Programming in R

• conditionals (if/else)
• loops (for, while)
• how to define your own functions
• basic programming flow

• section 19 (functions)
• 21.1-21.3

if (2 > 1) {
  print('fact')
}

[1] "fact"

if (2 < 1) {
  print('Math is Broken')
}

# Flip a coin 
if (runif(1) < 0.5) {
    print('heads')
} else {
    print('tails')
}

[1] "heads"

r <- runif(1)
# rock paper sissors
if (r < 1/3) {
    print('rock')
} else if (1/3 < r && r < 2/3) {
    print('paper')
} else{
    print('scissors')
}

[1] "scissors"

If you have lots of nested If-thens, a switch() function could be more handy. Here is an example from Section 19.4.2. What does it do?

function(x, y, op) {
  switch(op,
    plus = x + y,
    minus = x - y,
    times = x * y,
    divide = x / y,
    stop("Unknown op!")
  )
}

2*5 == 10

[1] TRUE

# oops
sqrt(2)^2 == 2

[1] FALSE

dplyr::near(sqrt(2)^2, 2)

[1] TRUE

c(1, 1, 1) == c(1, 2, 1)

[1]  TRUE FALSE  TRUE

default to && or || in a for loop

c(T, T, T) || c(T, F, T)

[1] TRUE

This operator appears to look at the first term of each vector only.

c(T, T, T) | c(T, F, T)

[1] TRUE TRUE TRUE

• for
• while

for (i in 1:10) {
    print("Oh Canada!")
    print(i)
}

[1] "Oh Canada!"
[1] 1
[1] "Oh Canada!"
[1] 2
[1] "Oh Canada!"
[1] 3
[1] "Oh Canada!"
[1] 4
[1] "Oh Canada!"
[1] 5
[1] "Oh Canada!"
[1] 6
[1] "Oh Canada!"
[1] 7
[1] "Oh Canada!"
[1] 8
[1] "Oh Canada!"
[1] 9
[1] "Oh Canada!"
[1] 10

nums <- vector("double", 10) # or rep(0, 10) or something else
for (i in 1:10) {
  nums[i] <- runif(1)
}

# nums <- c()
# for (i in 1:10) {
#   nums <- c(nums, runif(1))
# }

current_position <- 10
n_iter <- 0
while (current_position > 0){
    current_position <- current_position + rnorm(1)
    n_iter <- n_iter + 1
}
print(paste0('you lost all your money after ', n_iter, ' trips to the casino'))

[1] "you lost all your money after 1205 trips to the casino"

while (TRUE){
    print('Duke sucks')
}

Try to vectorize anything you can (once you learn what that means…)

sapply(1:10, function(x) x * 2)

• break long program up into small chunks
• more readable
• code reuse
  • catch errors
  • don't have to copy/paste
• only need to fix code in one place

You should consider writing a function whenever you’ve copied and pasted a block of code more than twice

power <- function(num, exponent){
    # returns num raised to the exponent
    num ^ exponent
}

power(2, 3)

[1] 8

power <- function(num, exponent=3){
    # returns num raised to the exponent
    num ^ exponent
}

power(2)

[1] 8

random_rps <- function(){
    # randomly returns one of rock, paper or scissors

    r <- runif(1)
    # rock paper sissors
    if (r < 1/3) {
        return('rock')
    } else if (1/3 < r && r < 2/3) {
        return('paper')
    } else{
        return('scissors')
    }
}

random_rps()

[1] "paper"

source('fun.R')

helper_fun()

[1] "Im not a very helpful helper function"

• vectors are homogeneous and sequential (1 dimensional)
• lists are heterogeneous and hierarchical

section 20 from r4ds

boolean, character, complex, raw, integer and double

c(1,2,3) # 1:3

[1] 1 2 3

# boolean
c(TRUE, FALSE, TRUE) 

[1]  TRUE FALSE  TRUE

# string
c('Iain', 'wishes', 'vectors', 'were', 'named', 'lists', 'instead')

[1] "Iain"    "wishes"  "vectors" "were"    "named"   "lists"   "instead"

typeof(rep(TRUE, 4))

[1] "logical"

What are the types of the following vectors

# a
c(1, 2, 'three')

# b
c(TRUE, TRUE, "FALSE")

# c
c(1, 2, 3.1)

as.integer(c('1', '2', '3'))

[1] 1 2 3

c(1, 2, TRUE)

[1] 1 2 1

sum(c(-2, -1, 1, 2) > 0)

[1] 2

v <- 11:20
v[3]

[1] 13

v[c(1,10)]

[1] 11 20

v[v %%2 == 0]

[1] 12 14 16 18 20

Create the numbers2words function from https://github.com/ateucher/useful_code/blob/master/R/numbers2words.r

Copy this code to an R-script, name it, save it, source it and run the function with a few different arguments.

numbers2words(312)

[1] "three hundred twelve"

substr('Iain', 2, 3)

[1] "ai"

How many numbers below 4869 are divisible by 5 and start with the letter “f”?

Use the two previous slides and submit your answer on Sakai.

How many numbers below 4869 are divisible by three and start with the letter n?

sum(1:4869 %% 3 == 0 & substr(numbers2words(1:4869), 1, 1) == 'n' )

[1] 39

Lists can contain objects of multiple types and are indexed by names (as opposed to index sequentially)

L <- list(number=1, letter='a', bool=TRUE)
L

$number
[1] 1

$letter
[1] "a"

$bool
[1] TRUE

To access elements of a list use [[]]

L[['number']]

[1] 1

you can use a single [] and this will return a list

L['number']

$number
[1] 1

see section 20.5.3 for the difference between [] and [[]]

a <- list(a = 1:3, b = "a string", c = pi, d = list(-1, -5))

Now look at a, a[2] and a[[2]] and note the differences.

LoL <- list(names = list('Iain', 'Brendan', 'Varun'),
            numbers=list(1:3, 1:5, 1:7))

LoL[['numbers']][[2]]

[1] 1 2 3 4 5