• if
• else
• ifelse()

if (logical argument){ # e.g., x > 1 or is.character(x)
  body
  }

• logical statement must be TRUE to proceed…
• otherwise nothing happens
• To access help for if you have to quote it: ?"if"

data(rivers) #length of 141 "major" North American rivers (miles)
head(rivers)

## [1] 735 320 325 392 524 450

is.numeric(rivers)

## [1] TRUE

hist(log10(rivers), breaks=30)
abline(v=log10(median(rivers)),col=3, lwd=3)

if (is.numeric(rivers)){
  median(rivers)
  }

## [1] 425

You can put it all on one line, but…

if (is.numeric(rivers)){ median(rivers) }

## [1] 425

if (!is.numeric(rivers)){
  median(rivers)
  }

• ! negates the logical
• is.numeric(rivers) is TRUE, so !is.numeric(rivers) is FALSE
• Nothing happens

• 2nd (optional) half of if else statements

# Pass character object
x<-letters
if (is.numeric(x)){
  median(rivers)
  } else {
  class(x)
}

## [1] "character"

# Pass numeric object
x<-1:10
if (is.numeric(x)){
  median(rivers)
  } else {
  class(x)
}

## [1] 425

The closing brace of if and the else keyword must be on the same line.

# WRONG -- R thinks the if statement is done
if (is.numeric(x)){
  median(rivers)
  }
else {
  class(x)
}

# RIGHT -- else follows the closing brace
if (is.numeric(x)){
  median(rivers)
  } else {
  class(x)
}

• if/else is for scalar decisions
• What if we need to apply a test to every element of a vector?

Single element example

x<-TRUE
ifelse(x, 1, 2)

## [1] 1

• ifelse(test, TRUE , FALSE)

Long vector example

lowercase<-TRUE
ifelse(lowercase, letters, LETTERS)

## [1] "a"

lowercase<-c(TRUE, FALSE, TRUE, FALSE)
ifelse(lowercase, letters, LETTERS)

## [1] "a" "B" "c" "D"

• Result of ifelse always the same shape (length usually) as the logical
• Evaluates each element one-at-a-time

river.bin<-ifelse(rivers > median(rivers), "Long", "Short")
river.bin[1:10] # Just look at the first 10

##  [1] "Long"  "Short" "Short" "Short" "Long"  "Long"  "Long"  "Short" "Long" 
## [10] "Long"

Make ordered factor:

river.bin<-factor(river.bin, 
  levels= rev(unique(river.bin)), ordered=T)
str(river.bin)

##  Ord.factor w/ 2 levels "Short"<"Long": 2 1 1 1 2 2 2 1 2 2 ...

# Scalar decision
if (is.numeric(rivers)) median(rivers)

# Vectorized classification
ifelse(rivers > median(rivers), "Long", "Short")

for (each element in a vector){
  if(an element meets this condition){
    do this stuff
    } else {
    do this other stuff
    }
}

Using a for loop and an if statement, print only the elements of par() with more than three elements.

1. Figure out the loop structure.
2. Figure out the conditional (using if).

1.Figure out the loop structure.

for(i in par()){
    print(i)
}

This prints every element of par(). Now we need to filter.

2.Figure out the conditional (using if).

for(i in par()){
  if (length(i) >3){
    print(i)
  }
}

## [1] 0 1 0 1
## [1] 1.02 0.82 0.82 0.42
## [1] 5.1 4.1 4.1 2.1
## [1] 1 1 1 1
## [1] 0 0 0 0
## [1] 0 1 0 1
## [1] 0 0 0 0
## [1] 0.1171429 0.9400000 0.1700000 0.8633333
## [1] 0 1 0 1

• if (test){ body } – runs body when test is a single TRUE
• else { body } – runs when test is FALSE; must follow } on the same line
• ifelse(test, yes, no) – vectorized; applies to every element
• if/else for control flow (one decision)
• ifelse() for data transformation (many decisions at once)