Exercise 1

Did you know R has actually lots of built-in datasets that we can use to practise? For example, the rivers data “gives the lengths (in miles) of 141 “major” rivers in North America, as compiled by the US Geological Survey” (you can find this description, and additonal information, if you enter help(rivers) in R. Also, for an overview of all built-in datasets, enter data().

Have a look at the rivers data by simply entering rivers at the R prompt. Create a vector v with 7 elements, containing the number of elements (length) in rivers, their sum (sum), mean (mean), median (median), variance (var), standard deviation (sd), minimum (min) and maximum (max).

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Exercise 2

For many functions, we can tweak their result through additional arguments. For example, the mean function accepts a trim argument, which trims a fraction of observations from both the low and high end of the vector the function is applied to.

1. What is the result of mean(c(-100, 0, 1, 2, 3, 6, 50, 73), trim=0.25)? Don’t use R, but try to infer the result from the explanation of the trim argument I just gave. Then check your answer with R.
2. Calculate the mean of rivers after trimming the 10 highest and lowest observations. Hint: first calculate the trim fraction, using the length function.

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Exercise 3

Some functions accept multiple vectors as inputs. For example, the cor function accepts two vectors and returns their correlation coefficient. The women data “gives the average heights and weights for American women aged 30-39”. It contains two vectors height and weight, which we access after entering attach(women) (we’ll discuss the details of attach in a later chapter).

1. Using the cor function, show that the average height and weight of these women are almost perfectly correlated.
2. Calculate their covariance, using the cov function.
3. The cor function accepts a third argument method which allows for three distinct methods (“pearson”, “kendall”, “spearman”) to calculate the correlation. Repeat part (a) of this exercise for each of these methods. Which is the method chosen by the default (i.e. without specifying the method explicitly?)

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Exercise 4

In the previous three exercises, we practised functions that accept one or more vectors of any length as input, but return a single value as output. We’re now returning to functions that return a vector of the same length as their input vector. Specifically, we’ll practise rounding functions. R has several functions for rounding. Let’s start with floor, ceiling, and trunc:

• floor(x) rounds to the largest integer not greater than x
• ceiling(x) rounds to the smallest integer not less than x
• trunc(x) returns the integer part of x

To appreciate the difference between the three, I suggest you first play around a bit in R with them. Just pick any number (with or without a decimal point, positive and negative values), and see the result each of these functions gives you. Then make it somewwat closer to the next integer (either above or below), or flip the sign, and see what happens. Then continue with the following exercise:

Below you will find a series of arguments (x), and results (y), that can be obtained by choosing one or more of the 3 functions above (e.g. y <- floor(x)). Which of the above 3 functions could have been used in each case? First, choose your answer without using R, then check with R.

1. x <- c(300.99, 1.6, 583, 42.10)
y <- c(300, 1, 583, 42)
2. x <- c(152.34, 1940.63, 1.0001, -2.4, sqrt(26))
y <- c(152, 1940, 1, 5, -2)
3. x <- -c(3.2, 444.35, 1/9, 100)
y <- c(-3, -444, 0, -100)
4. x <- c(35.6, 670, -5.4, 3^3)
y <- c(36, 670, -5, 27)

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Exercise 5

In addition to trunc, floor, and ceiling, R also has round and signif rounding functions. The latter two accept a second argument digits. In case of round, this is the number of decimal places, and in case of signif, the number of significant digits. As with the previous exercise, first play around a little, and see how these functions behave. Then continue with the exercise below:

Below you will find a series of arguments (x), and results (y), that can be obtained by choosing one, or both, of the 2 functions above (e.g. y <- round(x, digits=d)). Which of these functions could have been used in each case, and what should the value of d be? First, choose your answer without using R, then check with R.

1. x <- c(35.63, 300.20, 0.39, -57.8)
y <- c(36, 300, 0, -58)
2. x <- c(153, 8642, 10, 39.842)
y <- c(153.0, 8640.0, 10.0, 39.8)
3. x <- c(3.8, 0.983, -23, 7.1)
y <- c(3.80, 0.98, -23.00, 7.10)

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Exercise 6

Ok, let’s continue with a really interesting function: cumsum. This function returns a vector of the same length as its input vector. But contrary to the previous functions, the value of an element in the output vector depends not only on its corresponding element in the input vector, but on all previous elements in the input vector. So, its results are cumulative, hence the cum prefix. Take for example: cumsum(c(0, 1, 2, 3, 4, 5)), which returns: 0, 1, 3, 6, 10, 15. Do you notice the pattern?

Functions that are similar in their behavior to cumsum, are: cumprod, cummax and cummin. From just their naming, you might already have an idea how they work, and I suggest you play around a bit with them in R before continuing with the exercise.

1. The nhtemp data contain “the mean annual temperature in degrees Fahrenheit in New Haven, Connecticut, from 1912 to 1971”. (Although nhtemp is not a vector, but a timeseries object (which we’ll learn the details of later), for the purpose of this exercise this doesn’t really matter.) Use one of the four functions above to calculate the maximum mean annual temperature in New Haven observed since 1912, for each of the years 1912-1971.
2. Suppose you put $1,000 in an investment fund that will exhibit the following annual returns in the next 10 years: 9% 18% 10% 7% 2% 17% -8% 5% 9% 33%. Using one of the four functions above, show how much money your investment will be worth at the end of each year for the next 10 years, assuming returns are re-invested every year. Hint: If an investment returns e.g. 4% per year, it will be worth 1.04 times as much after one year, 1.04 * 1.04 times as much after two years, 1.04 * 1.04 * 1.04 times as much after three years, etc.

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Exercise 7

R has several functions for sorting data: sort takes a vector as input, and returns the same vector with its elements sorted in increasing order. To reverse the order, you can add a second argument: decreasing=TRUE.

1. Use the women data (exercise 3) and create a vector x with the elements of the height vector sorted in decreasing order.
2. Let’s look at the rivers data (exercise 1) from another perspective. Looking at the 141 data points in rivers, at first glance it seems quite a lot have zero as their last digit. Let’s examine this a bit closer. Using the modulo operator you practised in exercise 9 of the previous exercise set, to isolate the last digit of the rivers vector, sort the digits in increasing order, and look at the sorted vector on your screen. How many are zero?
3. What is the total length of the 4 largest rivers combined? Hint: Sort the rivers vector from longest to shortest, and use one of the cum... functions to show their combined length. Read off the appropriate answer from your screen.

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Exercise 8

Another sorting function is rank, which returns the ranks of the values of a vector. Have a look at the following output:

x <- c(100465, -300, 67.1, 1, 1, 0)
rank(x)
## [1] 6.0 1.0 5.0 3.5 3.5 2.0

1. Can you describe in your own words what rank does?
2. In exercise 3(c) you estimated the correlation between height and weight, using Spearman’s rho statistic. Try to replicate this using the cor function, without the method argument (i.e., using its default Pearson method, and using rank to first obtain the ranks of height and weight.

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Exercise 9

A third sorting function is order. Have a look again at the vector x introduced in the previous exercise, and the output of order applied to this vector:

x <- c(100465, -300, 67.1, 1, 1, 0)
order(x)
## [1] 2 6 4 5 3 1

1. Can you describe in your own words what order does? Hint: look at the output of sort(x) if you run into trouble.
2. Remember the time series of mean annual temperature in New Haven, Connecticut, in exercise 6? Have a look at the output of order(nhtemp):

order(nhtemp)
##  [1]  6 15 29  9 13  3  5 12  7 23  1 24 47 25 51 14 18 32 16 11 56  8 17
## [24] 28 45 52 31 37  4 22 36 39 54 19 34 26 49 30 33 53 55 21 27 58 10 50
## [47] 57 59 43 44 35  2 46 48 40 20 60 41 38 42

Given that the starting year for this series is 1912, in which years did the lowest and highest mean annual temperature occur?

3. What is the result of order(sort(x)), if x is a vector of length 100, and all of its elements are numbers? Explain your answer.

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Exercise 10

In exercise 1 of this set, we practised the max function, followed by the cummax function in exercise 6. In the final exercise of this set, we’re returning to this topic, and will practise yet another function to find a maximum. While the former two functions applied to a single vector, it’s also possible to find a maximum across multiple vectors.

1. First let’s see how max deals with multiple vectors. Create two vectors x and y, where x contains the first 5 even numbers greater than zero, and y contains the first 5 uneven numbers greater than zero. Then see what max does, as in max(x, y). Is there a difference with max(y, x)?
2. Now, try pmax(x, y), where p stands for “parallel”. Without using R, what do you think intuitively, what it will return? Then, check, and perhaps refine, your answer with R.
3. Now try to find the parallel minimum of x and y. Again, first try to write down the output you expect. Then check with R (I assume, you can guess the appropriate name of the function).
4. Let’s move from two to three vectors. In addition to x and y, add -x as a third vector. Write down the expected output for the parallel minima and maxima, then check your answer with R.
5. Finally, let’s find out how pmax handles vectors of different lenghts Write down the expected output for the following statements, then check your answer with R.

• pmax(x, 6)
• pmax(c(x, x), y)
• pmin(x, c(y, y), 3)

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