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Heike Zimmermann

2004-Oct-11 14:32 UTC

[R] logistic regression

Hello,

I have a problem concerning logistic regressions.  When I add a quadratic
term to my linear model, I cannot draw the line through my scatterplot
anymore, which is no problem without the quadratic term.
In this example my binary response variable is "incidence", the
explanatory
variable is "sun":> model0<-glm(incidence~1,binomial)
> model1<-glm(incidence~sun,binomial)
> anova(model0,model1,test="Chi")Analysis of Deviance Table

Model 1: incidence ~ 1
Model 2: incidence ~ sun
  Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
1       299     332.94                       
2       298     287.19   1    45.74 1.349e-11> qsun<-sun^2
> model2<-glm(incidence~sun+qsun,binomial)
> anova(model1,model2,test="Chi")Analysis of Deviance Table

Model 1: incidence ~ sun
Model 2: incidence ~ sun + qsun
  Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
1       298    287.194                       
2       297    280.623   1    6.571     0.010

So the second, non-linear, model explains more than the first. 
Now to create a graph I write:
> plot(sun,incidence)
> min(sun)
[1] 0> max(sun)
[1] 90> xsun<-seq(0,90,1)
>lines(xsun,predict(model2,type="response",data.frame(sun=xsun)))
Error in model.frame(formula, rownames, variables, varnames, extras,
extranames,  : 
        variable lengths differ> 
So this is the message I receive everytime I try to draw the fitted values
of my model. I know for sure that exactly the same command works in S-Plus
(with the same data). How is ist possible to do this in R?

Thank you in advance, Heike

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Sundar Dorai-Raj

2004-Oct-11 14:52 UTC

head link

[R] logistic regression

Heike Zimmermann wrote:
> Hello,
> 
> I have a problem concerning logistic regressions.  When I add a quadratic
> term to my linear model, I cannot draw the line through my scatterplot
> anymore, which is no problem without the quadratic term.
> In this example my binary response variable is "incidence", the
explanatory
> variable is "sun":
> 
>>model0<-glm(incidence~1,binomial)
>>model1<-glm(incidence~sun,binomial)
>>anova(model0,model1,test="Chi")
> 
> Analysis of Deviance Table
> 
> Model 1: incidence ~ 1
> Model 2: incidence ~ sun
>   Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       299     332.94                       
> 2       298     287.19   1    45.74 1.349e-11
> 
>>qsun<-sun^2
>>model2<-glm(incidence~sun+qsun,binomial)
>>anova(model1,model2,test="Chi")
> 
> Analysis of Deviance Table
> 
> Model 1: incidence ~ sun
> Model 2: incidence ~ sun + qsun
>   Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       298    287.194                       
> 2       297    280.623   1    6.571     0.010
> 
> So the second, non-linear, model explains more than the first. 
> Now to create a graph I write:
> 
> 
>>plot(sun,incidence)
>>min(sun)
> 
> [1] 0
> 
>>max(sun)
> 
> [1] 90
> 
>>xsun<-seq(0,90,1)
> 
> 
>>lines(xsun,predict(model2,type="response",data.frame(sun=xsun)))
> 
> 
> Error in model.frame(formula, rownames, variables, varnames, extras,
> extranames,  : 
>         variable lengths differ
> 
> 
> So this is the message I receive everytime I try to draw the fitted values
> of my model. I know for sure that exactly the same command works in S-Plus
> (with the same data). How is ist possible to do this in R?
> 
> Thank you in advance, Heike
> 
This is because your newdata=data.frame(sun=xsun) does not contain a 
variable called "qsun". Try the following instead:

model2 <- glm(incidence ~ sun + I(sun^2), binomial)
...
lines(xsun, predict(model2, type = "response", data.frame(sun =
xsun)))


--sundar

Prof Brian Ripley

2004-Oct-11 15:01 UTC

head link

[R] logistic regression

On Mon, 11 Oct 2004, Heike Zimmermann wrote:
> Hello,
> 
> I have a problem concerning logistic regressions.  When I add a quadratic
> term to my linear model, I cannot draw the line through my scatterplot
> anymore, which is no problem without the quadratic term.
> In this example my binary response variable is "incidence", the
explanatory
> variable is "sun":
> > model0<-glm(incidence~1,binomial)
> > model1<-glm(incidence~sun,binomial)
> > anova(model0,model1,test="Chi")
> Analysis of Deviance Table
> 
> Model 1: incidence ~ 1
> Model 2: incidence ~ sun
>   Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       299     332.94                       
> 2       298     287.19   1    45.74 1.349e-11
> > qsun<-sun^2
> > model2<-glm(incidence~sun+qsun,binomial)
> > anova(model1,model2,test="Chi")
> Analysis of Deviance Table
> 
> Model 1: incidence ~ sun
> Model 2: incidence ~ sun + qsun
>   Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       298    287.194                       
> 2       297    280.623   1    6.571     0.010
> 
> So the second, non-linear, model explains more than the first. 
> Now to create a graph I write:
> 
> > plot(sun,incidence)
> > min(sun)
> [1] 0
> > max(sun)
> [1] 90
> > xsun<-seq(0,90,1)
> 
>
>lines(xsun,predict(model2,type="response",data.frame(sun=xsun)))
Why are you predicting using a new data frame containing just one of the 
two predictors?
> 
> Error in model.frame(formula, rownames, variables, varnames, extras,
> extranames,  : 
>         variable lengths differ
> > 
> 
> So this is the message I receive everytime I try to draw the fitted values
> of my model. I know for sure that exactly the same command works in S-Plus
> (with the same data). 
If so, you have been trapped by a bug in S-PLUS, for you have at no point
supplied the quadratic term needed for prediction.  I suspect more likely
that the command is accepted but gives incorrect answers.  See the
warnings in MASS, for example.
> How is ist possible to do this in R?
NB: note the use of spaces for readability.

mydata <- data.frame(sun)
model2 <- glm(incidence ~ sun + I(sun^2), binomial, data=mydata)
plot(sun, incidence)
lines(xsun, predict(model2, type="response", data.frame(sun=0:90)))

PLEASE use a meaningful subject line.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

Dimitris Rizopoulos

2004-Oct-11 15:02 UTC

head link

[R] logistic regression

Hi Heike,

you'd  probably want

lines(xsun, predict(model2, type="response", 
newdata=data.frame(sun=xsun, qsun=xsun*xsun)))

I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/16/396887
Fax: +32/16/337015
Web: http://www.med.kuleuven.ac.be/biostat/
     http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Heike Zimmermann" <heikz at gmx.de>
To: <R-help at stat.math.ethz.ch>
Sent: Monday, October 11, 2004 4:32 PM
Subject: [R] logistic regression

> Hello,
>
> I have a problem concerning logistic regressions.  When I add a 
> quadratic
> term to my linear model, I cannot draw the line through my 
> scatterplot
> anymore, which is no problem without the quadratic term.
> In this example my binary response variable is "incidence", the 
> explanatory
> variable is "sun":
>> model0<-glm(incidence~1,binomial)
>> model1<-glm(incidence~sun,binomial)
>> anova(model0,model1,test="Chi")
> Analysis of Deviance Table
>
> Model 1: incidence ~ 1
> Model 2: incidence ~ sun
>  Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       299     332.94
> 2       298     287.19   1    45.74 1.349e-11
>> qsun<-sun^2
>> model2<-glm(incidence~sun+qsun,binomial)
>> anova(model1,model2,test="Chi")
> Analysis of Deviance Table
>
> Model 1: incidence ~ sun
> Model 2: incidence ~ sun + qsun
>  Resid. Df Resid. Dev  Df Deviance P(>|Chi|)
> 1       298    287.194
> 2       297    280.623   1    6.571     0.010
>
> So the second, non-linear, model explains more than the first.
> Now to create a graph I write:
>
>> plot(sun,incidence)
>> min(sun)
> [1] 0
>> max(sun)
> [1] 90
>> xsun<-seq(0,90,1)
>
>>lines(xsun,predict(model2,type="response",data.frame(sun=xsun)))
>
> Error in model.frame(formula, rownames, variables, varnames, extras,
> extranames,  :
>        variable lengths differ
>>
>
> So this is the message I receive everytime I try to draw the fitted 
> values
> of my model. I know for sure that exactly the same command works in 
> S-Plus
> (with the same data). How is ist possible to do this in R?
>
> Thank you in advance, Heike
>
> -- 
> +++ GMX DSL Premiumtarife 3 Monate gratis* + WLAN-Router 0,- EUR* 
> +++
> Clevere DSL-Nutzer wechseln jetzt zu GMX: 
> http://www.gmx.net/de/go/dsl
>
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