R help - Dec 2004 - Updating a formula w a portion of another formula

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

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)

In the above we used the fact that formulas are
represented internally as trees whose parts
can be extracted using indexing.  After
running the above, try entering the following to
get a better idea of how formulas are represented:

	as.list(old)
	as.list(tmp)
	as.list(template)
	as.list(template[[3]])

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)