[R] using a variable for a column name in a formula

[arun]

Sorry, a mistake in the code:
#should be "collapse" instead of "sep"
res <- lm(formula(paste(nnn,"~",paste(Others, collapse="+"))),data=X)


A.K.


On Sunday, October 13, 2013 5:55 PM, arun <smartpink111 at yahoo.com> wrote:


Hi,

May be:
set.seed(24)
 X <- data.frame(weight=sample(100:250,20,replace=TRUE),height=sample(140:190,20,replace=TRUE))
Others <- colnames(X)[!colnames(X)%in%"height"]
 nnn <- "height"
res <- lm(formula(paste(nnn,"~",paste(Others, sep="+"))),data=X)
 res1<- lm(height~.,data=X)

#or
res2<- lm(get(nnn)~get(Others),data=X) #needs some renaming of rownames


 identical(coef(summary(res)),coef(summary(res1)))
#[1] TRUE
A.K.


On Sunday, October 13, 2013 5:06 PM, David Epstein <David.Epstein at warwick.ac.uk> wrote:
lm(height ~ ., data=X)
works fine.

However
nnn <- "height" ;  lm(nnn ~ . ,data=X)
fails

How do I write such a formula, which depends on the value of a string variable like nnn above?

A typical application might be a program that takes a data frame containing only numerical data, and figures out which of the columns can be best predicted from all the other columns.

Thanks
David

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