In agoldst/litdata: R resources for Literary Data, Spring 2015

## ignore this fiddly bit and just leave it as is
knitr::opts_chunk$set(comment=NA, error=T)
library("stringr")

Groups

Please continue discussing with one another your mutual interests and ideas about groups for the end-of-semester work. One good way to think about this is to collaborate on some of these early homeworks.

Loop practice

Even more vector subscripting

Before you can subscript a vector, it has to exist. There are a couple of ways to create a "blank" or empty vector in R. On the last homework, you saw that c() yields a zero-length something. But this expression is not clear about what type of data you intend to store in the vector. R doesn't care, but your human code-readers do. Jockers has his naming convention (calling things .v and .t and so on). Another way is to use an expression like x <- numeric(). The strange function numeric() can be replaced with character(), logical(), and others too. What is the length of x in this case? Now try x[2] <- 10. Finally, try x["word"] <- 1. Generalize: what happens when you assign to an element of a vector that does not exist?

(Incidentally, try supplying a number as an argument to these constructor functions: what is character(3)?)

"For" is for counting

Set

words <- c("I", "am", "that", "I", "am")
count <- 0

Now write a for loop that tallies the number of times am occurs in words and stores the result in count. At the end of the loop, count should be 2:

## insert loop here

count == 2 # should be TRUE

Hint: each time through the loop, the value of count might increase by 1.

Now rewrite the loop so that this tally is stored not as the sole value of count but as an element of a vector, accessible by name. Instead of using a literal "am", let's use a variable, which we'll just set to be "am" at the start:

count <- numeric()
word <- "am"
count[word] <- 0
## your slightly modified loop here

count[word] == 2 # should be TRUE

Make sure you understand what is going on in the subscripting expression.

Boundary condition

What value do you get if you ask for count["and"]? This is a problem for tabulating, since we want our starting value for word counts to be zero. The handy function is.na(x) is TRUE is x is NA (actually, this function is vectorized---try it!)

Write an if block that tests whether count[word] is NA, and, if so, sets count[word] to 1; if it is not NA, your block should increase count[word] by 1. Demonstrate its correctness by copying the same block twice as directed below:

word <- "am"
## if block here

count[word] == 3 # should be TRUE

word <- "and"
## identical if block here

count[word] == 1 # should be TRUE

Tabulate

Now modify your loop so that count[w] gives the number of occurrences of word w in words for any w that occurs at least once. Hint: you don't have to include use any literal character strings. Just variables!

count <- numeric() # clear the slate
## your modified loop here

all(count[c("I", "am", "that")] == c(2, 2, 1)) # should be TRUE

If you're stuck, think about how you'd explain tabulation step by step to someone who only knew about adding 1 but who, weirdly, always liked to start counting like this: NA, 1, 2, 3...

Jockers

Work through the short third chapter of Jockers.

Practice 3.1

Go ahead and do it. This may be getting a bit repetitive, but the practice will help increase your comfort in R. See if you can make your word-counting code more concise than Jockers's. Use str_split, str_c instead of strsplit and paste.

As an enhancement, use an if...else statement in your code to print out a warning if the Austen text file cannot be found. Then try knitting with the file in place and without (submit the knitted version with the file).

Practice 3.2

Should be a quickie. I won't make you implement unique with a for loop, but you could!

Practice 3.3

## expression for the twelve shared words among the two top-10 lists

Practice 3.4

## expression for the top words in S&S not in the top-10 for MB

Extra

%in% is very important. Implementing this with a for loop (containing an if statement) is excellent practice but strictly optional. However, to firm up your intuition for it, do the following:

1. Fill in values below as directed in the comments:

x <- "Fortinbras"
names1 <- c() # make this a three element character vector
names2 <- c() # and this too, such that...
x %in% names1 # is TRUE
x %in% names2 # is FALSE

1. Now examine the values of match(x, names1) and match(x, names2). Use your findings to write a general expression for x %in% y using match. Demonstrate with a couple of examples.

## not evaluated: fill in your match expression
x %in% y == 

## add your demo here

1. But we're not done! %in% is vectorized in its left hand argument. For example,

c("Fortinbras", "Guildenstern") %in% c("Hamlet", "Ophelia", "Fortinbras")

That is, each element of the result is a logical value indicating whether the corresponding element of the left-hand side of %in% is an element of the right-hand side.

Give an expression for x %in% y in terms of match that is generally valid for x and y vectors of any length. Demonstrate with a few examples.

## not evaluated: fill in your match expression
x %in% y == 

## add your demo here

Comment on the relationship between this expression and the one you gave in the previous part.

agoldst/litdata documentation built on May 10, 2019, 7:34 a.m.