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using names() as a variable in a formula
3 messages · femke, Sundar Dorai-Raj, Duncan Murdoch
femke wrote:
Greetings List, I'm having some trouble with the use of the names function in a formula. Below is an example of something that works (i.e first line), and the second line is the same formula which doesn't. I want to be able to reference the column of the dataC table so I can run the variogram iteratively with a loop.
v<-variogram(A1~1,loc=~x+y, dataC) v<-variogram(names(dataC[3])~1,loc=~x+y, dataC)
Error in model.frame(formula, rownames, variables, varnames, extras, extranames, :
invalid variable type
Where dataC looks like:
x y A1 A2 A3
1 514030 4793587 0.0 7.9 0.1
2 517995 4792516 5.8 5.1 0.0
3 514232 4792210 0.0 6.5 0.0
I though initially that it might need some escape character if quotes are added, and tried the following, but it looks ok.
names(dataC[3])
[1] "A1"
mode(names(dataC[3]))
[1] "character"
v<-variogram(as.character(names(dataC[3]))~1,loc=~x+y, dataC)
Error in model.frame(formula, rownames, variables, varnames, extras, extranames, :
invalid variable type
v<-variogram(as.formula((names(dataC[3]))~1),loc=~x+y, dataC)
Error in model.frame(formula, rownames, variables, varnames, extras, extranames, :
invalid variable type
I'd greatly appreciate any suggestions for fixing this.
Thanks again,
femke
[[alternative HTML version deleted]]
You're not building a valid formula. Try this:
v <- list()
for(i in 3:5) {
gr <- names(dataC[i])
f <- formula(paste(gr, "1", sep = " ~ "))
v[[gr]] <- variogram(f, loc = ~ x + y, dataC)
}
BTW, since variogram is not in the base package it would also be helpful
in the future if you add that you are using the spatial package.
HTH,
Sundar
On Wed, 18 Feb 2004 15:44:59 -0500, "femke" <femke at geog.umd.edu> wrote
v<-variogram(A1~1,loc=~x+y, dataC)
What package is variogram in? The one in spatial takes different args.
v<-variogram(names(dataC[3])~1,loc=~x+y, dataC)
Doing this sort of thing is tricky. Your names(dataC[3]) is a
character string; you want something there that would be interpreted
as a name instead. However, as.name(names(dataC[3])) isn't enough.
I think you need to build up the formula using fairly low level stuff,
something like this:
formula <- as.call(list(as.name('~'), as.name(names(dataC[3])),
quote(1)))
and then do
variogram(formula, ....)
but a formula constructed this way doesn't work in lm(), so it may not
work in variogram either.
Duncan Murdoch