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design matrix construction question

4 messages · Ben Bolker, David Winsemius, Ken Knoblauch +1 more

Ben Bolker
# 
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?  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
#
On Nov 2, 2009, at 10:40 PM, Ben Bolker wrote:

            

      
      
      
    

        
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

            

      
      
      
    

        
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      Ken Knoblauch
    

    

      
      #
    

  


  

          

      
      
      
    

        
How about something like this?

dd$xf <- (dd$x > 0) + 0
model.matrix(~ z:xf - 1, dd)
  za:xf zb:xf
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"

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Greg Snow
    

    

      
      #
    

  


  

      
Try this:

dd$i <- with(dd,interaction(z,x>0))
contrasts(dd$i,2) <- rbind( c(0,0),c(0,0),c(1,0),c(0,1) )
model.matrix( ~i, dd )[,-1]