similar to: Accessing more than two coefficients in a plot

Displaying 20 results from an estimated 100 matches similar to: "Accessing more than two coefficients in a plot"

2003 Jul 30
2
Comparing two regression slopes
Hello, I've written a simple (although probably overly roundabout) function to test whether two regression slope coefficients from two linear models on independent data sets are significantly different. I'm a bit concerned, because when I test it on simulated data with different sample sizes and variances, the function seems to be extremely sensitive both of these. I am wondering if
2004 Aug 19
3
List dimention labels to plots of components
It is frustrating to see the labels I want in the dimensions of a list but not be able to extract those labels into titles for plots generated from component objects. If someone could set me straight, I would appreciate it. For your amusement, I have provided an example of the Byzantine code I am currently using to avoid loops: # Simulate ANOVA type test data sex<-c(rep(1,8),rep(0,8))
2005 Aug 12
1
as.formula and lme ( Fixed effects: Error in as.vector(x, "list") : cannot coerce to vector)
This is a continuing issue with the one on the list a long time ago (I couldn't find a solution to it from the web): -------------------------------------------------------------------------- > Using a formula converted with as.formula with lme leads > to an error message. Same works ok with lm, and with > lme and a fixed formula. > > # demonstrates problems with lme and
2006 Sep 03
2
lm, weights and ...
> lm2 <- function(...) lm(...) > lm2(mpg ~ wt, data=mtcars) Call: lm(formula = ..1, data = ..2) Coefficients: (Intercept) wt 37.285 -5.344 > lm2(mpg ~ wt, weights=cyl, data=mtcars) Error in eval(expr, envir, enclos) : ..2 used in an incorrect context, no ... to look in Can anyone explain why this is happening? (Obviously this is a manufactured example, but it
2012 Jul 06
2
Anova Type II and Contrasts
the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data. I created a dummy data set to explain my situation: df1 <- data.frame( Observation =
2011 May 10
2
Leyenda de las series en tsplot
Hola de nuevo: Sigo enfrascado con mi dichoso procedimiento para generar modelos de predicción de series temporales. Llegado un momento pretendo guardar un gráfico en el que se representara: -En a estarán los puntos obtenidos por el alisado (tanto en el pasado como las estimaciones a futuro) -En o estará la serie de datos original. -En inf los limites inferiores de los intervalos de confianza
2010 Apr 08
2
Overfitting/Calibration plots (Statistics question)
This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data
2009 Jan 14
1
Help with Plot/Legend
Dear R-Users I have 2 questions: Firstly, If I create a matplot and legend for multiple vectors and then tag another vector on using matlines (e.g. a 'total' of all vectors), is there anyway to add the new line to the legend without recreating it? I have created the plot this way because I would like to define the lty and lwd for the 'total' vector so that it can be
2001 Feb 23
1
as.formula and lme ( Fixed effects: Error in as.vector(x, "list") : cannot coerce to vector)
Using a formula converted with as.formula with lme leads to an error message. Same works ok with lm, and with lme and a fixed formula. # demonstrates problems with lme and as.formula demo<-data.frame(x=1:20,y=(1:20)+rnorm(20),subj=as.factor(rep(1:2,10))) demo.lm1<-lme(y~x,data=demo,random=~1|subj) print(summary(demo.lm1)) newframe<-data.frame(x=1:5,subj=rep(1,5))
2006 Mar 10
1
add trend line to each group of data in: xyplot(y1+y2 ~ x | grp...
Although this should be trivial, I'm having a spot of trouble. I want to make a lattice plot of the format y1+y2 ~ x | grp but then fit a lm to each y variable and add an abline of those models in different colors. If the xyplot followed y~x|grp I would write a panel function as below, but I'm unsure of how to do that with y1 and y2 without reshaping the data before hand. Thoughts
2011 Aug 22
3
Ignoring loadNamespace errors when loading a file
On a Unix machine I ran caret::rfe using the multicore package, and I saved the resulting object using save(lm2, file = "lm2.RData"). [Reproducible example below.] When I try to load("lm2.RData") on my Windows laptop, I get Error in loadNamespace(name) : there is no package called 'multicore' I completely understand the error and I would like to ignore it and
2002 Dec 20
1
Printing correlation matrices (lm/glm)
Hi Folks, I'm analysing some data which, in its simplest aspect, has 3 factors A, B, C each at 2 levels. If I do lm1 <- lm(y ~ A*B) say, and then summary(lm1, corr=T) I get the correlation matrix of the estimated coeffcients with numerical values for the correlations (3 coeffs in this case). Likewise with 'glm' instead of 'lm'. However, if I do lm2 <- lm(y ~
2007 Dec 07
1
AIC v. extractAIC
Hello, I am using a simple linear model and I would like to get an AIC value. I came across both AIC() and extractAIC() and I am not sure which is best to use. I assumed that I should use AIC for a glm and extractAIC() for lm, but if I run my model in glm the AIC value is the same if I use AIC() on an lm object. What might be going on? Did I interpret these functions incorrectly? Thanks,
2007 Aug 06
1
test the significances of two regression lines
R-help, I'm trying to test the significance of two regression lines , i.e. the significance of the slopes from two samples originated from the same population. Is it correct if I fit a liner model for each sample and then test the slope signicance with 'anova'. Something like this: lm1 <- lm(Y~ a1 + b1*X) # sample 1 lm2 <- lm(Y~ a2 + b2*X) # sample 2 anova(lm1, lm2)
2012 Jan 13
2
Help needed in interpreting linear models
Dear members of the R-help list, I have sent the email below to the R-SIG-ME list to ask for help in interpreting some R output of fitted linear models. Unfortunately, I haven't yet received any answers. As I am not sure if my email was sent successfully to the mailing list I am asking for help here: Dear members of the R-SIG-ME list, I am new to linear models and struggling with
2000 Feb 02
0
Bugs and comments. (PR#410)
Hi, Here are a few errors I found as well as a few comments. 1) In the man page of par: lty: The line type. Line types can either be specified as an integer (0=blank, 1=solid, 2=dashed, 3=dot- ted, 4=dotdash, 5=longdash, 6=twodash) or as one of the character strings `"blank"', `"solid"',
2008 Nov 24
3
Is this correct?
I have to answer the following question for a homework assignment. A researcher was interested in whether people taking part in sports at university made more money after graduating, taking into account the students' GPA. They sampled 200 alumni from a large university. The variables are: income (income 10 years after graduating), sports (1 if they did sports, 0 if they did not), and GPA (the
2018 Jul 18
2
Legendas en una gráfica de ggplot2
Buenas tardes, estoy haciendo una gráfica de múltiples lineas pero no he podido generar las legendas. Alguno de ustedes me podría colaborar. library(ggplot2) #### Con b=-2 t=seq (-4, 4, by=0.01) l=exp(t+2)/(1+(exp(t+2))) ##con b igual a -1 t=seq (-4, 4, by=0.01) o=exp(t+1)/(1+(exp(t+1))) ### Con b igual a 0.7 t=seq (-4, 4, by=0.01) i=exp(t-0.7)/(1+(exp(t-0.7))) ### Con b igual a 2 t=seq
2003 Feb 01
0
AIC.default (PR#2518)
There is a bug in AIC.default and AIC.lm, as illustrated below. (I've only checked this under 1.6.1, and can't easily check if it has already been reported since the site is down.) > lm1 <- lm(y ~ x, list(x=1:10, y=jitter(1:10))) > lm2 <- lm(y ~ x, list(x=1:10, y=jitter(1:10))) > AIC(lm1, lm2) df AIC lm1 3 -18.662493 lm2 3 -7.265906 > AIC(lm1, lm2, k = 2)
2008 Oct 05
1
Help on R Coding
Hi all, I am kind of stuck of using Predict function in R to make prediction for a model with continuous variable and categorial variables. i have no problem making the model, the model is e.g. cabbage.lm2<- lm(VitC ~ HeadWt + Date + Cult) HeadWt is a continuous variable, Date and Culte are factors. Date have three levels inside (d16,d20,d21), Cult has two levels(c39,c52). I need to