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] "tails"

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] "paper"

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 1297 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

numbers2words(312)

[1] "three hundred twelve"

substr('Iain', 2, 3)

[1] "ai"

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 [[]]

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

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

[1] 1 2 3 4 5