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('alternative fact')
}

# 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"

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

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

[1] TRUE TRUE TRUE

• for
• while

for (i in 1:10) {
  print(i)
}

[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
[1] 6
[1] 7
[1] 8
[1] 9
[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 101 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"

I basically know of two principles for treating complicated systems in simple ways; the first is the principle of modularity and the second is the principle of abstraction. I am an apologist for computational probability in machine learning, and particularly for graphical models and variational methods, because I believe that probability theory implements these two principles in deep and intriguing ways – namely through factorization and through averaging. Exploiting these two mechanisms as fully as possible seems to me to be the way forward in machine learning.

Michael I. Jordan Massachusetts Institute of Technology, 1997.

(quote borrowed from Ben Vigoda's thesis)

• modularity
  • break a complicated task into many smaller sub-tasks
• abstraction
  • you don't need to know the inner workings of every part of the system, just how things interact

• testing
• vectorization
• recursion
• debugging / profiling
• environments / scoping
• object oriented programming
• functional programming
• non-standard evaluation
• dynamic programming
• memory usage
• using Rcpp

• 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('I', 'wish', 'vectors', 'were', 'named', 'lists', 'instead')

[1] "I"       "wish"    "vectors" "were"    "named"   "lists"   "instead"

typeof(rep(TRUE, 4))

[1] "logical"

What are the types of the following vectors https://goo.gl/forms/Rcpz7rQxUQPmc3Cu1

# 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 <- 1:10
v[c(1,10)]

[1]  1 10

v[v %%2 == 0]

[1]  2  4  6  8 10

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?

https://goo.gl/forms/VBt4BOHQZPdNrPJn2

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