Gabor Grothendieck a écrit :
Try this:
ix <- 1:2
lm(as.matrix(freeny[ix]) ~., freeny[-ix])
clean and clever!!! Thanks a lot!! You really simplified the code!!!
Just for curiosity, do you see why parse(eval)) was not working twice in
same formula?
thanks a lot!!
Matthieu
On Sun, Nov 29, 2009 at 8:56 AM, Matthieu Stigler
<matthieu.stig...@gmail.com <mailto:matthieu.stig...@gmail.com>> wrote:
Thanks for answering so fast!!
lm(freeny)
:-)
Ok that's working for the one equation case :-) Was example case...
But now I want to have not only first column of freeny on the left
but both first? And I don't know their names a priori...
Thanks!
On Sun, Nov 29, 2009 at 8:49 AM, Matthieu Stigler
<matthieu.stig...@gmail.com
<mailto:matthieu.stig...@gmail.com>
<mailto:matthieu.stig...@gmail.com
<mailto:matthieu.stig...@gmail.com>>> wrote:
Hi
My goal is to do a (multiple) regression, just knowing that
my Y
variables will be the say k first variables of a matrix/data
frame. I thought I should do it with eval(parse)) but
encounter a
strange problem.
See:
lm(y~.-y, data=freeny) #that's what I want to do in the one
equation case
#Problem is I don't know name of the variable... only that
it is
the first one...
#so idea is to just take first name
a<-colnames(freeny)
#and then use eval(parse(text=a[1]))
#it works if I replace y on either the left or right side:
lm(eval(parse(text=a[1]))~.-y, data=freeny) #does the same
lm(y~.-eval(parse(text=a[1])), data=freeny)
#but not if I do this call twice:
lm(eval(parse(text=a[1]))~.-eval(parse(text=a[1])),
data=freeny)
#variable I wanted to remove (y) ist still there
Do you understand why I can call eval(parse) only once?
Should I
try a update workaround? Or have idea of any other
solution? Maybe
there is something much simpler I'm missing:-(
Thanks a lot!!!
Matthieu Stigler
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