`rivers`

, `women`

and `nhtemp`

), or created them by stringing together several numbers with the `c`

function (e.g. `c(1, 2, 3, 4)`

). R offers an extremely useful shortcut to create vectors of the latter kind, which is the colon `:`

operator. Instead of having to type:
`x <- c(1, 2, 3, 4)`

we can simply type

`x <- 1:4`

to create exactly the same vector. Obviously this is especially useful for longer sequences.

In fact, you will use sequences like this a lot in real-world applications of R, e.g. to select subsets of data points, records, or variables. The exercises in this set might come across as a little abstract, but trust me, these sequences are really the basic building blocks for your future R scripts. So let’s go ahead!

Before starting the exercises, please note this is the fourth set in a series of five: In the first three sets, we practised creating vectors, vector arithmetics, and various functions. You can find all sets in our ebook Start Here To Learn R – vol. 1: Vectors, arithmetic, and regular sequences. The book also includes all solutions (carefully explained), and the fifth and final set of the series. This final set focuses on the application of the concepts you learned in the first four sets, to real-world data.

One more thing: I would really appreciate your feedback on these exercises: Which ones did you like? Which ones were too easy or too difficult? Please let me know what you think here!

Try to shorten the notation of the following vectors as much as possible, using `:`

notation:

`x <- c(157, 158, 159, 160, 161, 162, 163, 164)`

`x <- c(15, 16, 17, 18, 20, 21, 22, 23, 24)`

`x <- c(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)`

`x <- c(-1071, -1072, -1073, -1074, -1075, -1074, -1073, -1072, -1071)`

`x <- c(1.5, 2.5, 3.5, 4.5, 5.5)`

(Solution)

The `:`

operator can be used in more complex operations along with arithmetic operators, and variable names. Have a look at the following expressions, and write down what sequence you think they will generate. Then check with R.

`(10:20) * 2`

`105:(30 * 3)`

`10:20*2`

`1 + 1:10/10`

`2^(0:5)`

(Solution)

Use the `:`

operator and arithmetic operators/functions from the previous chapter to create the following vectors:

`x <- c(5, 10, 15, 20, 25, 30)`

`x <- c(0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3)`

`x <- c(1/5, 2/6, 3/7, 4/8, 5/9, 6/10, 7/11, 8/12)`

`x <- (1, 4, 3, 8, 5, 12, 7, 16, 9, 20)`

(Hint: you have to use the recycle rule)

(Solution)

Another way to generate a sequence is the `seq`

function. Its first two arguments are `from`

and `to`

, followed by a third, which is `by`

. `seq(from=5, to=30, by=5)`

replicates part (a) of the previous exercise.

Note that you can omit the argument names `from`

, `to`

, and `by`

, if you stick to their positions, i.e., `seq(5, 30, 5)`

. Have a look at the following expressions, and write down what sequence you think they will generate. Then check with R.

`seq(from=20, to=80, by=20)`

`seq(from=-10, to=5, by=0.5)`

`seq(from=10, to=-3, by=-2)`

`seq(from=0.01, to=0.09, by=0.02)`

(Solution)

Compare the regular sequence of exercises 2(a) and 3(a) (both using the `:`

operator) with the same sequences using the `seq`

function with appropriate `by`

argument. Can you think of a more general rule how to convert any `seq(from, to, by)`

statement to a sequence generated with the `:`

operator?

In other words, rewrite `seq(from=x, to=y, by=z)`

to a statement using the `:`

operator. Hint: if this appears difficult, try to do this first by choosing some values for `x`

, `y`

, and `z`

, and see which pattern emerges.

(Solution)

The previous exercises in this set were aimed at generating sets of *increasing* or *decreasing* numbers. However, sometimes you just want a set of *equal* numbers. You can accomplish this with the `rep`

function (from “replicate”). Its first argument is the number or vector that will be replicated, and its second argument `times`

, … well I guess you can guess that one already. Now, let’s shorten the following statements, using `rep`

:

`x <- c(5, 5, 5, 5, 5, 5, 5)`

`x <- c(5, 6, 7)`

`y <- c(x, x, x, x, x)`

`x <- (10, 16, 71, 10, 16, 71, 10, 16, 71)`

(Solution)

`rep`

has a third very useful argument: `each`

. As we saw in the previous exercise (part b), vectors are replicated in their entirety by `rep`

.

However, you can also replicate “each” individual element. Consider for example:

`seq(c(1, 2, 3), times=2, each=3)`

.

This says: “replicate each element of the input vector `c(1, 2, 3)`

3 times, and then replicate the resulting vector 2 times.” Now, let’s shorten the following statements, using `rep`

:

`x <- c(5, 5, 5, 5, 8, 8, 8, 8, -3, -3, -3, -3, 0.34, 0.34, 0.34, 0.34)`

`x <- c(-0.1, -0.1, -0.9, -0.9, -0.6, -0.6)`

`x <- c(1, 1, 2, 2, 3, 3, 1, 1, 2, 2, 3, 3, 1, 1, 2, 2, 3, 3)`

(Solution)

We can actually write part c of te previous exercise even more compact by using `rep`

in combination with the `:`

operator. Do you see how?

In this exercise we’re using combinations of `rep`

, `:`

and `seq`

to create the following sequences:

`x <- c(97, 98, 99, 100, 101, 102, 97, 98, 99, 100, 101, 102, 97, 98, 99, 100, 101, 102)`

`x <- c(-5, -5, -5, -5, -6, -6, -6, -6, -7, -7, -7, -7, -8, -8, -8, -8)`

`x <- c(13, 13, 17, 17, 21, 21, 25, 25, 29, 29, 13, 13, 17, 17, 21, 21, 25, 25, 29, 29)`

`x <- c(1, 2, 3, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 3, 2, 1, 0)`

(Solution)

Suppose there would be no `each`

argument for `rep`

. Rewrite the following statement, without using the `each`

argument: `x <- rep(c(27, 31, 19, 14), each=v, times=w)`

(Solution)

Let’s finish this set off with an application. Let’s create a series of vectors for later use in a timeseries dataset. The idea is that each observation in this dataset can be identified by a *timestamp*, which is defined by four vectors:

- s (for seconds)
- m (minutes)
- h (hours)
- d (days)

For this exercise, we’ll limit the series to a full week of 7 days.

This is a somewhat more complicated problem than the previous ones in this exercise. Don’t worry however! Whenever you’re faced with a somewhat more complicated problem than you are used to, the best strategy is to break it down into smaller problems. So, we’ll simply start with the `s`

vector.

- Since
`s`

counts the number of seconds, we know it has to start at 1, run to 60, restart at 1, etc. As it should cover a full week, we also know we have to replicate this series many times. Can you calculate exactly how many times it has to replicate this series? Use the outcome of your calculation to create the full`s`

vector. - Now, let’s create the vector
`m`

. Think about how this vector differs from`s`

. What does this mean for the`times`

and`each`

arguments? - Now, let’s create vector
`h`

and`d`

using the same logic. Check that`s`

,`m`

,`h`

, and`d`

have equal length.

(Solution)

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Sofar, the functions we have practised (`log`

, `sqrt`

, `exp`

, `sin`

, `cos`

, and `acos`

) always return a vector with the same length as the input vector. In other words, the function is applied element by element to the elements of the input vector. Not all functions behave this way though. For example, the function `min(x)`

returns a single value (the minimum of all values in `x`

), regardless of whether x has length 1, 100 or 100,000.

Before starting the exercises, please note this is the third set in a series of five: In the first two sets, we practised creating vectors and vector arithmetics. In the fourth set (posted next week) we will practise regular sequences and replications.

You can find all sets right now in our ebook Start Here To Learn R – vol. 1: Vectors, arithmetic, and regular sequences. The book also includes all solutions (carefully explained), and the fifth and final set of the series. This final set focuses on the application of the concepts you learned in the first four sets, to real-world data.

One more thing: I would really appreciate your feedback on these exercises: Which ones did you like? Which ones were too easy or too difficult? Please let me know what you think here!

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`

).

(Solution)

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.

- 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. - Calculate the mean of
`rivers`

after trimming the 10 highest and lowest observations. Hint: first calculate the trim fraction, using the`length`

function.

(Solution)

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).

- Using the
`cor`

function, show that the average height and weight of these women are almost perfectly correlated. - Calculate their covariance, using the
`cov`

function. - 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?)

(Solution)

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.

`x <- c(300.99, 1.6, 583, 42.10)`

`y <- c(300, 1, 583, 42)`

`x <- c(152.34, 1940.63, 1.0001, -2.4, sqrt(26))`

`y <- c(152, 1940, 1, 5, -2)`

`x <- -c(3.2, 444.35, 1/9, 100)`

`y <- c(-3, -444, 0, -100)`

`x <- c(35.6, 670, -5.4, 3^3)`

`y <- c(36, 670, -5, 27)`

(Solution)

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.

`x <- c(35.63, 300.20, 0.39, -57.8)`

`y <- c(36, 300, 0, -58)`

`x <- c(153, 8642, 10, 39.842)`

`y <- c(153.0, 8640.0, 10.0, 39.8)`

`x <- c(3.8, 0.983, -23, 7.1)`

`y <- c(3.80, 0.98, -23.00, 7.10)`

(Solution)

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.

- 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. - 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.

(Solution)

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`

.

- Use the
`women`

data (exercise 3) and create a vector`x`

with the elements of the`height`

vector sorted in decreasing order. - 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? - 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.

(Solution)

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`

- Can you describe in your own words what
`rank`

does? - 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`

.

(Solution)

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`

- Can you describe in your own words what
`order`

does? Hint: look at the output of`sort(x)`

if you run into trouble. - 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?

- 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.

(Solution)

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.

- 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)`

? - 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. - 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).
- 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. - 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)`

(Solution)

This is the second set in a series of five: In the first set (posted last week) we practised the basics of vectors. In set three and four (upcoming) we will practise more vector arithmetics to e.g. calculate all kinds of statistics, carry out simulations, sort data, or calculate the distance between two cities.

If you can’t wait till all sets are posted: you can find them right now in my ebook Start Here To Learn R – vol. 1: Vectors, arithmetic, and regular sequences. The book also includes the fifth and final set of the series. This final set focuses on the application of the concepts you learned in the first four sets, to real-world data.

Let’s create the following vectors:

`u <- 4`

`v <- 8`

Use the elementary arithmetic operators `+`

, `-`

, `*`

, `/`

, and `^`

to:

- add
`u`

and`v`

- subtract
`v`

from`u`

- multiply
`u`

by`v`

- divide
`u`

by`v`

- raise
`u`

to the power of`v`

(Solution)

Now, suppose u and v are not scalars, but vectors with multiple elements:

`u <- c(4, 5, 6)`

`v <- c(1, 2, 3)`

Without using R, write down what you expect as the result of the same operations as in the previous exercise:

- add
`u`

and`v`

- subtract
`v`

from`u`

- multiply
`u`

by`v`

- divide
`u`

by`v`

- raise
`u`

to the power of`v`

(Solution)

We just saw how arithmetic operators work on vectors with the same length. But how about vectors that differ in length? Let’s find out… Consider the following vectors:

`u <- c(5, 6, 7, 8)`

`v <- c(2, 3, 4)`

Without using R, write down what you expect as the result of the same operations as in the previous exercise:

- add
`u`

and`v`

- subtract
`v`

from`u`

- multiply
`u`

by`v`

- divide
`u`

by`v`

- raise
`u`

to the power of`v`

Then check your answer with R. Which rule does R use, when it has to deal with vectors of different lengths?

(Solution)

When we want to carry out a series of arithmetic operations, we can either use a single expression, or a series of expressions. Consider two vectors `u`

and `v`

:

`u <- c(8, 9, 10)`

`v <- c(1, 2, 3)`

We can create a new vector w in a single line of code:

`w <- (2 * u + v) / 10`

or carry out each operation on a separate line:

`w <- 2 * u`

`w <- w + v`

`w <- w / 10`

Convert the following expressions to separate operations, and check that both approaches give the same result:

`w <- (u + 0.5 * v) ^ 2`

`w <- (u + 2) * (u - 5) + v`

`w <- (u + 2) / ((u - 5) * v)`

(Solution)

We can do the reverse as well. Convert the following multi-line operations to a single expression. Check that both approaches give the same result.

Part a:

`w <- u + v`

`w <- w / 2`

`w <- w + u`

Part b:

`w1 <- u^3`

`w2 <- u - v`

`w <- w1 / w2`

(Solution)

Exercise 6, 7, and 8 focus on mathematics. Sooner or later you might have to translate mathematical formulas into R, to perform simple or more elaborate mathematical calculations. The goal of these exercises is to practise just that: how to translate a mathematical expression to R code. So, we won’t delve into the mathematics behind these formulas, and their derivation.

Also, in some cases, these formulas have already been translated to R by others, and are available in so-called *contributed packages*. We will deal with using these packages at a later stage, and for now the goal is just to become familiar with implementing mathematical formulas in R yourself.

So, here’s the deal: If you really hate math, or know for sure you will never use math in R, then it’s ok to skip exercise 6, 7, and 8. Otherwise: Let’s go for it and enjoy!

Besides the arithmetic operators we have used so far, there are some more that we often use: `log`

, `exp`

, and `sqrt`

. We can also use the well-known constant *pi*, by simply typing `pi`

, instead of its value 3.1415927.

Let’s try to apply what we have learned so far to some well-known, somewhat more advanced formulas. Don’t let the math scare you. Just translate the formulas to R code, one operator at a time. Don’t hesitate to use multiple lines if that makes things easier, or add parentheses to make sure operations are carried out in the right order.

- Suppose the surface area of a circle equals 25, what is the radius?
- What is the probability density at
`x=0`

of a normally distributed random variable`x`

with mean (`mu`

) equal to zero, and standard devation (`sigma`

) equal to one (look up the formula online, e.g. https://en.wikipedia.org/wiki/Normal_distribution)?

(Solution)

Consider the following formula to calculate the number of mortgage payment terms: \[n=\frac{\ln \Bigg(\dfrac{i}{\dfrac{M}{P}-i}+1\Bigg)}{\ln(1+i)}\] In this equation, `M`

represents the monthly payment amount, `P`

the principle, and `i`

the (monthly) interest rate.

- Calculate the number of payment terms
`n`

for a mortgage with a principle balance of 200,000, monthly interest rate of 0.5%, and monthly payment amount of 2000. - Now construct a vector
`n`

of length 6 with the results of this calculation for a series of monthly payment amounts: 2000, 1800, 1600, 1400, 1200, 1000. - Does the last value of
`n`

surprise you? Can you explain it?

(Solution)

Suppose you have geographical data and want to calculate the distance between two places on earth, given by their latitude and longitude coordinates. Consider the coordinates for:

- Paris: 48.8566° N (latitude), 2.3522° E (longitude), and
- New York 40.7128° N (latitude), 74.0060° W (longitude)

If you’re up for a real challenge, lookup “Great-circle distance” on Wikipedia, and use the *spherical law of cosines* to find the distance (and stop reading right now!).

If this sounds like a pretty daunting task, don’t worry! I will walk you through this step-by-step in the remainder of this exercise.

Ok, here we go. We will use the following common abbreviations:

- latitude (\(\phi\))
`phi`

- longitude (\(\lambda\))
`lambda`

- Create 4 scalars
`phi.paris`

,`phi.ny`

,`lambda.paris`

,`lambda.ny`

, representing these coordinates. Because New York is located in the West, you have to enter this as a negative value (-74.0060). - Convert the 4 coordinates from degrees to radians, using the formula: \[radians = degrees \frac{\pi}{180}\]
- Calculate the central angle between both cities, using the spherical law of cosines: \[\Delta\sigma=\arccos(\sin \phi_1 \sin \phi_2 + \cos \phi_1 \cos \phi_2 \cos(\Delta\lambda))\] where: \(\Delta\sigma\) is just a scalar (name it anything you want in R), \(\phi_1\) is the latitude of Paris, \(\phi_2\) the latitude of New York, and \(\Delta\lambda\) the
*absolute*difference between both longitudes.Hint: For this calculation you need the following mathematical functions in R:`sin`

,`cos`

,`acos`

, and`abs`

. - Finally, to find the distance, multiply \(\Delta\sigma\) (i.e., the outcome you just calculated) by the radius of the earth (6371 km.)

(Solution)

Use the modulo operator (`%%`

) to find out for which of the following pairs, the second number is a multiple of the first. Your R code should contain the modulo operator just once!

530, 1429410

77, 13960

231, 2425

8, 391600

(Solution)

May I kindly ask you to share your thoughts on these exercises? This will allow me to further improve the quality of the exercises.

You can share your thoughts simply by adding a comment below. I am particularly interested in:

- Which exercises you liked least and most
- Which (if any) exercises were too hard, and should be simplified
- Which (if any) exercises were too easy and should be more challenging
- Overall comments on the quality of this set
- Which topics you’d like to see addressed in future sets

Because vectors are such a key concept, we’re going to practise their use and application slowly, step-by-step. For now we’ll just practise *numeric* vectors (and save other types, such as *character* vectors, for later).

In this set we’re practising the basics of vectors, i.e. how to create vectors and assign them to a name. It is the first set in a series of five: In the second set (posted next week) we will practise working with vectors. In set three and four we will practise vector arithmetics to e.g. calculate all kinds of statistics, carry out simulations, sort data, or calculate the distance between two cities.

If you can’t wait till all sets are posted: you can find them right now in our ebook Start Here To Learn R – vol. 1: Vectors, arithmetic, and regular sequences. The book also includes the fifth and final set of the series. This final set focuses on the application of the concepts you learned in the first four sets, to real-world data.

Let’s start really easy (don’t worry, we’ll quickly move to more challenging problems) with a vector containing just a single number, which we also call a scalar. Enter a vector in R, by just typing a random number, e.g. `100`

, at the prompt and hit the Enter key.

(Solution)

Great! You just created your first vector! Now, let’s first enter a vector with more than one number. E.g. a vector containing the numbers 1, 2, 3, 4, 5, in that order. If you enter these numbers just like this, R will respond with an error message. It throws an error, because it needs a little bit more information from our side that we actually want to store those numbers in a vector structure. We have to use the following notation for this:

`c(1, 2, 3, 4, 5)`

.

Now, enter a vector with the first 5 even numbers in R, and hit Enter.

(Solution)

Let’s now enter a much longer vector, containing the numbers 1 to 10, 10 times (use copy & paste). What do the numbers between square brackets in the R output mean?

(Solution)

You should be pretty familiar with entering vectors now. You might actually feel a little *bored* by typing all these numbers. Life would be pretty miserable if we would have to enter data this way over and over again in R. But fortunately, there is a neat solution! We can *assign* a vector to a variable name such that we can retrieve the data we have entered, conveniently by just typing the name of the variable.

Try to assign a vector containing the numbers 1, 2, 3, 4, 5 to a variable named `a`

, using the assignment operator (`<-`

), and see which of the statements below work.

Enter each of the 9 statements one at a time at the prompt, hit Enter, and try to retrieve the contents of `a`

, by typing `a`

at the prompt after you entered each statement:

`a<-c(1, 2, 3, 4, 5)`

`a <- c(50, 60, 70, 80, 90)`

`a -> c(20, 31, 42, 53, 64)`

`c(5, 6, 7, 9, 10) <- a`

`c(101, 102, 103, 104, 105) -> a`

`a < - c(11, 12, 13, 14, 15)`

`a < -c(100, 99, 88, 77, 66)`

`assign(a, c(1000, 2000, 3000, 4000, 5000))`

`assign('a', c(83, 16, 35, 58, 3))`

(Solution)

In an R script, you might have created dozens or even hundreds of vectors. In that case, naming them `a`

, `b`

, `c`

etc. is not ideal, because it will be difficult to keep track of what all those letters actually mean. This problem is easily mitigated by using longer, and meaningful, variable names.

Assign the following vectors to a meaningful variable name:

`c(2, 4, 6, 8, 10, 12, 14, 16, 20)`

`0`

`3.141593`

`c(1, 10, 100, 1000, 10000, 100000)`

(Solution)

Create vectors that correspond to the following variables names:

- bmi
- age
- daysPerMonth
- firstFivePrimeNumbers

(Solution)

So far, we have created vectors from a bunch of numbers. Instead of numbers, however, you can also enter other vectors, e.g. `c(vector1, vector2, vector3)`

, and string them together.

To practise this, let’s first create three vectors that each contain just 1 element with variable names `p`

, `q`

, and `r`

, and values 1, 2, and 3. Then, create a new vector that contains multiple elements, using the scalars we just created. I.e., create a vector `u`

of length 3, with the subsequent elements of `p`

, `q`

and `r`

.

(Solution)

To play with this a little more, let’s create a longer vector, using only the assignment operator (`<-`

), the `c()`

function, and the vector `u`

we just created. I.e., create a new vector `u`

with length 96 that contains the elements of `u`

as follows: 1, 2, 3, 1, 2, 3, …., 1, 2, 3

(Solution)

May I kindly ask you to share your thoughts on these exercises? This will allow me to further improve the quality of the exercises.

You can share your thoughts simply by adding a comment below. I am particularly interested in:

- Which exercises you liked least and most
- Which (if any) exercises were too hard, and should be simplified
- Which (if any) exercises were too easy and should be more challenging
- Overall comments on the quality of this set
- Which topics you’d like to see addressed in future sets

The shinydashboard package provides a well-designed dashboard theme for Shiny apps and allows for an easy assembly of a dashboard from a couple of basic building blocks. The package is widely used in commercial environments as well, due to its neat features for building convenient and robust layouts.

This exercise set will help you practice all of the main features of this great package. By completing the two parts of the exercise series, you’ll know that you’re ready to start building well-designed Shiny apps. We will make some minimal use of the built-in data-set `datasets::CO2`

(specific description of the data-set is irrelevant, but you can check it out by typing `?datasets::CO2`

). Each exercise is adding some more features/functionalities to the code of the previous exercise, so be sure to not discard the code until after you’re done with all of the exercises. Answers to these exercises are available here.

In the solutions page, you’ll first find only the relevant components of each exercise. Then, at the end of the page, you will find the entire Shiny app code that contains all of the different components together. It is advised to use the following template to get started with the exercises:

header <- dashboardHeader(...)

sidebar <- dashboardSidebar(...)

body <- dashboardBody(...)

ui <- dashboardPage(...)

server <- function(input, output, session) {}

shinyApp(ui = ui, server = server)

For other parts of the series, follow the tag shinydashboard.

**Exercise 1**

Set the title of the dashboard to be “Practicing shinydashboard.”

The font-family should be “Tahoma” and the font-weight should be “bold.”

**Exercise 2**

Since the title that we chose is a bit long, set the header and sidebar width to 300 pixels.

**Exercise 3**

Select a black skin to the dashboard.

**Exercise 4**

Select a title for the browser tab, which is different than the dashboard title.

**Exercise 5**

Add two menu items to the sidebar: one named “data” and one named “about.”

Change the font-size of the menu items to 20 pixels.

**Exercise 6**

Add icons of your choice next to the two menu items that you added.

**Exercise 7**

Add a yellow badge which says “New” next to the “data” menu item.

**Exercise 8**

Under the “about” menu item, add two sub-items: one named “licences” and one named “contact us.”

**Exercise 9**

Add a search box at the top of the sidebar with the label “What are you looking for?”

Hint: use `sidebarSearchForm()`

.

**Exercise 10**

Add a `selectInput`

under the search box you just added.

The input ID should be `plant`

and the choices should be `unique(CO2$Plant)`

.

- Spatial Data Analysis: Introduction to Raster Processing (Part 1)
- Parallel Computing Exercises: Snow and Rmpi (Part-3)
- Advanced Techniques With Raster Data: Part 1 – Unsupervised Classification
- Become a Top R Programmer Fast with our Individual Coaching Program
- Explore all our (>4000) R exercises
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In this exercise set, we will practice using the forcats factor manipulation package by Hadley Wickham. In the last exercise set, we saw that it is entirely possible to deal with factors in base R, but also that things can get a bit involved and un-intuitive. Forcats simplifies many common factor manipulation tasks and worth mastering if you cannot avoid using factors in your work. Also, studying the package and its source code can give you ideas for writing your own custom function to simplify everyday tasks that you think can be dealt with in a better way.

Solutions are available here.

**Exercise 1**

Load the gapminder data-set from the gapminder package, as well as forcats. Check what the levels of the continent factor variable are and their frequency in the data.

**Exercise 2**

Notice that one continent, Antarctica, is missing – add it as the last level of six.

**Exercise 3**

Actually, you change your mind. There is no permanent human population on Antarctica. Drop this (unused) level from your factor.

**Exercise 4**

Again, modify the continent factor, making it more precise. Add two new levels: instead of Americas, add North America and South America. The countries in the following vector should be classified as South America and the rest as North America.

c("Argentina", "Bolivia", "Brazil", "Chile", "Colombia", "Ecuador",

"Paraguay", "Peru", "Uruguay", "Venezuela")

**Exercise 5**

Arrange the levels of the continent factor in alphabetical order.

**Exercise 6**

Re-order the continent levels again so that they appear in order of total population in 2007.

**Exercise 7**

Reverse the order of the factors.

**Exercise 8**

Make continent, again, an unordered factor. Set North America as the first level, therefore interpreted as a reference group in modeling functions such as `lm()`

.

**Exercise 9**

Turn the following messy vector into a factor with two levels: “Female” and “Male” using the factor function. Use the labels argument in the factor() function.

`gender <- c("f", "m ", "male ","male", "female", "FEMALE", "Male", "f", "m")`

**Exercise 10**

Gender can be considered sensitive data. Convert the gender variable into a factor that takes the integer values “1” and “2”, where one integer represents female and the other male, but make the choice randomly.

]]>Factor variables in R can be mind-boggling. Often, you can just avoid them and use characters vectors instead – just don’t forget to set `stringsAsFactors=FALSE`

. They are, however, very useful in some circumstances, such as statistical modelling and presenting data in graphs and tables. Relying on factors but misunderstanding them has been known to *“eat up hours of valuable time in any given analysis”,* as one member of the community put it. It is therefore a good investment to get them straight as soon as possible on your R journey.

The intent behind these exercises is to help you find and fill in the cracks and holes in your relationship with factor variables.

Solutions are available here.

**Exercise 1**

Load the gapminder data-set from the gapminder package. Save it to an object called `gp`

. Check programmatically how many factors it contains and how many levels each factor has.

**Exercise 2**

Notice that one continent, Antarctica, is missing from the corresponding factor – add it as the last level of six.

**Exercise 3**

Actually, you change your mind. There is no permanent human population on Antarctica. Drop this (unused) level from your factor. Can you find three ways to do this, then you are an expert.

**Exercise 4**

Again, modify the continent factor, making it more precise. Add two new levels instead of Americas, North-America and South-America. The countries in the following vector should be classified as South-America and the rest as North-America.

c("Argentina", "Bolivia", "Brazil", "Chile", "Colombia", "Ecuador",

"Paraguay", "Peru", "Uruguay", "Venezuela")

**Exercise 5**

Get the levels of the factor in alphabetical order.

**Exercise 6**

Re-order the continent levels again so that they appear in order of total population in 2007.

**Exercise 7**

Reverse the order of the factor and define continents as an ordered factor.

**Exercise 8**

Make the continent an unordered factor again and set North-America as the first level, thus interpreted as a reference group in modelling functions such as `lm()`

.

**Exercise 9**

Turn the following messy vector into a factor with two levels: Female and Male, using the factor function. Use the labels argument in the factor() function (ps: you can save some time by applying tolower() and trimws() before you apply factor()).

`gender <- c("f", "m ", "male ","male", "female", "FEMALE", "Male", "f", "m")`

**Exercise 10**

Use the fact that factors are built on top of integers and create a dummy (binary) variable `male`

that takes the value 1 if the gender has the value “Male.”

**INTRODUCTION**

With flexdashboard, you can easily create interactive dashboards for R. What is amazing about it is that with R Markdown, you can publish a group of related data visualizations as a dashboard.

Additionally, it supports a wide variety of components, including htmlwidgets; base, lattice, and grid graphics; tabular data; gauges and value boxes and text annotations.

It is flexible and easy to specify rows and column-based layouts. Components are intelligently re-sized to fill the browser and adapted for display on mobile devices.

In combination with Shiny, you can create a high quality dashboard with interactive visualizations.

Before proceeding, please follow our short tutorial.

Look at the examples given and try to understand the logic behind them. Then, try to solve the exercises below by using R without looking at the answers. Then, check the solutions to check your answers.

**Exercise 1**

Create a new flexdashboard R Markdown file from the R console.

**Exercise 2**

Create the very initial dashboard interface in a single column.

**Exercise 3**

Add the space that you will put your first chart in.

**Exercise 4**

Add the space that you will put your second chart in. The two charts should be stacked vertically.

**Exercise 5**

Add a third chart with the same logic.

**Exercise 6**

Transform your layout to scrolling.

**Exercise 7**

Displays THE 3 charts split across two columns.

**Exercise 8**

Change the width of these two columns.

**Exercise 9**

Define two rows, instead of columns. The first has a single chart and the second has two charts.

**Exercise 10**

Change the height of these two columns.

]]>The goal of patchwork is to make it simple to combine separate ggplots into the same graphic. It tries to solve the same problem as gridExtra::grid.arrange() and cowplot::plot_grid, but using an API that incites exploration and iteration.

Before proceeding, please follow our short tutorial.

Look at the examples given and try to understand the logic behind them. Then, try to solve the exercises below by using R without looking at the answers. Then, check the solutions to check your answers.

**Exercise 1**

Create a scatter-plot object of `mtcars`

between `mpg`

and `disp`

.

**Exercise 2**

Create a box-plot object of `mtcars`

between `gear`

and `disp`

grouped by `gear`

.

**Exercise 3**

Compose those two objects into one graph.

**Exercise 4**

Repeat the previous process but in one plotting operation.

**Exercise 5**

Display the composed graph in one column with the two graphs, one below the other.

**Exercise 6**

Set the graph on top to have two times the size of the graph at the bottom.

**Exercise 7**

Add space between your plots.

**Exercise 8**

Create two objects of your choice (`p3`

,` p4`

) and display all of your four objects in nested mode by putting `p4`

on top` p1`

and `p2`

and `p3`

at the bottom (nested one below the other.)

**Exercise 9**

Now, put the two nested plots next to each other.

**Exercise 10**

Finally, display `p4`

in one column and the rest of your objects in another.

- Parallel Computing Exercises: Foreach and DoParallel (Part-2)
- Parallel Computing Exercises: Snow and Rmpi (Part-3)
- Building Shiny App exercises part 4
- Become a Top R Programmer Fast with our Individual Coaching Program
- Explore all our (>4000) R exercises
- Find an R course using our R Course Finder directory

In the age of Rmarkdown and Shiny, or when making any custom output from your data, you want your output to look consistent and neat. Also, when writing your output, you often want it to obtain a specific (decorative) format defined by the html or LaTeX engine. These exercises are an opportunity to refresh our memory on functions, such as paste, sprintf, formatC and others that are convenient tools to achieve these ends. All of the solutions rely partly on the ultra flexible sprintf(), but there are no-doubt many ways to solve the exercises with other functions. Feel free to share your solutions in the comment section.

Example solutions are available here.

**Exercise 1**

Print out the following vector as prices in dollars (to the nearest cent):

`c(14.3409087337707, 13.0648270623048, 3.58504267621646, 18.5077076398145,`

. Example:

16.8279241011882)`$14.34`

**Exercise 2**

Using these numbers, `c(25, 7, 90, 16)`

, make a vector of filenames in the following format: `file_025.txt`

. Left pad the numbers so they are all three digits.

**Exercise 3**

Actually, if we are only dealing with numbers less than one hundred, `file_25.txt`

would have been enough. Change the code from the last exercise so that the padding is pro-grammatically decided by the biggest number in the vector.

**Exercise 4**

Print out the following haiku on three lines, right aligned, with the help of cat: `c("Stay the patient course.", "Of little worth is your ire.", "The network is down.")`

.

**Exercise 5**

Write a function that converts a number to its hexadecimal representation. This is a useful skill when converting bmp colors from one representation to another. Example output:

tohex(12) [1] "12 is c in hexadecimal"

**Exercise 6**

Take a string and pro-grammatically surround it with the html header tag `h1`

.

**Exercise 7**

Back to the poem from exercise 4, let R convert to html unordered list so that it would appear like the following in a browser:

- Stay the patient course
- Of little worth is your ire
- The network is down

**Exercise 8**

Here is a list of the current top 5 movies on imdb.com in terms of rating `c("The Shawshank Redemption", "The Godfather", "The Godfather: Part II", "The Dark Knight", "12 Angry Men", "Schindler's List")`

. Convert them into a list compatible with the written text.

Example output:

`[1] "The top ranked films on imdb.com are The Shawshank Redemption, The Godfather, The Godfather: Part II, The Dark Knight, 12 Angry Men and Schindler's List"`

**Exercise 9**

Now, you should be able to solve this quickly: write a function that converts a proportion to a percentage that takes as input number of decimal places. An input of 0.921313 and 2 decimal places should return `"92.13%"`

.

**Exercise 10**

Improve the function from the last exercise so that the percentage consistently takes 10 characters by doing some left padding. Raise an error if the percentage already happens to be longer than 10.

(Image by Daniel Friedman).

]]>