similar to: design package (plot problems)

Dear All, using the example from the help of summary.rms library(rms) n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10) blood.pressure <- rnorm(n, 120, 15) cholesterol <- rnorm(n, 200, 25) sex <- factor(sample(c('female','male'), n,TRUE)) label(age) <- 'Age'

Dear R-helpers, I'm having a problem in using plot.design in Design Library. Tho following example code produce the error: > n <- 1000 # define sample size > set.seed(17) # so can reproduce the results > age <- rnorm(n, 50, 10) > blood.pressure <- rnorm(n, 120, 15) > cholesterol <- rnorm(n, 200, 25) > sex <-

Hi, When running z <- plot(fit, age=NA, cholesterol=NA, perim=boundaries, method='image') Legend(z, fun=plogis, at=qlogis(c(.01,.05,.1,.2,.3,.4,.5)), zlab='Probability') And after pointing the cursor to the plot() screen in R, I obtain the following message: Using function "locator(2)" to place opposite corners of image.legend Error in

I have a plot produced by function bplot (package = rms) that is really a lattice plot (class="trellis"). It is similar to this plot produced by a very minor modification of the first example on the bplot help page: requiere(rms) n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10) blood.pressure <- rnorm(n, 120,

dear£º in the example£¨nomogram£©£¬I don't understand the meanings of the program which have been marked by red line.And how to compile the program(L <- .4*(sex=='male') + .045*(age-50) + (log(cholesterol - 10)-5.2)*(-2*(sex=='female') + 2*(sex=='male'))). n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10)

dear:teacher i have a problem which about the polr()(package "MASS"), if the response must have 3 or more levels? and how to fit the polr() to 2 levels? thank you. turly yours [[alternative HTML version deleted]]

Dear Rlist, How can the Legend of the plot.Design() function can be visualized? Following the documentation in R, I did the following (see below), only the 'Legend' function doesn't visualize the legend of the plot (method='image') of the lrmfit. I tried to change par( margin setting) but this didn’t solve it. How can this be solved? Thanks a lot, Jan

dear: I am a user of R project.And now ,I have a problem. I want to know what is the name of the datasets in the web page--"Draw a Nomogram Representing a Regression Fit" which come from the R-home(http://www.r-project.org/ Package rms version 3.0-0). And can you supply the dataset to me? wish for your help,thank you!

Hello. One of my teaching assistants was experimenting and encountered unexpected behaviour with the lrtest function in the rms package. It appears that when you have a pair of non-nested models that employ an RCS, the error checking for non-nested models appears not to work. Here is a reproducible example. > library(rms) Loading required package: Hmisc Loading required package: lattice

Hi Group, I have a question about obtaining the bias-corrected c-index using validate from the Design library. As an example, consider the example from help page: library(Design) ?validate.lrm n <- 1000 age <- rnorm(n, 50, 10) blood.pressure <- rnorm(n, 120, 15) cholesterol <- rnorm(n, 200, 25) sex <- factor(sample(c('female','male'),

Dear R-helpers I have a question related to logistic GAM models. Consider the following example: # Load package library(mgcv) # Simulation of dataset n <- 1000 set.seed(0) age <- rnorm(n, 50, 10) blood.pressure <- rnorm(n, 120, 15) cholesterol <- rnorm(n, 200, 25) sex <- factor(sample(c('female','male'), n,TRUE)) L <-

Dear all, I have performed a simple logistic regression using the lrm function from the Design library. Now I want to plot the summary, or make a nomogram. I keep getting a datadist error: options(datadist= m.full ) not created with datadist. I have tried to specify datadist beforhand (although I don't know why it should be done): ddist<-datadist(d) ##where d is my dataset

dear professor: thank you for your help,witn your help i develop the nomogram successfully. after that i want to do the internal validation to the model.i ues the bootpred to do it,and then i encounter problem again,just like that.(´íÎóÓÚerror to :complete.cases(x, y, wt) : ²»ÊÇËùÓеIJÎÊý¶¼Ò»Ñù³¤(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have

I have fitted a logistic regression model > failed.lr2$call lrm(formula = failed ~ Age + task2 + Age:task2, data = time.long, na.action = na.omit) using the Design package functions and would like to generate a nomogram from this model. the datadist information is generated and stored in > ddist time.long$Age time.long$task2 Low:effect 45

dear professor: I have a problem about the nomogram.I have got the result through analysing the dataset "exp2.sav" through multinominal logistic regression by SPSS 17.0. and I want to deveop the nomogram through R-Projject,just like this : > n<-100 > set.seed(10) > T.Grade<-factor(0:3,labels=c("G0", "G1", "G2","G3")) >

Hi everyone, I'm doing a logistic regression with an ordinal variable. I'd like to set the contrasts on the ordinal variable. However, when I set the contrasts, they work for ordinary linear regression (lm), but not logistic regression (lrm): ddist = datadist(bin.time, exp.loc) options(datadist='ddist') contrasts(exp.loc) = contr.treatment(3, base = 3, contrasts = TRUE) lrm.loc =

Hi, Design isn't strictly an R base package, but maybe someone can explain the following. When lrm is called within a function, it can't find the dataset dd: > library(Design) > age <- rnorm(30, 50, 10) > cholesterol <- rnorm(30, 200, 25) > ch <- cut2(cholesterol, g=5, levels.mean=TRUE) > fit <- function(ch, age) + { + d <- data.frame(ch, age) +

Hi! Does anyone know how to do the test for goodness of fit of a logistic model (in rms package) after running fit.mult.impute? I am using the rms and Hmisc packages to do a multiple imputation followed by a logistic regression model using lrm. Everything works fine until I try to run the test for goodness of fit: residuals(type=c("gof")) One needs to specify y=T and x=T in the fit. But

Dear all, I search the mail list about this topic and learn that no simple way is available to get "lsmeans" in R as in SAS. Dr.John Fox and Dr.Frank E Harrell have given very useful information about "lsmeans" topic. Dr. Frank E Harrell suggests not to think about lsmeans, just to think about what predicted values wanted and to use the predict

Hi, Running the following example from ?Predict() throws an error I have never seen before: set.seed(1) x1 <- runif(300) x2 <- runif(300) ddist <- datadist(x1,x2); options(datadist='ddist') y <- exp(x1+ x2 - 1 + rnorm(300)) f <- ols(log(y) ~ pol(x1,2) + x2) p1 <- Predict(f, x1=., conf.type='mean') Error in paste(nmc[i], "=", if (is.numeric(x))