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

[Jan van der Laan]

With lrm.fit you are fitting a completely different model. One of the 
things lrm does, is preparing the input for lrm.fit which in this case 
means that dummy variables are generated for categorical variables such 
as 'KILLIP'.

The error message means that model did not converge after the maximum 
number of iterations. One possible solution is to try to increase the 
maximum number of iterations, e.g.:

fit1 <- lrm(DAY30~AGE+HYP+KILLIP+HRT+ANT, data = gusto2, maxit = 100)

HTH,

Jan



On 14-09-17 09:30, Bonnett, Laura 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 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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> 
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