• Arrays
• Matrices
• Lists
• Dataframes
• Structures of structures

Many data structures in R are made by adding bells and whistles to vectors, so "vector structures"

Most useful: arrays

x <- c(7, 8, 10, 45)
x.arr <- array(x,dim=c(2,2))
x.arr

##      [,1] [,2]
## [1,]    7   10
## [2,]    8   45

dim says how many rows and columns; filled by columns

Can have \(3, 4, \ldots n\) dimensional arrays; dim is a length-\(n\) vector

Some properties of the array:

dim(x.arr)

## [1] 2 2

is.vector(x.arr)

## [1] FALSE

is.array(x.arr)

## [1] TRUE

typeof(x.arr)

## [1] "double"

str(x.arr)

##  num [1:2, 1:2] 7 8 10 45

attributes(x.arr)

## $dim
## [1] 2 2

typeof() returns the type of the elements

str() gives the structure: here, a numeric array, with two dimensions, both indexed 1–2, and then the actual numbers

Exercise: try all these with x

Can access a 2-D array either by pairs of indices or by the underlying vector:

x.arr[1,2]

## [1] 10

x.arr[3]

## [1] 10

Omitting an index means "all of it":

x.arr[c(1:2),2]

## [1] 10 45

x.arr[,2]

## [1] 10 45

Using a vector-style function on a vector structure will go down to the underlying vector, unless the function is set up to handle arrays specially:

which(x.arr > 9)

## [1] 3 4

Many functions do preserve array structure:

y <- -x
y.arr <- array(y,dim=c(2,2))
y.arr + x.arr

##      [,1] [,2]
## [1,]    0    0
## [2,]    0    0

Others specifically act on each row or column of the array separately:

rowSums(x.arr)

## [1] 17 53

We will see a lot more of this idea

Census data for California and Pennsylvania on housing prices, by Census "tract"

calif_penn <- read.csv("http://www.stat.cmu.edu/~cshalizi/uADA/13/hw/01/calif_penn_2011.csv")
penn <- calif_penn[calif_penn[,"STATEFP"]==42,]
coefficients(lm(Median_house_value ~ Median_household_income, data=penn))

##             (Intercept) Median_household_income 
##           -26206.564325                3.651256

Fit a simple linear model, predicting median house price from median household income

Census tracts 24–425 are Allegheny county

Tract 24 has a median income of $14,719; actual median house value is $34,100 — is that above or below what's?

34100 < -26206.564 + 3.651*14719

## [1] FALSE

Tract 25 has income $48,102 and house price $155,900

155900 < -26206.564 + 3.651*48102

## [1] FALSE

What about tract 26?

We could just keep plugging in numbers like this, but that's

• boring and repetitive
• error-prone (what if I forget to change the median income, or drop a minus sign from the intercept?)
• obscure if we come back to our work later (what are these numbers?)

penn.coefs <- coefficients(lm(Median_house_value ~ Median_household_income, data=penn))
penn.coefs

##             (Intercept) Median_household_income 
##           -26206.564325                3.651256

allegheny.rows <- 24:425
allegheny.medinc <- penn[allegheny.rows,"Median_household_income"]
allegheny.values <- penn[allegheny.rows,"Median_house_value"]
allegheny.fitted <- penn.coefs["(Intercept)"]+penn.coefs["Median_household_income"]*allegheny.medinc

plot(x=allegheny.fitted, y=allegheny.values,
     xlab="Model-predicted median house values",
     ylab="Actual median house values",
     xlim=c(0,5e5),ylim=c(0,5e5))
abline(a=0,b=1,col="grey")

Factory makes cars and trucks, using labor and steel

• a car takes 40 hours of labor and 1 ton of steel
• a truck takes 60 hours and 3 tons of steel
• resources: 1600 hours of labor and 70 tons of steel each week

In R, a matrix is a specialization of a 2D array

factory <- matrix(c(40,1,60,3),nrow=2)
is.array(factory)

## [1] TRUE

is.matrix(factory)

## [1] TRUE

could also specify ncol, and/or byrow=TRUE to fill by rows.

Element-wise operations with the usual arithmetic and comparison operators (e.g., factory/3)

Compare whole matrices with identical() or all.equal()

Gets a special operator

six.sevens <- matrix(rep(7,6),ncol=3)
six.sevens

##      [,1] [,2] [,3]
## [1,]    7    7    7
## [2,]    7    7    7

factory %*% six.sevens # [2x2] * [2x3]

##      [,1] [,2] [,3]
## [1,]  700  700  700
## [2,]   28   28   28

What happens if you try six.sevens %*% factory?

Numeric vectors can act like proper vectors:

output <- c(10,20)
factory %*% output

##      [,1]
## [1,] 1600
## [2,]   70

output %*% factory

##      [,1] [,2]
## [1,]  420  660

R silently casts the vector as either a row or a column matrix

Transpose:

t(factory)

##      [,1] [,2]
## [1,]   40    1
## [2,]   60    3

Determinant:

det(factory)

## [1] 60

The diag() function can extract the diagonal entries of a matrix:

diag(factory)

## [1] 40  3

It can also change the diagonal:

diag(factory) <- c(35,4)
factory

##      [,1] [,2]
## [1,]   35   60
## [2,]    1    4

Re-set it for later:

diag(factory) <- c(40,3)

diag(c(3,4))

##      [,1] [,2]
## [1,]    3    0
## [2,]    0    4

diag(2)

##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1

solve(factory)

##             [,1]       [,2]
## [1,]  0.05000000 -1.0000000
## [2,] -0.01666667  0.6666667

factory %*% solve(factory)

##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1

Solving the linear system \(\mathbf{A}\vec{x} = \vec{b}\) for \(\vec{x}\):

available <- c(1600,70)
solve(factory,available)

## [1] 10 20

factory %*% solve(factory,available)

##      [,1]
## [1,] 1600
## [2,]   70

We can name either rows or columns or both, with rownames() and colnames()

These are just character vectors, and we use the same function to get and to set their values

Names help us understand what we're working with

Names can be used to coordinate different objects

rownames(factory) <- c("labor","steel")
colnames(factory) <- c("cars","trucks")
factory

##       cars trucks
## labor   40     60
## steel    1      3

available <- c(1600,70)
names(available) <- c("labor","steel")

output <- c(20,10)
names(output) <- c("trucks","cars")
factory %*% output # But we've got cars and trucks mixed up!

##       [,1]
## labor 1400
## steel   50

factory %*% output[colnames(factory)]

##       [,1]
## labor 1600
## steel   70

all(factory %*% output[colnames(factory)] <= available[rownames(factory)])

## [1] TRUE

Notice: Last lines don't have to change if we add motorcycles as output or rubber and glass as inputs (abstraction again)

Take the mean: rowMeans(), colMeans(): input is matrix, output is vector. Also rowSums(), etc.

summary(): vector-style summary of column

colMeans(factory)

##   cars trucks 
##   20.5   31.5

summary(factory)

##       cars           trucks     
##  Min.   : 1.00   Min.   : 3.00  
##  1st Qu.:10.75   1st Qu.:17.25  
##  Median :20.50   Median :31.50  
##  Mean   :20.50   Mean   :31.50  
##  3rd Qu.:30.25   3rd Qu.:45.75  
##  Max.   :40.00   Max.   :60.00

apply(), takes 3 arguments: the array or matrix, then 1 for rows and 2 for columns, then name of the function to apply to each

rowMeans(factory)

## labor steel 
##    50     2

apply(factory,1,mean)

## labor steel 
##    50     2

What would apply(factory,1,sd) do?

Sequence of values, not necessarily all of the same type

my.distribution <- list("exponential",7,FALSE)
my.distribution

## [[1]]
## [1] "exponential"
## 
## [[2]]
## [1] 7
## 
## [[3]]
## [1] FALSE

Most of what you can do with vectors you can also do with lists

Can use [ ] as with vectors
or use [[ ]], but only with a single index
[[ ]] drops names and structures, [ ] does not

is.character(my.distribution)

## [1] FALSE

is.character(my.distribution[[1]])

## [1] TRUE

my.distribution[[2]]^2

## [1] 49

What happens if you try my.distribution[2]^2? What happens if you try [[ ]] on a vector?

Add to lists with c() (also works with vectors):

my.distribution <- c(my.distribution,7)
my.distribution

## [[1]]
## [1] "exponential"
## 
## [[2]]
## [1] 7
## 
## [[3]]
## [1] FALSE
## 
## [[4]]
## [1] 7

Chop off the end of a list by setting the length to something smaller (also works with vectors):

length(my.distribution)

## [1] 4

length(my.distribution) <- 3
my.distribution

## [[1]]
## [1] "exponential"
## 
## [[2]]
## [1] 7
## 
## [[3]]
## [1] FALSE

We can name some or all of the elements of a list

names(my.distribution) <- c("family","mean","is.symmetric")
my.distribution

## $family
## [1] "exponential"
## 
## $mean
## [1] 7
## 
## $is.symmetric
## [1] FALSE

my.distribution[["family"]]

## [1] "exponential"

my.distribution["family"]

## $family
## [1] "exponential"

Lists have a special short-cut way of using names, $ (which removes names and structures):

my.distribution[["family"]]

## [1] "exponential"

my.distribution$family

## [1] "exponential"

Creating a list with names:

another.distribution <- list(family="gaussian",mean=7,sd=1,is.symmetric=TRUE)

Adding named elements:

my.distribution$was.estimated <- FALSE
my.distribution[["last.updated"]] <- "2011-08-30"

Removing a named list element, by assigning it the value NULL:

my.distribution$was.estimated <- NULL

Lists give us a way to store and look up data by name, rather than by position

A really useful programming concept with many names: key-value pairs, dictionaries, associative arrays, hashes

If all our distributions have components named family, we can look that up by name, without caring where it is in the list

Dataframe = the classic data table, \(n\) rows for cases, \(p\) columns for variables

Lots of the really-statistical parts of R presume data frames penn from last time was really a dataframe

Not just a matrix because columns can have different types

Many matrix functions also work for dataframes (rowSums(), summary(), apply())

but no matrix multiplying dataframes, even if all columns are numeric

a.matrix <- matrix(c(35,8,10,4),nrow=2)
colnames(a.matrix) <- c("v1","v2")
a.matrix

##      v1 v2
## [1,] 35 10
## [2,]  8  4

a.matrix[,"v1"]  # Try a.matrix$v1 and see what happens

## [1] 35  8

a.data.frame <- data.frame(a.matrix,logicals=c(TRUE,FALSE))
a.data.frame

##   v1 v2 logicals
## 1 35 10     TRUE
## 2  8  4    FALSE

a.data.frame$v1

## [1] 35  8

a.data.frame[,"v1"]

## [1] 35  8

a.data.frame[1,]

##   v1 v2 logicals
## 1 35 10     TRUE

colMeans(a.data.frame)

##       v1       v2 logicals 
##     21.5      7.0      0.5

We can add rows or columns to an array or data-frame with rbind() and cbind(), but be careful about forced type conversions

rbind(a.data.frame,list(v1=-3,v2=-5,logicals=TRUE))

##   v1 v2 logicals
## 1 35 10     TRUE
## 2  8  4    FALSE
## 3 -3 -5     TRUE

rbind(a.data.frame,c(3,4,6))

##   v1 v2 logicals
## 1 35 10        1
## 2  8  4        0
## 3  3  4        6

So far, every list element has been a single data value

List elements can be other data structures, e.g., vectors and matrices:

plan <- list(factory=factory, available=available, output=output)
plan$output

## trucks   cars 
##     20     10

Internally, a dataframe is basically a list of vectors

List elements can even be other lists
which may contain other data structures
including other lists
which may contain other data structures…

This recursion lets us build arbitrarily complicated data structures from the basic ones

Most complicated objects are (usually) lists of data structures

eigen() finds eigenvalues and eigenvectors of a matrix
Returns a list of a vector (the eigenvalues) and a matrix (the eigenvectors)

eigen(factory)

## eigen() decomposition
## $values
## [1] 41.556171  1.443829
## 
## $vectors
##            [,1]       [,2]
## [1,] 0.99966383 -0.8412758
## [2,] 0.02592747  0.5406062

class(eigen(factory))

## [1] "eigen"

With complicated objects, you can access parts of parts (of parts…)

factory %*% eigen(factory)$vectors[,2]

##             [,1]
## labor -1.2146583
## steel  0.7805429

eigen(factory)$values[2] * eigen(factory)$vectors[,2]

## [1] -1.2146583  0.7805429

eigen(factory)$values[2]

## [1] 1.443829

eigen(factory)[[1]][[2]] # NOT [[1,2]]

## [1] 1.443829

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

colnames(states)

##  [1] "Population" "Income"     "Illiteracy" "Life.Exp"   "Murder"    
##  [6] "HS.Grad"    "Frost"      "Area"       "abb"        "region"    
## [11] "division"

states[1,]

##         Population Income Illiteracy Life.Exp Murder HS.Grad Frost  Area
## Alabama       3615   3624        2.1    69.05   15.1    41.3    20 50708
##         abb region           division
## Alabama  AL  South East South Central

• 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. 
##   37.80   48.05   53.25   53.11   59.15   67.30

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

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.3780  0.4805  0.5325  0.5311  0.5915  0.6730

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

What percentage of literate adults graduated HS?

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

## [1] 42.18590 67.71574 59.16497 40.67278 63.29626 64.35045

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

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

## [1] 42.18590 67.71574 59.16497 40.67278 63.29626 64.35045

Lots of functions take data arguments, and look variables up in that data frame:

plot(Illiteracy~Frost, data=states)

\(R^2 =0.45\), \(p \approx {10}^{-7}\)

• Arrays add multi-dimensional structure to vectors
• Matrices act like you'd hope they would
• Lists let us combine different types of data
• Dataframes are hybrids of matrices and lists, for classic tabular data
• Recursion lets us build complicated data structures out of the simpler ones