For the record: lattice behaves slightly differently, it requires an
as.expression with bquote.
Dieter
plotExp - function(what) {
plot.new()
lab = bquote(Estimated~t[50]~ from ~.(what) ) ;
text(0.5,0.2,lab)
}
plotExp(tgv)
library(lattice)
plotLattice - function(what) {
lab =
Disclaimer: I have read plotmath, but maybe it's too late today:
How do I get the two labels to be the same:
plot.new()
lab =expression(paste(Estimated , t[50], from tgv))
text(0.5,0.5,lab)
# Should look the same as above. I could not get the substitute right:
what = tgv
lab
hi,
try this
lab =bquote(paste(Estimated , t[50], from ,.(what)))
HTH,
baptiste
On 27 August 2010 20:19, Dieter Menne dieter.me...@menne-biomed.de wrote:
plot.new()
lab =expression(paste(Estimated , t[50], from tgv))
text(0.5,0.5,lab)
# Should look the same as above. I could not get the
On Aug 27, 2010, at 2:19 PM, Dieter Menne wrote:
Disclaimer: I have read plotmath, but maybe it's too late today:
How do I get the two labels to be the same:
plot.new()
lab =expression(paste(Estimated , t[50], from tgv))
text(0.5,0.5,lab)
# Should look the same as above. I could not get the
Thanks to both of you. I noted that my example was over-simplified. Looks
like I need to correct the environment when nested in a function, but I have
to catch the last bus now.
Dieter
plotExp = function(what) {
plot.new()
lab =expression(paste(Estimated , t[50], from tgv))
On Aug 27, 2010, at 2:28 PM, baptiste auguie wrote:
hi,
try this
lab =bquote(paste(Estimated , t[50], from ,.(what)))
bquote doesn't need the paste() if you use plotmath separators:
lab =bquote(Estimated ~t[50]~from~.(what))
text(0.5,0.2,lab)
HTH,
baptiste
On 27 August 2010 20:19,
On Aug 27, 2010, at 2:51 PM, Dieter Menne wrote:
Thanks to both of you. I noted that my example was over-simplified.
Looks
like I need to correct the environment when nested in a function,
but I have
to catch the last bus now.
Dieter
plotExp = function(what) {
plot.new()
lab
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