[R] Replacing plot symbols w/ subject IDs in xyplot()

David Afshartous dafshartous at med.miami.edu
Fri Feb 29 16:17:08 CET 2008



For plot():
plot(x,y, type = "n")
text(x,y, lab= junk.frm$ID)

This accomplishes the goal of using subject IDs as plot symbols, but I can't
seem to get the same result below w/ xyplot().  Is there something different
I have to do for xyplot()?

xyplot(y ~ x, type = "n")
text(x,y , lab=junk.frm$ID)
# doesn't work 



On 2/28/08 5:09 PM, "Bert Gunter" <gunter.berton at gene.com> wrote:

> ?text
> 
> e.g 
> 
> plot(x,y,type="n") ##
> text(x,y,lab= whatever)
> 
> This replaces the plot symbols with the ID's. If you just wish to plot them
> next, remove the type="n" specification and just offset the x's in text a
> hair.
> 
> -- Bert Gunter
> Genentech
> 
> 
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
> Behalf Of David Afshartous
> Sent: Thursday, February 28, 2008 1:15 PM
> To: r-help at r-project.org
> Subject: [R] Replacing plot symbols w/ subject IDs in xyplot()
> 
> 
> 
> All, 
> 
> How does one replace plot symbols with say subject IDs when using xyplot? Or
> superimpose them next to plot symbols?  I searched the archives under
> various key words but haven't had much.  Any suggestions or links much
> appreciated.  Sample code below.
> 
> David
> 
> 
> 
> 
> 
> junk.frm = data.frame(ID = rep(1:16, each = 2), x, y, z = rep(c("D", "P"),
> 16))
> y = c( 0.4,  0.6, -0.1,  0.3,  0.3, -0.2,  0.7,  0.7,  0.2,  0.0,  0.9,
> -0.1,  0.6, -1.1,  0.8, -1.0,  0.4,  0.1,  0.7, -0.2, -0.1, -0.1,  2.2,
> 0.7,  1.1,  0.2, -0.2, -0.9,  0.4,  0.1, -0.3, -0.4)
> x = c(4.1000,  4.9600,  1.2000,  3.9000,  3.1875,  1.9000,  1.8625,
> 0.7650,  1.5750,  2.4700,  1.6250,  1.5500,  2.3125,  1.3125,  1.0600,
> -0.5500,  1.1000,  0.0200, -0.0375,  3.4600,  2.5250,  2.0950,  0.8000,
> 1.6050, -0.4150, -0.7300,  1.1550,  1.4850,  2.2000,  2.2500,  0.6000,
> 2.1000)
> xyplot(y ~ x , data = junk.frm[junk.frm$z =="D",], type = c("g", "p",
> "smooth"), pch = 20)
> 
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