---
title: "Lab 3: Text Manipulation"
author: "Statistical Computing, 36-350"
date: "Week of Tuesday January 30, 2018"
---

Name:  
Andrew ID:  
Collaborated with:  

This lab is to be done in class (completed outside of class if need be). You can collaborate with your classmates, but you must identify their names above, and you must submit **your own** lab as an knitted HTML file on Canvas, by Thursday 10pm, this week.

**This week's agenda**: basic string manipulations; more vectorization; practice reading in and summarizing real text data (Shakespeare); just a little bit of regular expressions.

Some string basics
===

- **1a.** Define two strings variables, equal to "Statistical Computing" and 'Statistical Computing', and check whether they are equal. What do you conclude about the use of double versus single quotation marks for creating strings in R? Give an example that shows why might we prefer to use double quotation marks as the standard (think of apostrophes).

- **1b.** The functions `tolower()` and `toupper()` do as you'd expect: they convert strings to all lower case characters, and all upper case characters, respectively. Apply them to the strings below, as directed by the comments, to observe their behavior.

```{r}
"I'M NOT ANGRY I SWEAR"         # Convert to lower case
"Mom, I don't want my veggies"  # Convert to upper case
"Hulk, sMasH"                   # Convert to upper case
"R2-D2 is in prime condition, a real bargain!" # Convert to lower case
```

- **1c.** Consider the string vector `presidents` of length 5 below, containing the last names of past US presidents. Define a string vector `first.letters` to contain the first letters of each of these 5 last names. Hint: use `substr()`, and take advantage of vectorization; this should only require one line of code. Define `first.letters.scrambled` to be the output of `sample(first.letters)` (the `sample()` function can be used to perform random permutations, we'll learn more about it later in the course). Lastly, reset the first letter of each last name stored in `presidents` according to the scrambled letters in `first.letters.scrambled`. Hint: use `substr()` again, and take advantage of vectorization; this should only take one line of code. Display these new last names.

```{r}
presidents = c("Clinton", "Bush", "Reagan", "Carter", "Ford")
```

- **1d.** Now consider the string `phrase` defined below. Using `substr()`, replace the first four characters in `phrase` by "Provide". Print `phrase` to the console, and describe the behavior you are observing. Using `substr()` again, replace the last five characters in `phrase` by "kit" (don't use the length of `phrase` as magic constant in the call to `substr()`, instead, compute the length using `nchar()`). Print `phrase` to the console, and describe the behavior you are observing.

```{r}
phrase = "Give me a break"
```

- **1e.** Consider the string `ingredients` defined below. Using `strsplit()`, split this string up into a string vector of length 5, with elements "chickpeas", "tahini", "olive oil", "garlic", and "salt." Using `paste()`, combine this string vector into a single string "chickpeas + tahini + olive oil + garlic + salt". Then produce a final string of the same format, but where the ingredients are sorted in alphabetical (increasing) order.

```{r}
ingredients = "chickpeas, tahini, olive oil, garlic, salt"
```

Shakespeare's complete works
===

[Project Gutenberg](http://www.gutenberg.org) offers over 50,000 free online books, especially old books (classic literature), for which copyright has expired. We're going to look at the complete works of [William Shakespeare](https://en.wikipedia.org/wiki/William_Shakespeare), taken from the Project Gutenberg website. 

To avoid hitting the Project Gutenberg server over and over again, we've grabbed a text file from them that contains the complete works of William Shakespeare and put it on our course website. Visit http://www.stat.cmu.edu/~ryantibs/statcomp-S18/data/shakespeare.txt in your web browser and just skim through this text file a little bit to get a sense of what it contains (a whole lot!). 

Reading in text, basic exploratory tasks
===

- **2a.** Read in the Shakespeare data linked above into your R session with `readLines()`. Make sure you are reading the data file directly from the web (rather than locally, from a downloaded file on your computer). Call the result `shakespeare.lines`. This should be a vector of strings, each element representing a "line" of text. Print the first 5 lines. How many lines are there? How many characters in the longest line? What is the average number of characters per line? How many lines are there with zero characters (empty lines)? Hint: each of these queries should only require one line of code; for the last one, use an on-the-fly Boolean comparison and `sum()`.

- **2b.** Remove the lines in `shakespeare.lines` that have zero characters. Hint: use Boolean indexing. Check that the new length of `shakespeare.lines` makes sense to you.

- **2c.** Collapse the lines in `shakespeare.lines` into one big string, separating each line by a space in doing so, using `paste()`. Call the resulting string `shakespeare.all`. How many characters does this string have? How does this compare to the sum of characters in `shakespeare.lines`, and does this make sense to you?

- **2d.** Split up `shakespeare.all` into words, using `strsplit()` with `split=" "`. Call the resulting string vector (note: here we are asking you for a vector, not a list) `shakespeare.words`. How long is this vector, i.e., how many words are there? Using the `unique()` function, compute and store the unique words as `shakespeare.words.unique`. How many unique words are there?  

- **2e.** Plot a histogram of the number of characters of the words in `shakespeare.words.unique`. You will have to set a large value of the `breaks` argument (say, `breaks=50`) in order to see in more detail what is going on. What does the bulk of this distribution look like to you? Why is the x-axis on the histogram extended so far to the right (what does this tell you about the right tail of the distribution)?

- **2f.** Reminder: the `sort()` function sorts a given vector into increasing order; its close friend, the `order()` function, returns the indices that put the vector into increasing order. Both functions can take `decreasing=TRUE` as an argument, to sort/find indices according to decreasing order. See the code below for an example.
    ```{r}
    set.seed(0)
    (x = round(runif(5, -1, 1), 2))
    sort(x, decreasing=TRUE)
    order(x, decreasing=TRUE)
    ```
    Using the `order()` function, find the indices that correspond to the top 5 longest words in `shakespeare.words.unique`. Then, print the top 5 longest words themselves. Do you recognize any of these as actual words? **Challenge**: try to pronounce the fourth longest word! What does it mean?
    
Computing word counts
===

- **3a.** Using `table()`, compute counts for the words in `shakespeare.words`, and save the result as `shakespeare.wordtab`. How long is `shakespeare.wordtab`, and is this equal to the number of unique words (as computed above)? Using named indexing, answer: how many times does the word "thou" appear? The word "rumour"? The word "gloomy"? The word "assassination"?

- **3b.** How many words did Shakespeare use just once? Twice? At least 10 times? More than 100 times? 

- **3c.** Sort `shakespeare.wordtab` so that its entries (counts) are in decreasing order, and save the result as `shakespeare.wordtab.sorted`. Print the 25 most commonly used words, along with their counts. What is the most common word? Second and third most common words?

- **3d.** What you should have seen in the last question is that the most common word is the empty string "". This is just an artifact of splitting `shakespeare.all` by spaces, using `strsplit()`. Redefine `shakespeare.words` so that all empty strings are deleted from this vector. Then recompute `shakespeare.wordtab` and `shakespeare.wordtab.sorted`. Check that you have done this right by printing out the new 25 most commonly used words, and verifying (just visually) that is overlaps with your solution to the last question.

- **3e.** As done at the end of the lecture notes, produce a plot of the word counts (y-axis) versus the ranks (x-axis) in `shakespeare.wordtab.sorted`. Set `xlim=c(1,1000)` as an argument to `plot()`; this restricts the plotting window to just the first 1000 ranks, which is helpful here to see the trend more clearly. Do you see **Zipf's law** in action, i.e., does it appear that $\mathrm{Frequency} \approx C(1/\mathrm{Rank})^a$ (for some $C,a$)? **Challenge**: either programmatically, or manually, determine reasonably-well-fitting values of $C,a$ for the Shakespeare data set; then draw the curve $y=C(1/x)^a$ on top of your plot as a red line to show how well it fits.

A tiny bit of regular expressions
===

- **4a.** There are a couple of issues with the way we've built our words in `shakespeare.words`. The first is that capitalization matters; from Q3c, you should have seen that "and" and "And" are counted as separate words. The second is that many words contain punctuation marks (and so, aren't really words in the first place); to see this, retrieve the count corresponding to "and," in your word table `shakespeare.wordtab`.

    The fix for the first issue is to convert `shakespeare.all` to all lower case characters. Hint: recall `tolower()` from Q1b. The fix for the second issue is to use the argument `split="[[:space:]]|[[:punct:]]"` in the call to `strsplit()`, when defining the words. In words, this means: *split on spaces or on punctuation marks*; more precisely, it uses what we call a **regular expression** for the `split` argument. Carry out both of these fixes to define new words `shakespeare.words.new`. Then, delete all empty strings from this vector, and compute word table from it, called `shakespeare.wordtab.new`. 

- **4b.** Compare the length of `shakespeare.words.new` to that of `shakespeare.words`; also compare the length of `shakespeare.wordtab.new` to that of `shakespeare.wordtab`. Explain what you are observing.

- **4c.** Compute the unique words in `shakespeare.words.new`, calling the result `shakespeare.words.new.unique`. Then repeat the queries in Q2e and Q2f on `shakespeare.words.new.unique`. Comment on the histogram---is it different in any way than before? How about the top 5 longest words? 

- **4d.** Sort `shakespeare.wordtab.new` so that its entries (counts) are in decreasing order, and save the result as `shakespeare.wordtab.sorted.new`. Print out the 25 most common words and their counts, and compare them (informally) to what you saw in Q3d. Also, produce a plot of the new word counts, as you did in Q3e. Does Zipf's law look like it still holds?

- **Challenge.** When we use `split="[[:space:]]|[[:punct:]]"` in the call to `strsplit()`, a word like "non-Gaussian" gets split into separate words "non" and "Gaussian". E.g.,
    ```{r}
    strsplit("'Hey!' My distribution: it's (super) non-Gaussian!",
             split="[[:space:]]|[[:punct:]]")[[1]]
    ```
    which is not really desirable, since we don't want to consider "it", "s", "non", and "Gaussian" all as separate words. Of course, Shakespeare wouldn't have been talking about Gaussians, but the same idea applies. In fact, you should have seen in your solution to Q4d that "s" was one of the 25 most common words used by Shakespeare in his complete works (no doubt, due to the splitting of words like "it's").
    
    Modify the regular expression `"[[:space:]]|[[:punct:]]"`, to be used for the `split` argument, so that this doesn't happen. Specifically, design a regular expression that splits on spaces, or on punctuation marks that are at the left or right boundary of a word, meaning:
      * they are at the beginning or end of the sentence; or
      * follow a space or are followed by a space; or
      * follow another punctuation mark or are followed by another punctuation mark.
      
      Using your new regular expression (as the `split` argument, in the call to `strsplit()`), repeat Q4a through Q4d, to compute words and a word table for the Shakespeare data. Comment on what has changed in your latest answers. 
      
      Hint #1: you will have to read up a bit on regular expressions; there are many possible solutions here using regular expressions. Hint #2: you might find it useful to try out your ideas on the example in the last code chunk; in this example, you want to keep "it's" and "non-Gaussian" intact, but remove all other punctuation marks. 