R version 2.15.0 (2012-03-30)
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
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> #
> # A hard-core test of losses and priors
> #  Simple data set where I know what the answers must be
> #
> library(rpart)
> aeq <- function(x,y, ...) all.equal(as.vector(x), as.vector(y), ...)
> 
> dummy <- c(3,1,4,1,5,9,2,6,5,3,5,8,9,7,9)/5
> pdata <- data.frame(y=factor(rep(1:3, 5)),
+                     x1 = 1:15,
+                     x2 = c(1:6, 1:6, 1:3),
+                     x3 = (rep(1:3, 5) + dummy)*10)
> 
> pdata$x3[c(1,5,10)] <- NA
> pdata$y[15] <- 1     # make things unbalanced
> 
> pfit <- rpart(y ~ x1 + x2 + x3, pdata,
+               cp=0, xval=0, minsplit=5, maxdepth=1,
+               parms=list(prior=c(.2, .3, .5), 
+                          loss =matrix(c(0,2,2,2,0,6,1,1,0), 3,3,byrow=T)))
> 
> #
> # See section 12.1 of the report for these numbers
> #
> 
> ntot   <- c(6,5,4)
> phat   <- c(6,5,4)/15   # observed class probabilities
> prior  <- c(.2, .3, .5) # priors
> aprior <- c(4,12,5)/21  # altered priors
> lmat   <- matrix(c(0,1,2, 2,0,1, 2,6,0), ncol=3)  #loss matrix
> 
> gini <- function(p) 1-sum(p^2)
> loss <- function(n, class)  sum(n * lmat[,class])
> 
> phat <- function(n, ntot=c(6,5,4), prior=c(.2, .3, .5)) {
+     n*prior/ntot
+     }
> 
> # Are the losses correct?
> # Class counts for the two children are (4,4,0) and (2,1,4), when
> #   using surrogates
> aeq(pfit$frame$dev/15, c(loss(prior,2), loss(phat(c(4,4,0)),2), 
+                                  loss(phat(c(2,1,4)),3)))
[1] TRUE
> # Node probabilities?
> aeq(pfit$frame$yval2[,8] , 
+     c(1, sum(phat(c(4,4,0))), sum(phat(c(2,1,4)))))
[1] TRUE
> 
> aeq(pfit$frame$yval2[,5:7] , rbind(prior, 
+                                    phat(c(4,4,0))/ sum(phat(c(4,4,0))), 
+                                    phat(c(2,1,4))/ sum(phat(c(2,1,4)))))
[1] TRUE
> 
> # Now the node and class probs, under altered priors
> phat2 <- function(n, ntot=c(6,5,4), prior=aprior) {
+     n*prior/ntot
+     }
> 
> # Use these to create the gini losses, base data, and for the best
> #  splits on variables 1, 2, 3
> gfun <- function(n) {  #The gini loss for a node, given the counts
+     temp <- phat2(n)
+     sum(temp) * gini(temp/sum(temp))
+     }
>            
> # These are in order x3, x2, x1 (best split to worst)
> # Note that for x3, missing values cause the "parent" to be viewed as
> #   having 12 obs instead of 15.
> # Each line is gini(parent) - gini(children)
> aeq(pfit$splits[1:3, 3],
+     15* c(gfun(c(4,4,4)) - (gfun(c(3,4,0)) +  gfun(c(1,0,4))),
+           gfun(c(6,5,4)) - (gfun(c(6,5,2)) +  gfun(c(0,0,2))),
+           gfun(c(6,5,4)) - (gfun(c(4,4,4)) +  gfun(c(2,1,0)))))
[1] TRUE
> 
> proc.time()
   user  system elapsed 
  0.211   0.059   0.315