Gabor Grothendieck <ggrothendieck <at> myway.com> writes:
:
: Daniel Almirall <dalmiral <at> umich.edu> writes:
:
: :
: : R-list,
: :
: : Suppose I have
: :
: : oldfmla <- y ~ x
: :
: : I would like to update it to y ~ x + z which I know I can get using
: :
: : newfmla <- update(oldfmla, ~ . + z)
: :
: : However, what if I have
: :
: : fmlatmp <- ~ z
: :
: : Can I combine oldfmla and fmlatmp to get y ~ x + z some how?
: :
: : Clearly,
: :
: : newfmla <- update(oldfmla, ~ . + fmlatmp)
: :
: : will not work.
: :
: : Thanks in advance,
: : Danny
:
: Assuming we have:
:
: old <- y ~ x
: tmp <- ~ z
:
: Then type in this:
:
: template <- . ~ . + X
: template[[3]][[3]] <- tmp[[2]]
: update(old, template)
:
One can alternately use substitute avoiding determination of
the precise indices of X. Note that the eval is required
since substitute returns a call object and eval turns it
back into a formula (as per the Note near the bottom of
?substitute):
tmp2 <- eval(substitute(. ~ . + X, list(X = tmp[[2]])))
update(old, tmp2)
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