Hi !!
I am trying to analyze some of my data using linear discriminant analysis.
I worked out the following example code in Venables and Ripley
It does not seem to be happy with it.
============================
library(MASS)
library(stats)
data(iris3)
ir<-rbind(iris3[,,1],iris3[,,2],iris3[,,3])
ir.species<-factor(c(rep("s",50),rep("c",50),rep("v",50)))
ir.lda<-lda(log(ir),ir.species)
ir.ld<-predict(ir.lda,dimen=2)$x
eqscplot(ir.ld, type="n", xlab = "First linear discriminant", ylab =
"second linear discriminant")
text(ir.ld, labels= as.character(ir.species[-143]), col =3
+codes(ir.species),cex =0.8)
======================================
eqscplot does not plot anything and it gives me an error
saying codes is defunct. Have I missed anything there.
Thanks../Murli
lda
3 messages · T. Murlidharan Nair, Brian Ripley, Sundar Dorai-Raj
On Tue, 2 Nov 2004, T. Murlidharan Nair wrote:
Hi !! I am trying to analyze some of my data using linear discriminant analysis. I worked out the following example code in Venables and Ripley
It does not seem to be happy with it.
What is `it'? If you mean R, which version, and which version of the VR bundle?
============================ library(MASS) library(stats)
That line is definitely not in `Venables and Ripley'
data(iris3)
ir<-rbind(iris3[,,1],iris3[,,2],iris3[,,3])
ir.species<-factor(c(rep("s",50),rep("c",50),rep("v",50)))
ir.lda<-lda(log(ir),ir.species)
ir.ld<-predict(ir.lda,dimen=2)$x
eqscplot(ir.ld, type="n", xlab = "First linear discriminant", ylab =
"second linear discriminant")
text(ir.ld, labels= as.character(ir.species[-143]), col =3
+codes(ir.species),cex =0.8)
======================================
eqscplot does not plot anything and it gives me an error
saying codes is defunct. Have I missed anything there.
I have no idea why eqscplot is misbehaving (your example works up to the last line for me), but the R scripts which the book refers you to do work. See p.12 (and the R posting guide asking you to read the relevant section of the book).
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
T. Murlidharan Nair wrote:
Hi !!
I am trying to analyze some of my data using linear discriminant analysis.
I worked out the following example code in Venables and Ripley
It does not seem to be happy with it.
============================
library(MASS)
library(stats)
data(iris3)
ir<-rbind(iris3[,,1],iris3[,,2],iris3[,,3])
ir.species<-factor(c(rep("s",50),rep("c",50),rep("v",50)))
ir.lda<-lda(log(ir),ir.species)
ir.ld<-predict(ir.lda,dimen=2)$x
eqscplot(ir.ld, type="n", xlab = "First linear discriminant", ylab =
"second linear discriminant")
text(ir.ld, labels= as.character(ir.species[-143]), col =3
+codes(ir.species),cex =0.8)
======================================
eqscplot does not plot anything and it gives me an error saying codes is
defunct. Have I missed anything there.
Thanks../Murli
Murli,
eqscplot gives you nothing because you specified `type="n"'. You are not
plotting anything because your call to "text" never completed.
When I do this I get:
> R.version.string
[1] "R version 2.0.0, 2004-10-09"
> text(ir.ld, labels= as.character(ir.species[-143]), col =3
+codes(ir.species),cex =0.8)
Error: 'codes' is defunct.
See help("Defunct")
This means "codes" is no longer available. Use ?as.integer instead.
--sundar
P.S. Please do read the posting guide and tell us what version of R you
are using, etc.