# Spearman correlation "rs" with ties or no ties
rs<-function(x,y) {
d<-rank(x)-rank(y)
tx<-as.numeric(table(x))
ty<-as.numeric(table(y))
Lx<-sum((tx^3-tx)/12)
Ly<-sum((ty^3-ty)/12)
N<-length(x)
SX2<- (N^3-N)/12 - Lx
SY2<- (N^3-N)/12 - Ly
rs<- (SX2+SY2-sum(d^2))/(2*sqrt(SX2*SY2))
return(rs)
}
# Aplicacion
cor(y[,1],y[,2],method="spearman")
[1] 0.2319084
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Raymond Wan
Sent: Monday, May 28, 2007 10:29 PM
To: r-help at stat.math.ethz.ch
Subject: [R] R's Spearman
Hi all,
I am trying to figure out the formula used by R's Spearman rho (using
cor(method="spearman")) because I can't seem to get the same value as by
calculating "by hand". Perhaps I'm using "cor" wrong, but I don't know
where. Basically, I am running these commands:
> y=read.table(file="tmp",header=TRUE,sep="\t")
> y
IQ Hours
1 106 7
2 86 0
3 97 20
4 113 12
5 120 12
6 110 17
> cor(y[1],y[2],method="spearman")
Hours
IQ 0.2319084
[it's an abbreviated example of one I took from Wikipedia]. I
calculated by hand (apologies if the table looks strange when pasted
into e-mail):
IQ Hours rank(IQ) rank(hours) diff diff^2
1 106 7 3 2 1 1
2 86 0 1 1 0 0
3 97 20 2 6 -4 16
4 113 12 5 3.5 1.5 2.25
5 120 12 6 3.5 2.5 6.25
6 110 17 4 5 -1 1
26.5
rho= 0.242857
where rho = (1 - ((6 * 26.5) / 6 * (6^2 - 1))). I kept modifying the
table and realized that the difference in result comes from ties. i.e.,
if I remove the tie in rows 4 and 5, I get the same result from both cor
and calculating by hand. Perhaps I'm handling ties wrong...does anyone
know how R does it or perhaps I need to change how I'm using it?
Thank you!
Ray