Data Frames and Control

• Making and working with data frames
• Conditionals: switching between different calculations
• Iteration: Doing something over and over
• Vectorizing: Avoiding explicit iteration

• Vectors: series of values all of the same type
  v[5], v[“name”]
• Arrays: multi-dimensional generalization of vectors a[5,6,2], a[,6,], a[rowname, colname, layername]
• Matrices: special 2D arrays with matrix math
  m[5,6], m[,6], m[,colname]
• Lists: series of values of mixed types
  l[[3]], l$name
• Dataframes: hybrid of matrix and list

• 2D tables of data
• Each case/unit is a row
• Each variable is a column
• Variables can be of any type (numbers, text, Booleans, …)
• Both rows and columns can get names

library(datasets)
states <- data.frame(state.x77, abb=state.abb, region=state.region, division=state.division)

data.frame() is combining here a pre-existing matrix (state.x77), a vector of characters (state.abb), and two vectors of qualitative categorical variables (factors; state.region, state.division)

Column names are preserved or guessed if not explicitly set

• By row and column index

states[49,3]

[1] 0.7

• By row and column names

states["Wisconsin","Illiteracy"]

[1] 0.7

• All of a row:

states["Wisconsin",]

          Population Income Illiteracy Life.Exp Murder HS.Grad Frost  Area
Wisconsin       4589   4468        0.7    72.48      3    54.5   149 54464
          abb        region           division
Wisconsin  WI North Central East North Central

Exercise: what class is states["Wisconsin",]?

• All of a column:

head(states[,3])

[1] 2.1 1.5 1.8 1.9 1.1 0.7

head(states[,"Illiteracy"])

[1] 2.1 1.5 1.8 1.9 1.1 0.7

head(states$Illiteracy)

[1] 2.1 1.5 1.8 1.9 1.1 0.7

• Rows matching a condition:

states[states$division=="New England", "Illiteracy"]

[1] 1.1 0.7 1.1 0.7 1.3 0.6

states[states$region=="South", "Illiteracy"]

 [1] 2.1 1.9 0.9 1.3 2.0 1.6 2.8 0.9 2.4 1.8 1.1 2.3 1.7 2.2 1.4 1.4

Parts or all of the dataframe can be assigned to:

summary(states$HS.Grad)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   37.8    48.0    53.2    53.1    59.2    67.3 

states$HS.Grad <- states$HS.Grad/100
summary(states$HS.Grad)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.378   0.480   0.532   0.531   0.592   0.673 

states$HS.Grad <- 100*states$HS.Grad

What percentage of literate adults graduated HS?

head(100*(states$HS.Grad/(100-states$Illiteracy)))

[1] 42.19 67.72 59.16 40.67 63.30 64.35

with() takes a data frame and evaluates an expression “inside” it:

with(states, head(100*(HS.Grad/(100-Illiteracy))))

[1] 42.19 67.72 59.16 40.67 63.30 64.35

Have the computer decide what to do next

• Mathematically: \[ |x| = \left\{ \begin{array}{cl} x & \mathrm{if}~x\geq 0 \\ -x &\mathrm{if}~ x < 0\end{array}\right. ~,~ \psi(x) = \left\{ \begin{array}{cl} x^2 & \mathrm{if}~|x|\leq 1\\ 2|x|-1 &\mathrm{if}~ |x| > 1\end{array}\right. \]
  Exercise: plot \( \psi \) in R
• Computationally:

if the country code is not "US", multiply prices by current exchange rate

Simplest conditional:

if (x >= 0) {
  x
} else {
  -x
}

Condition in if needs to give one TRUE or FALSE value

else clause is optional

one-line actions don't need braces

if (x >= 0) x else -x

if can nest arbitrarily deeply:

if (x^2 < 1) {
  x^2
} else {
  if (x >= 0) {
    2*x-1
  } else {
     -2*x-1
  }
}

Can get ugly though

& work | like + or *: combine terms element-wise

Flow control wants one Boolean value, and to skip calculating what's not needed

&& and || give one Boolean, lazily:

(0 > 0) && (all.equal(42%%6, 169%%13))

[1] FALSE

This never evaluates the complex expression on the right

Use && and || for control, & and | for subsetting

Repeat similar actions multiple times:

table.of.logarithms <- vector(length=7,mode="numeric")
table.of.logarithms

[1] 0 0 0 0 0 0 0

for (i in 1:length(table.of.logarithms)) {
  table.of.logarithms[i] <- log(i)
}
table.of.logarithms

[1] 0.0000 0.6931 1.0986 1.3863 1.6094 1.7918 1.9459

for (i in 1:length(table.of.logarithms)) {
  table.of.logarithms[i] <- log(i)
}

for increments a counter (here i) along a vector (here 1:length(table.of.logarithms)) and loops through the *body until it runs through the vector

“iterates over the vector”

N.B., there is a better way to do this job!

Can contain just about anything, including:

• if() clauses
• other for() loops (nested iteration)

c <- matrix(0, nrow=nrow(a), ncol=ncol(b))
if (ncol(a) == nrow(b)) {
  for (i in 1:nrow(c)) {
    for (j in 1:ncol(c)) {
      for (k in 1:ncol(a)) {
        c[i,j] <- c[i,j] + a[i,k]*b[k,j]
      }
    }
  }
} else {
  stop("matrices a and b non-conformable")
}

while (max(x) - 1 > 1e-06) {
  x <- sqrt(x)
}

Condition in the argument to while must be a single Boolean value (like if)

Body is looped over until the condition is FALSE so can loop forever

Loop never begins unless the condition starts TRUE

for() is better when the number of times to repeat (values to iterate over) is clear in advance

while() is better when you can recognize when to stop once you're there, even if you can't guess it to begin with

Every for() could be replaced with a while()
Exercise: show this

R has many ways of avoiding iteration, by acting on whole objects

• It's conceptually clearer
• It leads to simpler code
• It's faster (sometimes a little, sometimes drastically)

How many languages add 2 vectors:

c <- vector(length(a))
for (i in 1:length(a)) {  c[i] <- a[i] + b[i]  }

How R adds 2 vectors:

a+b

or a triple for() loop for matrix multiplication vs. a %*% b

• Clarity: the syntax is about what we're doing
• Concision: we write less
• Abstraction: the syntax hides how the computer does it
• Generality: same syntax works for numbers, vectors, arrays, … - Speed: modifying big vectors over and over is slow in R; work gets done by optimized low-level code

Many functions are set up to vectorize automatically

abs(-3:3)

[1] 3 2 1 0 1 2 3

log(1:7)

[1] 0.0000 0.6931 1.0986 1.3863 1.6094 1.7918 1.9459

See also apply() from last week

We'll come back to this in great detail later

ifelse(x^2 > 1, 2*abs(x)-1, x^2)

1st argument is a Boolean vector, then pick from the 2nd or 3rd vector arguments as TRUE or FALSE

• Dataframes
• if, nested if, switch
• Iteration: for, while
• Avoiding iteration with whole-object (“vectorized”) operations

0 counts as FALSE; other numeric values count as TRUE; the strings “TRUE” and “FALSE” count as you'd hope; most everything else gives an error

Advice: Don't play games here; try to make sure control expressions are getting Boolean values

Conversely, in arithmetic, FALSE is 0 and TRUE is 1

mean(states$Murder > 7)

[1] 0.48

Simplify nested if with switch(): give a variable to select on, then a value for each option

switch(type.of.summary,
       mean=mean(states$Murder),
       median=median(states$Murder),
       histogram=hist(states$Murder),
       "I don't understand")

Set type.of.summary to, succesively, “mean”, “median”, “histogram”, and “mode”, and explain what happens

repeat {
  print("Help! I am Dr. Morris Culpepper, trapped in an endless loop!")
}

repeat {
  if (watched) { next() }
  print("Help! I am Dr. Morris Culpepper, trapped in an endless loop!")
  if (rescued) { break() }
}

break() exits the loop; next() skips the rest of the body and goes back into the loop

both work with for() and while() as well

Exercise: how would you replace while() with repeat()?