persp question
On 11/03/2008 2:40 PM, markleeds at verizon.net wrote:
someone sent in a question earlier about doing
something in 3D so i took a stab at it purely
for educational purposes ( i'm not even sure that I understood the question actually ).
Unfortunately, persp gives me an error that I don't understand because it says "object y not found". I'm sending y in as a parameter to persp similar to what ?persp shows in one of oits examples so I must not be understanding something. the code is below. thanks.
DF <- read.table(textConnection("station month bas
190 5 0.00
190 7 1.563
190 10 0.000
190 11 0.000
202 4 18.750
202 5 18.750
202 7 6.250
202 10 4.80
202 11 3.125
198 4 18.750
198 10 3.20
198 11 12.500
205 4 0.000
205 5 0.000
205 10 0.000
205 11 0.00"),header=TRUE,stringsAsFactors=FALSE)
#print(DF)
#print(str(DF))
temp1 <- seq(1,max(DF$month),length.out=max(DF$month))
temp2 <- seq(min(DF$station),max(DF$station),by=1)
print(temp1)
print(temp2)
persp(x = seq(1,max(DF$month),length.out=max(DF$month)),
y = seq(min(DF$station),max(DF$station),by=1),
z = DF$bas,
xlim=range(x), ylim=range(y), zlim=range(z))
The values you pass to persp() are expressions to be evaluated in your workspace. Since you don't have variables x, y or z defined, you can't specify range(x), range(y) or range(z) as arguments. (Things are different for the defaults for the parameters: those are evaluated in the local frame set up for the call.) Duncan Murdoch