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STATS 782 
THE UNIVERSITY OF AUCKLAND 
SEMESTER 2, 2018 
Campus: City 
STATISTICS 
Statistical Computing 
(Time allowed: TWO Hours) 
INSTRUCTIONS 
• This examination is in two parts, Part A and Part B. 
• Attempt ALL questions in Part A. 
• Attempt ALL questions in Part B. 
• Unless asked for, avoid using explicit loops if ever possible. 
• The total marks for this examination are 100 marks. 
Page 1 of 15 
STATS 782 
PART A 
1. [10 Marks] Write down the evaluation results of the following R expressions, 
which are not necessarily meaningful in practice. [2 marks each] 
(a) 0:1 * 1:4 
(b) if ({1;2} >= 3) {4;5} else {6;7} 
(c) NA | TRUE ! FALSE 
(d) mean(0:9 > 7) 
(e) {s = 12; while(s > 5) s = s / 2; s} 
Answer: 
a. 0 2 0 4 
b. 7 
c. TRUE 
d. 0.2 
e. 3 
2. [6 Marks] Use :, seq(), rep() and some other commonly-used operators/- 
functions, but definitely not c() or a loop, to create the sequences below. [2 marks 
each] 
(a) 2 4 6 8 10 12 14 16 18 20 
(b) 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 
(c) "a1" "b1" "c1" "a2" "b2" "c2" "a3" "b3" "c3" 
Answer: 
a. 1:10 * 2 # or, seq(2, 20, by=2) 
b. rep(1:5, 1:5) 
c. paste0(letters[1:3], rep(1:3,each=3)) 
3. [8 Marks] Write an R function named ldss() that finds and returns the (first) 
longest decreasing subsequence of a given sequence, where decreasing does not have 
to be strict. Taking the following sequence as an example, 
> x = c(69, 16, 92, 83, 13, 28, 68, 37, 19, 19) 
your function should perform as follows 
> ldss(x) # longest decreasing subsequence 
[1] 68 37 19 19 
Page 2 of 15 
STATS 782 
Answer: 
## The three lines in the function below find, respectively: 
## (1) The index of the last element in each decreasing subsequence 
## (2) The length of each decreasing subsequence, and then the index 
## of the decreasing subsequence before the longest 
## (3) The longest decreasing subsequence 
ldss = function(x) { 
i = c(0, which(diff(x) > 0), length(x)) 
im = which.max(diff(i)) 
x[(i[im]+1):i[im+1]] 
4. [8 Marks] Write an R function named nnc() such that, for each of the centres 
(given in an R vector center), finds and returns the number of x-values (given in 
an R vector x) that are nearest to it. Taking the following data as an example, 
> x = c(48, 13, 33, 31, 21, 16, 50, 17, 29, 78) 
> centre = c(20, 60, 100) 
your function should perform as follows 
> nnc(x, centre) 
centre count 
[1,] 20 7 
[2,] 60 3 
[3,] 100 0 
The result means that 7 of the 10 values in x are nearest to center at 20 (of all three 
centres), 3 of them are nearest to centre at 60, and no x-value is nearest to centre 
at 100. Note that vectors x and center can be arbitrarily long. 
Answer: 
nnc = function(x, center) { 
xc = abs(outer(x, center, "-")) 
i = apply(xc, 1, which.min) 
cbind(center=center, count=tabulate(i, nbins=length(center))) 
 
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