On Sep 21, 2010, at 1:28 PM, Mestat wrote:
Hi Denis,
Check it out my code... This is not my real data...
I would like to manage the size of the legend... Set the legend
smaller than
it is, because on my real data, the legend is over the values...
Thanks for the help...
x<-runif(100)
y<-runif(100)
color<-
c
(rep
(1,15
),rep
(2,20),rep(3,15),rep(4,10),rep(5,10),rep(6,15),rep(7,10),rep(8,5))
plot
(x,y,col=c("black","blue","green","red","red","green","blue","black")
[color],pch=c(22,3,3,3,19,19,19,23)[color],main='
',xlab='Sample Mean',ylab='Bootstrap Variance Estimators')
legend("topleft", legend=c("All 3 outliers","Presence of 2 outliers
(6,263
and 6,720)","Presence of 2 outliers (3,471 and 6,720)","Presence of 2
outliers (3,471 and 6,263)","Presence of outlier 6,720","Presence of
outlier
6,263","Presence of outlier 3,471","No outlier"), col =
c("black","blue","green","red","red","green","blue","black"), pch =
c(23,19,19,19,3,3,3,22), lty=c(1,1), xjust=0, yjust=0, bty="n")
Learn to use spaces between arguments and indent meaningfully!!
Best I could do with base graphics. Lattice would give you more
control I suspect:
plot
(x,y,col=c("black","blue","green","red","red","green","blue","black")
[color],
pch=c(22,3,3,3,19,19,19,23)[color],main='',
xlab='Sample Mean',
ylab='Bootstrap Variance Estimators')
legend("topleft", legend=expression(scriptstyle("All 3 outliers"),
scriptstyle("Presence of 2 outliers
(6,263 and 6,720)"),
scriptstyle("Presence of 2 outliers
(3,471 and 6,720)"),
scriptstyle("Presence of 2 outliers
(3,471 and 6,263)"),
scriptstyle("Presence of outlier 6,720"),
scriptstyle("Presence of outlier 6,263"),
scriptstyle("Presence of outlier 3,471"),
scriptstyle("No outlier")
),
col =
c("black","blue","green","red","red","green","blue","black"),
pch = c(23,19,19,19,3,3,3,22), lty=c(1,1), xjust=0,
yjust=0, bty="n")
Need to avoid embedding carriage returns in the expression arguments
since they get carried through to the plot ((and ISTR that \n does not.)
--
David Winsemius, MD
West Hartford, CT
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