similar to: help with nomogram function

Dear R-users, I have used the nomogram function from the rms package for a logistic regresison model made with lrm(). Everything works perfectly (r version 2.15.1 on a mac). My question is this: if my final model is not the one created by lrm, but I internally validated the model and 'shrunk' the regression coefficients and computed a new intercept, how can I build a nomogram using that

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])

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")) >

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

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 all, I am using the publically available GustoW dataset. The exact version I am using is available here: https://drive.google.com/open?id=0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk I would like to produce a nomogram for 5 covariates - AGE, HYP, KILLIP, HRT and ANT. I have successfully fitted a logistic regression model using the "glm" function as shown below. library(rms) gusto <-

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!

I have a matrix, I would like to use it like I use where query on an SQL table Let's say my matrix is > wok who task 1 joe task1 2 joe task2 3 joe task3 4 jack task1 5 jack task2 I want to extract a matrix that contains only rows where who = "joe" and task != "task2" The result would be the following matrix 1 joe task1 2 joe task3 How can I do

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]]

Hi, I have been able to generate association rules for Market Basket Analysis using the following codes: **************************************************************************** ******************************************* library("arules") rules <- read.csv("write1.csv",na.strings=c(".", "NA", "", "?"),header=TRUE)

> On Sep 14, 2017, at 12:30 AM, Bonnett, Laura <L.J.Bonnett at liverpool.ac.uk> wrote: > > Dear all, > > I am using the publically available GustoW dataset. The exact version I am using is available here: https://drive.google.com/open?id=0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk > > I would like to produce a nomogram for 5 covariates - AGE, HYP, KILLIP, HRT and ANT. I have

Maybe a useful addition to the predict functions would be to return the values of the predictor variables. It just (unless there are problems) requires an extra line. I have inserted an example below. "predict.glm" <- function (object, newdata = NULL, type = c("link", "response", "terms"), se.fit = FALSE,

The loop is correct, you just need to make sure that your result is computed and stored as the n-th element that is returned by the loop. Pick up any manual of R, and looping will be explained there. Also, I would recommend that you draw a random number for every iteration of the loop. Defining the random vectors outside the loop make sense to me only if they are the same length as n.

Fixed 'maxiter' in the help file. Thanks. Please give the original source of that dataset. That dataset is a tiny sample of GUSTO-I and not large enough to fit this model very reliably. A nomogram using the full dataset (not publicly available to my knowledge) is already available in http://biostat.mc.vanderbilt.edu/tmp/bbr.pdf Use lrm, not lrm.fit for this. Adding maxit=20 will

You did not follow the posting guide, did you use pure ascii email, and used illegal characters in your source code. This caused extra work. Once I cleaned up your characters and made the example self-contained, the labels worked fine for me. Here's the cleaned-up code: library(rms) x1 <- runif(20) x2 <- runif(20) y <- sample(0:1, 20, TRUE) label(x1) <- 'A' label(x2)

Hi Sarah Thank you very much for your pointers. Is there a way to specify the date string as POSIXct when reading in? I have tried the following (very inelegant way) and still have no luck. --- Begin Code --- # 1st read in the spreadsheet with stringsAsFactors set to FALSE > abc <- read.csv("gntr1.csv", header=TRUE, stringsAsFactors=FALSE) > abc code tasks.labels

Hi, Your attempt didn't work because you didn't do what I suggested, which was make sure that your data frame matched the example given in the help for gantt.chart. Here's what you have: > str(cdfg) 'data.frame': 3 obs. of 3 variables: $ c1d1: Factor w/ 3 levels "task 1","task 3",..: 1 3 2 $ c2d1: POSIXct, format: "2018-04-01"

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 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 <-