# Day 1

# Simple arithmetic using R
2 * 5

1 - 1 + 3

4 + 5 / 5

# Parentheses supercede other operations
4 + (5 + 5)

4 + (5 + 5)
# This line won't be executed

# scientific notation
2/10000
1e9

# introduction to functions
# Think of a 3-part summary: what input does it take?
# what does it do to the input?
# what does it return?
log(45)
sin(3.14)
exp(2)

# nested functions; R evaluates from the
# inside out
log(exp(3))

# comparisons between values
2 > 3 # greater than
2 < 3 # less than
2 >= 3 # greater than or equal to
2 <= 3 # less than or equal to

# R is waiting!
(1 + 4)
# Click into the console and press the ESC button to 
# cancel the operation and start fresh.
  
# more comparisons
4 == 3 # is equal to
4 == 4 
4 != 3

# The Assignment operator: <-
x <- 4 + 3

# How to look at a variable that we have assigned:
x

bookshelves <- 4 == 3
bookshelves

# What happens if you put an exclamation point in
# front of a variable name?
!bookshelves

!ryan

x

# !x why does this return FALSE if x is 7?
x
as.logical(x)
!x

y <- 0
y
!y

# Using the = operator for assignment
myVariable = 5
myVariable

# we can reassign variables
x
x <- 5
x

x <- x + 4
x

# naming conventions
surveyID <- 5
surveyID

# periods in names
survey.ID <- 12
survey.ID
survey_ID <- 23
survey_ID


# How to get help
?log # brings up the help file for a particular function

??base

# Passing arguments to a function
log(3)
log(x = 3)
log(x = 10)

log(x = 3)
log(x = 3, base = 10)

# You must pass all required arguments
log()
log(base = 10)

# Rearranging arguments when calling a function
log(x = 3, base = 10)
log(base = 10, x = 3)

# R *can* take a bunch of unnamed arguments, and
# it will assume that they are in the order
# specified in the help file
# But this makes your code harder to read and
# understand
log(3, 10)

exp(1)

# Example of use of log() from the help file
?log
log(exp(3))
log10(1e7) # = 7

# MCQ
log(x = 1000, base = 10)
log10(1000)
log(base = 10, x = 1000)
log(10, 1000)

# When R wants to tell you something
ryan + 3
# An error means the code totally broke

# Message means that the operation worked,
# but R thinks that perhaps that's not what
# you wanted it to do.
log_of_a_negative <- log(-2)
log_of_a_negative

# These are ways that R might tell you something,
# but they don't necessarily indicate anything is 
# amiss (even though the message() looks red)
message("Hello, world!")
cat("Hi again, world")

print("Hello there, world!")


# Vectors and data frames
# vector: a collection of objects of the same type

1:10

numbers <- 1:10
numbers

fruit <- c("apple", "orange", "lemon")
fruit

odds <- c(3, 5, 7)
odds

# Vectorization 
odds + 1

odds <- odds + 2
odds
odds <- odds - 4
odds

log(3)
log(odds)

# Intro to loading data.
load("data/continents.RDA")
continents

# the head() function shows the first six rows
head(continents)
?head

head(continents, n = 3)

# the str() function
str(continents)

# summary()
summary(continents)

# what is the mean population across all of the continents?

1:10
1:1000

# Accessing a data frame
head(continents)

# Use the $ operator to access individual columns
continents$population

# vectorization with data frames (same as with vectors!)
continents$population / 2

log(x = continents$population)
log(x = continents$population, base = 10)

# vectors on vectors
continents$population / continents$area_km2

pop_density <- continents$population / continents$area_km2
pop_density

# add a new column to a data frame
continents$population_density <- continents$population / continents$area_km2

continents

load("data/continents.RDA")
continents

# anatomy of a vector
odds

odds[2]

fruit
fruit[3] # 3rd element
fruit[2:3] # 2nd and 3rd elements
fruit[-2] # whole vector without the second element

fruit[c(1, 3)]
fruit[10]

# subsetting with data frames
continents
continents$population
odds
continents$population[3]
continents$population[2:3]
continents$population[odds]

continents$population > 1e9

# subsetting with logical
continents$population[continents$population > 1e9]

# Subsetting a Dataframe: [ROWS, COLUMNS]

continents[2, 3]
continents[ , 3]
continents[3, ]
continents[1:3, ]
continents$population > 1e9

continents[continents$population > 1e9, ]

continents$population > 1e9
which(continents$population > 1e9)

continents[continents$continent == "Africa", ]