[18814] in s-news-athena
RE: [S] predict function
daemon@ATHENA.MIT.EDU (Buttrey, Samuel)
Mon Aug 2 12:30:21 1999
Message-Id: <DB17FBA383E4D211A9AF00A0C99E0169CAD9B6@monterey.nps.navy.mil>
From: "Buttrey, Samuel" <sebuttre@monterey.nps.navy.mil>
To: "'Lionel Siau'" <cyls98@btinternet.com>, S-news@wubios.wustl.edu
Date: Mon, 2 Aug 1999 09:30:12 -0700
Mime-Version: 1.0
Content-Type: text/plain
fan@katan.cybercom.net and Lionel Siau between them raise an interesting
point. To answer the first question, if you're going to use a matrix for
your "new.data" in predict, then it should match up exactly with what the
original matrix would have looked like. In fan@katan's example:
a <-1:10
b<-a+rnorm(10)
mod <- lm(b~a)
# what's the prediction when x=11?
predict(mod, 11) # this doesn't work for me at all
predict (mod, c(1, 11)) # adding the intercept explicitly produces
what you need.
Better:
predict (mod, data.frame (a = 11)) # supplying the names gets rid of
the need for the intercept
To answer Lionel's question, yes, *in linear models* you need supply only
the required columns. But in trees, you apparently need to supply a data
frame with all the original columns -- even if those column aren't actually
used in building the tree. For example, consider this data frame in which
only c will be used to model d:
> test <- data.frame (a = rnorm (10), b = rexp (10), c = 1:10, d = 1:10 +
rnorm (10, sd = .3))
> test.tree <- tree (d ~ ., data = test)
> summary (test.tree)
...
Variables actually used in tree construction:
[1] "c" ...
So I should be able to predict using just c, I might have thought. But no.
> predict (test.tree, data.frame (c = 2:6))
Error in predict.tree(test.tree, data.frame(c = 2:6)): Length of variable 1
is 10 != length of row names (5)
Dumped
If I make up and a and b and stick them in there, it works:
> predict (test.tree, data.frame (c = 2:6, a = letters[1:5], b =
letters[1:5]))
1 2 3 4 5
2.95272 2.95272 2.95272 2.95272 8.012537
> predict (test.tree, data.frame (c = 2:6, a = letters[1:5], b =
letters[1:5]))
1 2 3 4 5
2.95272 2.95272 2.95272 2.95272 8.012537
> predict (test.tree, data.frame (c = 2:6, a = 100:104, b = log(100:104)))
1 2 3 4 5
2.95272 2.95272 2.95272 2.95272 8.012537
>
How 'bout that?
Sam Buttrey
buttrey@nps.navy.mil
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