Name:
Andrew ID:
Collaborated with:
On this homework, you can collaborate with your classmates, but you must identify their names above, and you must submit your own homework as an knitted HTML file on Canvas, by Sunday 10pm, this week.
## For reproducibility --- don't change this!
set.seed(01302018)(x = runif(10, -1, 1))## [1] -0.5703595 -0.4517724 0.4088408 -0.1856029 -0.2185913 -0.2291011
## [7] -0.3781738 0.7277174 -0.6395923 -0.9342617# Increment the negative entries by 0.1
(y = sample(c(TRUE,FALSE), 8, replace=TRUE)) ## [1] TRUE FALSE TRUE FALSE FALSE TRUE TRUE FALSE# Replace the FALSE entries by NA
(z = c("Hey", "you", "there", "what's", "going", "on")) ## [1] "Hey" "you" "there" "what's" "going" "on"# Paste an exclamation mark "!" at the end of the entries with at most 3 charactersprod(), and the binary operator ^. The second uses the functions exp(), mean(), and log(). Using the vector x defined below, implement both strategies, calling the results m1 and m2, respectively. Each computation should require one line of code. Check that the results m1 and m2 match using all.equal().n = 10
x = runif(n) # Generate 10 numbers uniformly between 0 and 11c. Rerun the code you wrote for the last question, but in the first line set n=10000, so that x is a vector of 10,000 numbers, distributed uniformly at random between 0 and 1. Do m1 and m2 match? What is the value of m1 now? Challenge: can you explain what is happening to m1 here, and why we would therefore prefer the strategy used to compute m2?
1d. Another reason to prefer the second strategy for computing the geometric mean—in which we use the exp(), mean(), and log() functions—is that it can be readily extended to computing the geometric means of rows/columns of matrices in R. Demonstrate this by computing the geometric means of each column of the matrix x defined below with just one line of code. (Do not use a for() loop here; restrict yourself to just three function calls, still.)
x = matrix(runif(40), 10, 4)1e. Nested for() loops work just like the usual (unnested) ones you’ve already been considering; nesting just means using a for() loop within the body of another for() loop. E.g., consider
x = matrix(0, 5, 5)
for (i in 1:5) {
for (j in 1:5) {
x[i,j] = i + j^2
}
}
x## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 5 10 17 26
## [2,] 3 6 11 18 27
## [3,] 4 7 12 19 28
## [4,] 5 8 13 20 29
## [5,] 6 9 14 21 30which populates the entries of the matrix x by first filling out all of its first row, then all of its second row, and so on. (To see this, look at the index variables in the for() loops, and step through their progression: first we set i=1 in the outer for() loop, then we set j=1, j=2, and so on in the inner for() loop, until j=5; then we move on to i=2, …)
Write a nested for() loop to multiply the two matrices a and b defined below, storing the result in the matrix c. You will have to remember how matrix multiplication works! And you must only use arithmetic operations in your solution. Hint: your solution should have a nesting of three for() loops (the example above had a nesting of two for() loops). Check using all.equal() that your result c matches a %*% b, which is R’s built-in way of multiplying a and b.
a = matrix(rnorm(15), 5, 3)
b = matrix(rnorm(12), 3, 4)
c = matrix(0, 5, 4)On to the more fun stuff! As in lab, we’re going to look at William Shakespeare’s complete works, taken from Project Gutenberg. The Shakespeare data file is up on our course website, and to load it into your R session, as a string vector called shakespeare.lines:
shakespeare.lines =
readLines("http://www.stat.cmu.edu/~ryantibs/statcomp-S18/data/shakespeare.txt")2a. Some lines in shakespeare.lines are empty, i.e., they are just equal to “”. How many such lines are there? Remove all empty lines from shakespeare.lines. Also, trim all “extra” white space characters in the lines of shakespeare.lines using the trimws() function. Note: if you are unsure about what trimws() does, try it out on some simple strings/some simple vectors of strings.
2b. Visit http://www.stat.cmu.edu/~ryantibs/statcomp-S18/data/shakespeare.txt in your web browser and just skim through this text file. Near the top you’ll see a table of contents. Note that “THE SONNETS” is the first play, and “VENUS AND ADONIS” is the last. Using which(), 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.
2c. Define n = 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.
2d. Use a for() 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.
2e. Define titles.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.
2f. In Q2d, 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 Q2d, 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 Q2e. Print out the new vectors titles.start and titles.end to the console—they should be free of NA values.
3a. Let’s look at two of Shakespeare’s most famous tragedies. Programmatically find the index at which “THE TRAGEDY OF HAMLET, PRINCE OF DENMARK” occurs in the titles 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?
shakespeare.lines.hamlet. That is:
shakespeare.lines.hamlet into one big string, separated by spaces;split="[[:space:]]|[[:punct:]]" in the call to strsplit();3c. Repeat the same task as in Q3b, but on shakespeare.lines.romeo. Comment on any similarities/differences you see in the answers.
Challenge. Using a for() 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?