subject:"Re\: \[R\] Help understanding why glm and lrm.fit runs with my data, but lrm does not"

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

2017-09-18 Thread Bonnett, Laura

Many thanks for the assistance.  I am using a small sample of GUSTO-1 as a 
teaching demonstration.  The Gusto-1 dataset in various smaller subsets is 
available from this website: 
http://clinicalpredictionmodels.org/doku.php?id=rcode_and_data:start  which is 
associated with the Clinical Prediction Models book by Steyerberg.

Many thanks again for your assistance.

Kind regards,
Laura

From: harre...@gmail.com [mailto:harre...@gmail.com] On Behalf Of Frank Harrell
Sent: 14 September 2017 17:22
To: David Winsemius 
Cc: Bonnett, Laura ; r-help@r-project.org
Subject: Re: [R] Help understanding why glm and lrm.fit runs with my data, but 
lrm does not

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 probably make it work on 
the small dataset but still not clear on why you are using this dataset.

Frank



Frank E Harrell Jr

Professor

School of Medicine


Department of Biostatistics

Vanderbilt University


On Thu, Sep 14, 2017 at 10:48 AM, David Winsemius 
> wrote:

> On Sep 14, 2017, at 12:30 AM, Bonnett, Laura 
> > wrote:
>
> Dear all,
>
> I am using the publically available GustoW dataset.  The exact version I am 
> using is available here: 
> https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fdrive.google.com%2Fopen%3Fid%3D0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk=02%7C01%7Cf.harrell%40vanderbilt.edu%7Cadb58b13c3994f89209708d4fb8807f0%7Cba5a7f39e3be4ab3b45067fa80faecad%7C0%7C0%7C636410009046132507=UZgX3%2Ba%2FU2Eeh8ybHMI6JnF0Npd2XJPXAzlmtEhDgOY%3D=0
>
> 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: 5nk: 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:
>
>   AGEHYP KILLIPHRTANT
> 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 DiscriminationRank Discrim.
>  Ratio Test   Indexes   Indexes
> Obs  2188LR chi2 233.59R2   0.273C   0.846
>  0   2053d.f. 5g1.642Dxy 0.691
>  1135Pr(> chi2) <0.0001gr   5.165gamma   0.696
> max |deriv| 4e-09  gp   0.079tau-a   0.080
>Brier0.048
>
>   Coef S.E.   Wald Z Pr(>|Z|)
> Intercept -13.8515

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

2017-09-14 Thread Frank Harrell

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 probably make it work
on the small dataset but still not clear on why you are using this dataset.

Frank


--
Frank E Harrell Jr  Professor  School of Medicine

Department of *Biostatistics*  *Vanderbilt University*

On Thu, Sep 14, 2017 at 10:48 AM, David Winsemius 
wrote:

>
> > On Sep 14, 2017, at 12:30 AM, Bonnett, Laura <
> l.j.bonn...@liverpool.ac.uk> wrote:
> >
> > Dear all,
> >
> > I am using the publically available GustoW dataset.  The exact version I
> am using is available here: https://na01.safelinks.
> protection.outlook.com/?url=https%3A%2F%2Fdrive.google.com%2Fopen%3Fid%
> 3D0B4oZ2TQA0PAoUm85UzBFNjZ0Ulk=02%7C01%7Cf.harrell%40vanderbilt.edu%
> 7Cadb58b13c3994f89209708d4fb8807f0%7Cba5a7f39e3be4ab3b45067fa80fa
> ecad%7C0%7C0%7C636410009046132507=UZgX3%2Ba%
> 2FU2Eeh8ybHMI6JnF0Npd2XJPXAzlmtEhDgOY%3D=0
> >
> > 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: 5nk: 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:
> >
> >   AGEHYP KILLIPHRTANT
> > 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 DiscriminationRank
> Discrim.
> >  Ratio Test   Indexes   Indexes
> > Obs  2188LR chi2 233.59R2   0.273C
>  0.846
> >  0   2053d.f. 5g1.642Dxy
>  0.691
> >  1135Pr(> chi2) <0.0001gr   5.165gamma
>  0.696
> > max |deriv| 4e-09  gp   0.079tau-a
>  0.080
> >Brier0.048
> >
> >   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.
>
> Try this:
>
> lrm  # look at code, do a search on "fail"
> ?lrm.fit  # read the structure of the returned value of lrm.fit
>
> my.fit <- lrm.fit(x =

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

2017-09-14 Thread David Winsemius


> On Sep 14, 2017, at 12:30 AM, 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: 5nk: 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:
> 
>   AGEHYP KILLIPHRTANT
> 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 DiscriminationRank Discrim.
>  Ratio Test   Indexes   Indexes
> Obs  2188LR chi2 233.59R2   0.273C   0.846
>  0   2053d.f. 5g1.642Dxy 0.691
>  1135Pr(> chi2) <0.0001gr   5.165gamma   0.696
> max |deriv| 4e-09  gp   0.079tau-a   0.080
>Brier0.048
> 
>   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.

Try this:

lrm  # look at code, do a search on "fail"
?lrm.fit  # read the structure of the returned value of lrm.fit

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

print(my.fit$fail)  # the error message you got from the lrm call means 
convergence failed

Documentation bug: The documentation of the cause of the 'fail'- value 
incorrectly gives the name of this parameter as 'maxiter' in the  Value section.

-- 
David.



> 
> 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.bonn...@liverpool.ac.uk
> 
> 
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.'   
-Gehm's Corollary to Clarke's

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

2017-09-14 Thread 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: 5nk: 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:

AGEHYP KILLIPHRTANT
  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 DiscriminationRank Discrim.
   Ratio Test   Indexes   Indexes
  Obs  2188LR chi2 233.59R2   0.273C   0.846
   0   2053d.f. 5g1.642Dxy 0.691
   1135Pr(> chi2) <0.0001gr   5.165gamma   0.696
  max |deriv| 4e-09  gp   0.079tau-a   0.080
 Brier0.048

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.bonn...@liverpool.ac.uk



[[alternative HTML version deleted]]

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

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

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

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

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