On Nov 2, 2009, at 10:40 PM, Ben Bolker wrote:
with the following simple data frame
dd = structure(list(z = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L
), .Label = c("a", "b"), class = "factor"), x = c(0.3, 0.2, 0.1,
0, 0, 0, 0.2, 0.3)), .Names = c("z", "x"), row.names = c(NA,
-8L), class = "data.frame")
I would like know if it's possible to use model.matrix()
to construct the following design matrix in some sensible way:
za zb
1 1 0
2 1 0
3 1 0
4 0 0
5 0 0
6 0 0
7 0 1
8 0 1
In other words, I want column 1 to be (z=="a" & x>0) and
column 2 to be (z=="b" & x>0). I can construct this matrix
using
sweep(model.matrix(~z-1,dd),1,dd$x>0,"*")
and then stick it into lm.fit -- but is there a more
elegant way to do this in general?
Elegance? You decide. But at least more economical from a keystroke
perspective:
model.matrix(~z-1,dd)*(dd$x>0)
#------
za zb
1 1 0
2 1 0
3 1 0
4 0 0
5 0 0
6 0 0
7 0 1
8 0 1
attr(,"assign")
[1] 1 1
attr(,"contrasts")
attr(,"contrasts")$z
[1] "contr.treatment
I haven't found a formula combining
(z-1) and I(x>0) that works, although I can imagine there is one.
thanks
Ben Bolker
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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