Homework Solutions to Problems 1--2.
===================================
[NB solutions are in Splus 6.0]

(1) 
> V1 <- rep(c(0,1,1),40)
  V2 <- sin(2*pi*sqrt(1:120))
### A better version (used below) would be:   
###       V2 <- round(sin(2*pi*sqrt(1:120)),1e-12)

  V3 <- ifelse(V2>=0, "H", "L")
  newmat <- cbind(Group = V1, Y = V2, Type = V3) 
  newfram <- cbind.data.frame(Group = V1, Y = V2, Type = V3,
      stringsAsFactors=F) 
### Note that the "stringsAsFactors=F" optional argument is
###   needed to avoid automatic conversion of character data
###   to factor-type. 
> unlist(lapply(newfram,class))
     Group         Y        Type 
 "integer" "numeric" "character"

> newfram[1:5,]
  Group  Y Type 
1     0  0    H
2     1  1    H
3     1 -1    L
4     0  0    H
5     1  1    H
> newmat[1:5,]
     Group    Y Type 
[1,] "0"   "0"  "H" 
[2,] "1"   "1"  "H" 
[3,] "1"   "-1" "L" 
[4,] "0"   "0"  "H" 
[5,] "1"   "1"  "H" 

(2) 

> exmpfram <- read.table(file="exampfram.txt", stringsAsFactors=F, 
    skip=1)
> dim(exmpfram)
[1] 216  13
> exmpfram <- exmpfram[,3:9]
> names(exmpfram) <- c("IDNUM","DTH","EVTTIME","TRTGP","LOGBILI",
    "AGEVAR","CIRRH")
> exmpfram$CIRRH <- exmpfram$CIRRH == "T"
> class(exmpfram$CIRRH)             
[1] "logical"

> exmpfram2 <- scan(file="exampfram.txt", skip=1)
### This is a giant vector, of type "character" and length 216*13=2808
### We are interested in pulling out the elements 3:8 in each row,
### which now correspond to elements with indices 3:8 plus 
### multiples 0:215 of 13.

> inds <- rep(3:9,216) + rep(13*(0:215),rep(7,216))
  inds2 <- c(outer(3:9, 13*(0:215), "+"))
> sum(abs(inds-inds2))
[1] 0
> exmpfram2 <- data.frame(matrix(exmpfram2[inds], ncol=7, byrow=T,
     dimnames=list(NULL,  names(exmpfram))), stringsAsFactors=F)
> exmpfram2[,7] <- as.logical(exmpfram2[,7])
> for(i in c(1:3,5:6)) exmpfram2[,i] <- as.numeric(exmpfram2[,i])
> unlist(lapply(exmpfram2,class))
     IDNUM       DTH   EVTTIME       TRTGP   LOGBILI    AGEVAR     CIRRH 
 "numeric" "numeric" "numeric" "character" "numeric" "numeric" "logical"

> c(sum(exmpfram == exmpfram2), sum(unlist(lapply(exmpfram2,class))==
      unlist(lapply(exmpfram,class))))
[1] 1512  7
### solves part (c), since all 216*7 entries match (including type)

> t(apply(exmpfram[,c(1:3,5:6)],2,range))     ### solution to part (d)
           [,1]      [,2] 
  IDNUM 1.00000 248.00000
    DTH 0.00000   1.00000
EVTTIME 0.00000 139.44000
LOGBILI 0.69897   2.72263
 AGEVAR 1.64900 330.30000   ### same matrix is produced by:
> cbind(apply(exmpfram[,c(1:3,5:6)],2,min),
      apply(exmpfram[,c(1:3,5:6)],2,max))