Text representations and Data Frames

Reading: Matloff Chapter 11.1 (for strings), 6.1 (for factors), 5.1 (for data frames)

The character type

• Strings have type character in R.

• No special type for single characters vs. strings:

typeof('a')

## [1] "character"

typeof("ABC")

## [1] "character"

• As with the numeric type, single strings are vectors of length 1:

length('a')

## [1] 1

length("ABC")

## [1] 1

• If you actually want the number of characters in a string, you need the nchar function:

nchar('a')

## [1] 1

nchar("ABC")

## [1] 3

Creating strings

You can use either double or single quotes.

"The Leopard"

## [1] "The Leopard"

'Burt Lancaster'

## [1] "Burt Lancaster"

Displaying strings

• print works.

• Sometimes slightly nicer to use cat, which prints the string to the console instead of giving the representation with subscripts that print does.

print("The Leopard")

## [1] "The Leopard"

"The Leopard"

## [1] "The Leopard"

cat("The Leopard")

## The Leopard

Escape characters

The \ is an escape character.

It tells R to interpret whatever comes after literally instead of as a special character.

For example, if you need quotes inside your string, you need to escape the quotation character:

cat("Giuseppe Tomasi di Lampedusa's \"The Leopard\"")

## Giuseppe Tomasi di Lampedusa's "The Leopard"

cat('Giuseppe Tomasi di Lampedusa\'s "The Leopard"')

## Giuseppe Tomasi di Lampedusa's "The Leopard"

cat("We use the '\\' character to escape")

## We use the '\' character to escape

String Manipulation

We can't use [] or [[]] to get at the internal parts of a string because strings are atomic in R.

We need to use substr.

Syntax is substr(string, start, stop):

lancaster = "Burt Lancaster"
substr(lancaster, 1, 4)

## [1] "Burt"

We can use substr for replacement as well:

lancaster

## [1] "Burt Lancaster"

substr(lancaster, 1, 4) = "Bill"
lancaster

## [1] "Bill Lancaster"

What happens if the replacement string isn't the same length as the piece of the string you want to replace?

Combining strings

paste is the workhorse function for string combination.

Simplest way to use it:

paste(s1, s2, ..., sn, sep)

This will paste together s1, s2, up to sn, with sep in between each one.

For example:

paste("Chico", "Harpo", "Groucho", sep = ", ")

## [1] "Chico, Harpo, Groucho"

The arguments to paste can be vectors, and the function is vectorized so we get recycling:

paste(c("Chico", "Harpo", "Groucho"), "Marx", sep = " ")

## [1] "Chico Marx"   "Harpo Marx"   "Groucho Marx"

paste("Marx", c("Chico", "Harpo", "Groucho"), sep = ", ")

## [1] "Marx, Chico"   "Marx, Harpo"   "Marx, Groucho"

Final argument: collapse.

Syntax paste(vector, collapse)

Will create one string (vector of length 1, not the length of the input vector) by pasting the elements of vector together, with the argument from collapse in between them.

paste(c("Chico", "Harpo", "Grouco"), collapse = ", ")

## [1] "Chico, Harpo, Grouco"

Finaly, we can specify both sep and collapse together.

Think of this as first calling paste with collapse = NULL, then calling paste with non-null collapse on the result:

paste(c("Chico", "Harpo", "Groucho"), "Marx", sep = " ", collapse = " and ")

## [1] "Chico Marx and Harpo Marx and Groucho Marx"

## note the equivalence:
marx.brothers = paste(c("Chico", "Harpo", "Groucho"), "Marx", sep = " ", collapse = NULL)
paste(marx.brothers, collapse = " and ")

## [1] "Chico Marx and Harpo Marx and Groucho Marx"

Splitting strings

Primariy function is strsplit.

Syntax: strsplit(s, split)

• s is a character vector (can have length greater than 1), and the function vectorizes.

• split gives the string we want to split on: every element of s will be split into pieces separated by split.

Given these parameters, what do you expect the output from strsplit to look like? Which of the data structures we've seen so far can accommodate everything we need?

A slightly more realistic example:

I have some fasta files that I'm going to use as input, perform some manipulations on, and then write some output based on each one. I want the output files to have the same prefix but have the extension .txt instead of .fasta:

file.names = c("ighv_human.fasta", "ighd_mouse.fasta", "ighj_human.fasta", "ighv_mouse.fasta", "ighd_human.fasta", "ighj_mouse.fasta")
split.files = strsplit(file.names, ".", fixed = TRUE) ## fixed = TRUE has to do with regular expressions, which we'll talk about on Thursday
output.names = character(length = length(file.names))
for(i in 1:length(split.files)) {
    prefix = split.files[[i]][[1]]
    output.names[i] = paste(prefix, ".txt", sep = "")
}
output.names

## [1] "ighv_human.txt" "ighd_mouse.txt" "ighj_human.txt" "ighv_mouse.txt"
## [5] "ighd_human.txt" "ighj_mouse.txt"

Factors

• A way of representing qualitative variables.

• Variable that takes string values, but we know in advance which strings are valid.

factor(c("a", "b", "b", "z"))

## [1] a b b z
## Levels: a b z

Factors vs. strings

• Factors represented differently from strings internally.

• Factors are an integer vector plus a set of names corresponding to each of the levels.

This representation is a bit more parsimonious in general (but not always...):

object.size(state.name)

## 3496 bytes

object.size(state.name.fac)

## 4080 bytes

object.size(rep(state.name, 1000))

## 403096 bytes

object.size(rep(state.name.fac, 1000))

## 203880 bytes

Factor manipulation

Problems we might need to deal with:

• Too many levels

• Too few levels

• Levels with the wrong names

How do we actually do this?

droplevels(state.name.fac) ## droplevels function does what it says it does

##  [1] Alabama        Alaska         Arizona        Arkansas      
##  [5] The PR of PR   Colorado       Connecticut    Delaware      
##  [9] Florida        Georgia        Hawaii         Idaho         
## [13] Illinois       Indiana        Iowa           Kansas        
## [17] Kentucky       Louisiana      Maine          Maryland      
## [21] Massachusetts  Michigan       Minnesota      Mississippi   
## [25] Missouri       Montana        Nebraska       Nevada        
## [29] New Hampshire  New Jersey     New Mexico     New York      
## [33] North Carolina North Dakota   Ohio           Oklahoma      
## [37] Oregon         Pennsylvania   Rhode Island   South Carolina
## [41] South Dakota   Tennessee      Texas          Utah          
## [45] Vermont        Virginia       Washington     West Virginia 
## [49] Wisconsin      Wyoming       
## 50 Levels: Alabama Alaska Arizona Arkansas Colorado ... The PR of PR

factor(state.name.fac, levels = levels(state.name.fac)[-cal.index])

##  [1] Alabama        Alaska         Arizona        Arkansas      
##  [5] The PR of PR   Colorado       Connecticut    Delaware      
##  [9] Florida        Georgia        Hawaii         Idaho         
## [13] Illinois       Indiana        Iowa           Kansas        
## [17] Kentucky       Louisiana      Maine          Maryland      
## [21] Massachusetts  Michigan       Minnesota      Mississippi   
## [25] Missouri       Montana        Nebraska       Nevada        
## [29] New Hampshire  New Jersey     New Mexico     New York      
## [33] North Carolina North Dakota   Ohio           Oklahoma      
## [37] Oregon         Pennsylvania   Rhode Island   South Carolina
## [41] South Dakota   Tennessee      Texas          Utah          
## [45] Vermont        Virginia       Washington     West Virginia 
## [49] Wisconsin      Wyoming       
## 50 Levels: Alabama Alaska Arizona Arkansas Colorado ... The PR of PR

factor(state.name.fac)

##  [1] Alabama        Alaska         Arizona        Arkansas      
##  [5] The PR of PR   Colorado       Connecticut    Delaware      
##  [9] Florida        Georgia        Hawaii         Idaho         
## [13] Illinois       Indiana        Iowa           Kansas        
## [17] Kentucky       Louisiana      Maine          Maryland      
## [21] Massachusetts  Michigan       Minnesota      Mississippi   
## [25] Missouri       Montana        Nebraska       Nevada        
## [29] New Hampshire  New Jersey     New Mexico     New York      
## [33] North Carolina North Dakota   Ohio           Oklahoma      
## [37] Oregon         Pennsylvania   Rhode Island   South Carolina
## [41] South Dakota   Tennessee      Texas          Utah          
## [45] Vermont        Virginia       Washington     West Virginia 
## [49] Wisconsin      Wyoming       
## 50 Levels: Alabama Alaska Arizona Arkansas Colorado ... The PR of PR

Ordered factors

No detail here, but for reference: ordered factors exist, are created with the ordered = TRUE flag, some functions will have nicer default behavior if you tell them that ordinal variables are ordinal.

Data frames

Why do we need data frames at all?

• A common setup is that we have some samples, and some observations on those samples. The observations are of different types: some are numbers, and some are qualitative variables.

• We can't use a matrix because matrices require all of the elements to be of the same type.
  
  Note that this isn't just an arbitrary rule about matrices: if we think about it from the computer's point of view, the amount of space required for a matrix containing all the same type of observations is fixed, but if the type of the observation can change, we don't know how much space to allocate.

• We need a data structure that mostly acts like a matrix (rectangular, we can extract subsets easily, and so on), but that allows for different data types in different columns.

Data frames in R are lists with some extra rules:

• Each element of the list needs to be a vector

• Each of the vectors needs to be of the same length

This gives essentially a heterogeneous analog of a matrix.

Data frame creation

kids = c("Jack", "Jill")
ages = c(12, 10)
df = data.frame(kids, ages)
df = data.frame(Kids = kids, Ages = ages)
typeof(df)

## [1] "list"

class(df)

## [1] "data.frame"

Referring to elements of a data frame

Remember that data frames are lists, wich each element of the list corresponding to one column.

This means we can use the [[]] notation to get a column:

df[[1]]

## [1] Jack Jill
## Levels: Jack Jill

df[,1]

## [1] Jack Jill
## Levels: Jack Jill

df[,1,drop = FALSE]

##   Kids
## 1 Jack
## 2 Jill

df[1,"Ages"]

## [1] 12

df$Age

## [1] 12 10

That's all, folks

• Please see me if you have any questions about how to submit the assignment or R scripts.

• Next time: Regular expressions!