Name:
Andrew ID:
Collaborated with:
This lab is to be done in class (completed outside of class time 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 PDF file on Gradescope, by Friday 9pm, this week.
This week’s agenda: basic string manipulations; practice reading in and summarizing real text data (Shakespeare); practice with iteration; just a little bit of regular expressions.
# YOUR CODE GOES HEREtolower() 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."I'M NOT ANGRY I SWEAR" # Convert to lower case## [1] "I'M NOT ANGRY I SWEAR""Mom, I don't want my veggies" # Convert to upper case## [1] "Mom, I don't want my veggies""Hulk, sMasH" # Convert to upper case## [1] "Hulk, sMasH""R2-D2 is in prime condition, a real bargain!" # Convert to lower case## [1] "R2-D2 is in prime condition, a real bargain!"# YOUR CODE GOES HEREpresidents 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.presidents = c("Clinton", "Bush", "Reagan", "Carter", "Ford")# YOUR CODE GOES HEREphrase 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.phrase = "Give me a break"# YOUR CODE GOES HEREingredients 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.ingredients = "chickpeas, tahini, olive oil, garlic, salt"# YOUR CODE GOES HEREProject Gutenberg 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, 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/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!).
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().# YOUR CODE GOES HEREshakespeare.lines (i.e., lines with zero characters). Check that that the new length of shakespeare.lines makes sense to you.# YOUR CODE GOES HEREshakespeare.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?# YOUR CODE GOES HEREshakespeare.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?# YOUR CODE GOES HEREshakespeare.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)?# YOUR CODE GOES HERE2f. 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.
set.seed(0)
(x = round(runif(5, -1, 1), 2))## [1] 0.79 -0.47 -0.26 0.15 0.82sort(x, decreasing=TRUE)## [1] 0.82 0.79 0.15 -0.26 -0.47order(x, decreasing=TRUE)## [1] 5 1 4 3 2Using 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?
# YOUR CODE GOES HEREtable(), 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”?# YOUR CODE GOES HERE# YOUR CODE GOES HEREshakespeare.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?# YOUR CODE GOES HEREshakespeare.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.# YOUR CODE GOES HEREshakespeare.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.# YOUR CODE GOES HERE4a. 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.
# YOUR CODE GOES HEREshakespeare.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.# YOUR CODE GOES HEREshakespeare.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?# YOUR CODE GOES HEREshakespeare.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?# YOUR CODE GOES HEREshakespeare.lines. Take a look at lines 19 through 23 of this vector: you should see a bunch of spaces preceding the text in lines 21, 22, and 23. Redefine shakespeare.lines by setting it equal to the output of calling the function trimws() on shakespeare.lines. Print out lines 19 through 23 again, and describe what’s happened.# YOUR CODE GOES HEREwhich(), find the indices of the lines in shakespeare.lines that equal “THE SONNETS”, report the index of the first such occurence, and store it as toc.start. Similarly, find the indices of the lines in shakespeare.lines that equal “VENUS AND ADONIS”, report the index of the first such occurence, and store it as toc.end.# YOUR CODE GOES HEREn = toc.end - toc.start + 1, and create an empty string vector of length n called titles. Using a for() loop, populate titles with the titles of Shakespeare’s plays as ordered in the table of contents list, with the first being “THE SONNETS”, and the last being “VENUS AND ADONIS”. Print out the resulting titles vector to the console. Hint: if you define the counter variable i in your for() loop to run between 1 and n, then you will have to index shakespeare.lines carefully to extract the correct titles. Think about the following. When i=1, you want to extract the title of the first play in shakespeare.lines, which is located at index toc.start. When i=2, you want to extract the title of the second play, which is located at index toc.start + 1. And so on.# YOUR CODE GOES HEREfor() loop to find out, for each play, the index of the line in shakespeare.lines at which this play begins. It turns out that the second occurence of “THE SONNETS” in shakespeare.lines is where this play actually begins (this first ocurrence is in the table of contents), and so on, for each play title. Use your for() loop to fill out an integer vector called titles.start, containing the indices at which each of Shakespeare’s plays begins in shakespeare.lines. Print the resulting vector titles.start to the console.# YOUR CODE GOES HEREtitles.end to be an integer vector of the same length as titles.start, whose first element is the second element in titles.start minus 1, whose second element is the third element in titles.start minus 1, and so on. What this means: we are considering the line before the second play begins to be the last line of the first play, and so on. Define the last element in titles.end to be the length of shakespeare.lines. You can solve this question either with a for() loop, or with proper indexing and vectorization. Challenge: it’s not really correct to set the last element in titles.end to be length of shakespeare.lines, because there is a footer at the end of the Shakespeare data file. By looking at the data file visually in your web browser, come up with a way to programmatically determine the index of the last line of the last play, and implement it.# YOUR CODE GOES HERE5f. In Q5d, you should have seen that the starting index of Shakespeare’s 38th play “THE TWO NOBLE KINSMEN” was computed to be NA, in the vector titles.start. Why? If you run which(shakespeare.lines == "THE TWO NOBLE KINSMEN") in your console, you will see that there is only one occurence of “THE TWO NOBLE KINSMEN” in shakespeare.lines, and this occurs in the table of contents. So there was no second occurence, hence the resulting NA value.
But now take a look at line 118,463 in shakespeare.lines: you will see that it is “THE TWO NOBLE KINSMEN:”, so this is really where the second play starts, but because of colon “:” at the end of the string, this doesn’t exactly match the title “THE TWO NOBLE KINSMEN”, as we were looking for. The advantage of using the grep() function, versus checking for exact equality of strings, is that grep() allows us to match substrings. Specifically, grep() returns the indices of the strings in a vector for which a substring match occurs, e.g.,
grep(pattern="cat",
x=c("cat", "canned goods", "batman", "catastrophe", "tomcat"))## [1] 1 4 5so we can see that in this example, grep() was able to find substring matches to “cat” in the first, fourth, and fifth strings in the argument x. Redefine titles.start by repeating the logic in your solution to Q5d, but replacing the which() command in the body of your for() loop with an appropriate call to grep(). Also, redefine titles.end by repeating the logic in your solution to Q5e. Print out the new vectors titles.start and titles.end to the console—they should be free of NA values.
# YOUR CODE GOES HEREtitles vector. Use this to find the indices at which this play starts and ends, in the titles.start and titles.end vectors, respectively. Call the lines of text corresponding to this play shakespeare.lines.hamlet. How many such lines are there? Do the same, but now for the play “THE TRAGEDY OF ROMEO AND JULIET”, and call the lines of text corresponding to this play shakespeare.lines.romeo. How many such lines are there?# YOUR CODE GOES HEREshakespeare.lines.hamlet. (This should mostly just involve copying and pasting code as needed.) That is, to be clear: * collapse shakespeare.lines.hamlet into one big string, separated by spaces; * convert this string into all lower case characters; * divide this string into words, by splitting on spaces or on punctuation marks, using split="[[:space:]]|[[:punct:]]" in the call to strsplit(); * remove all empty words (equal to the empty string ""), and report how many words remain; * compute the unique words, report the number of unique words, and plot a histogram of their numbers of characters; * report the 5 longest words; * compute a word table, and report the 25 most common words and their counts; * finally, produce a plot of the word counts verus rank.# YOUR CODE GOES HEREshakespeare.lines.romeo. (Again, this should just involve copying and pasting code as needed. P.S. Isn’t this getting tiresome? You’ll be happy when we learn functions, next week!) Comment on any similarities/differences you see in the answers.# YOUR CODE GOES HEREfor() loop and the titles.start, titles.end vectors constructed above, answer the following questions. What is Shakespeare’s longest play (in terms of the number of words)? What is Shakespeare’s shortest play? In which play did Shakespeare use his longest word (in terms of the number of characters)? Are there any plays in which “the” is not the most common word?# YOUR CODE GOES HERE