On 02/08/2011 10:48 AM, Thaler,Thorn,LAUSANNE,Applied Mathematics wrote:
> mm<- function(datf) {
> lm(y ~ x, data = datf)
> }
> mydatf<- data.frame(x = rep(1:2, 10), y = rnorm(20, rep(1:2, 10)), z
=
> rnorm(20))
>
> l<- mm(mydatf)
> update(l, . ~ . + z) # This fails, z is not found
G
> mm <- function(datf) {
>lm(y ~ x, data = datf)
> }
> mydatf <- data.frame(x = rep(1:2, 10), y = rnorm(20, rep(1:2, 10)), z
=
> rnorm(20))
>
> l <- mm(mydatf)
> update(l, . ~ . + z) # This fails, z is not found
Good point. So let me rephrase the initial problem:
1.) An lm object is fitted
On 02/08/2011 9:41 AM, Duncan Murdoch wrote:
It looks to me as though your proposal would allow update to remove
variables, but would give erroneous results when adding them. For example:
mm<- function(datf) {
lm(y ~ x, data = datf)
}
mydatf<- data.frame(x = rep(1:2, 10), y = rnorm(20, rep(
It looks to me as though your proposal would allow update to remove
variables, but would give erroneous results when adding them. For example:
mm <- function(datf) {
lm(y ~ x, data = datf)
}
mydatf <- data.frame(x = rep(1:2, 10), y = rnorm(20, rep(1:2, 10)), z =
rnorm(20))
l <- mm(mydatf)
Dear all,
Suppose the following code:
--8<--
mm <- function(datf) {
lm(y ~ x, data = datf)
}
mydatf <- data.frame(x = rep(1:2, 10), y = rnorm(20, rep(1:2, 10)))
l <- mm(mydatf)
-->8--
If I want to update l now without providing the data argument
