and will yield slower slopes in the variance-mean regression over the areas. I wonder if these slopes are diferents. I've looked trought the support list for methods for comparing slopes and found the

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 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

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

I''ve got this: file { ''/opt/sugarsync/tomcat/tomcat-home/current'': ensure => "tomcat-$config[''tomcat_version_server'']"; where $config[''tomcat_version_server''] was set with extlookup (the yaml one), by loading: --- tomcat_config: tomcat_version_server: 6.0.20-1 tomcat_version_libs: 1.0-1

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

Hi all, I'm trying to apply loess regression to my data and then use the fitted model to get the *standardized/studentized residuals. I understood that for linear regression (lm) there are functions to do that:* * * fit1 = lm(y~x) stdres.fit1 = rstandard(fit1) studres.fit1 = rstudent(fit1) I was wondering if there is an equally simple way to get the standardized/studentized residuals for a

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

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared