[R] Help understanding why glm and lrm.fit runs with my data, but lrm does not

Bonnett, Laura L.J.Bonnett at liverpool.ac.uk
Thu Sep 14 09:30:13 CEST 2017


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 <- spss.get("GustoW.sav")
fit <- glm(DAY30~AGE+HYP+factor(KILLIP)+HRT+ANT,family=binomial(link="logit"),data=gusto,x=TRUE,y=TRUE)

However, my review of the literature and other websites suggest I need to use "lrm" for the purposes of producing a nomogram.  When I run the command using "lrm" (see below) I get an error message saying:
Error in lrm(DAY30 ~ AGE + HYP + KILLIP + HRT + ANT, gusto2) :
  Unable to fit model using "lrm.fit"

My code is as follows:
gusto2 <- gusto[,c(1,3,5,8,9,10)]
gusto2$HYP <- factor(gusto2$HYP, labels=c("No","Yes"))
gusto2$KILLIP <- factor(gusto2$KILLIP, labels=c("1","2","3","4"))
gusto2$HRT <- factor(gusto2$HRT, labels=c("No","Yes"))
gusto2$ANT <- factor(gusto2$ANT, labels=c("No","Yes"))
var.labels=c(DAY30="30-day Mortality", AGE="Age in Years", KILLIP="Killip Class", HYP="Hypertension", HRT="Tachycardia", ANT="Anterior Infarct Location")
label(gusto2)=lapply(names(var.labels),function(x) label(gusto2[,x])=var.labels[x])

ddist = datadist(gusto2)
options(datadist='ddist')

fit1 <- lrm(DAY30~AGE+HYP+KILLIP+HRT+ANT,gusto2)

Error in lrm(DAY30 ~ AGE + HYP + KILLIP + HRT + ANT, gusto2) :
  Unable to fit model using "lrm.fit"

Online solutions to this problem involve checking whether any variables are redundant.  However, the results for my data suggest  that none are.
redun(~AGE+HYP+KILLIP+HRT+ANT,gusto2)

Redundancy Analysis

redun(formula = ~AGE + HYP + KILLIP + HRT + ANT, data = gusto2)

n: 2188         p: 5    nk: 3

Number of NAs:   0

Transformation of target variables forced to be linear

R-squared cutoff: 0.9   Type: ordinary

R^2 with which each variable can be predicted from all other variables:

   AGE    HYP KILLIP    HRT    ANT
 0.028  0.032  0.053  0.046  0.040

No redundant variables

I've also tried just considering "lrm.fit" and that code seems to run without error too:
lrm.fit(cbind(gusto2$AGE,gusto2$KILLIP,gusto2$HYP,gusto2$HRT,gusto2$ANT),gusto2$DAY30)

Logistic Regression Model

 lrm.fit(x = cbind(gusto2$AGE, gusto2$KILLIP, gusto2$HYP, gusto2$HRT,
     gusto2$ANT), y = gusto2$DAY30)

                       Model Likelihood     Discrimination    Rank Discrim.
                          Ratio Test           Indexes           Indexes
 Obs          2188    LR chi2     233.59    R2       0.273    C       0.846
  0           2053    d.f.             5    g        1.642    Dxy     0.691
  1            135    Pr(> chi2) <0.0001    gr       5.165    gamma   0.696
 max |deriv| 4e-09                          gp       0.079    tau-a   0.080
                                            Brier    0.048

           Coef     S.E.   Wald Z Pr(>|Z|)
 Intercept -13.8515 0.9694 -14.29 <0.0001
 x[1]        0.0989 0.0103   9.58 <0.0001
 x[2]        0.9030 0.1510   5.98 <0.0001
 x[3]        1.3576 0.2570   5.28 <0.0001
 x[4]        0.6884 0.2034   3.38 0.0007
 x[5]        0.6327 0.2003   3.16 0.0016

I was therefore hoping someone would explain why the "lrm" code is producing an error message, while "lrm.fit" and "glm" do not.  In particular I would welcome a solution to ensure I can produce a nomogram.

Kind regards,
Laura

Dr Laura Bonnett
NIHR Post-Doctoral Fellow

Department of Biostatistics,
Waterhouse Building, Block F,
1-5 Brownlow Street,
University of Liverpool,
Liverpool,
L69 3GL

0151 795 9686
L.J.Bonnett at liverpool.ac.uk



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