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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

- Spatial Data Analysis: Introduction to Raster Processing (Part 1)
- Advanced Techniques With Raster Data: Part 1 – Unsupervised Classification
- Spatial Data Analysis: Introduction to Raster Processing: Part-3
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- Explore all our (>4000) R exercises
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`10 150 30 45 20.3`

And here’s another one:

`-5 -4 -3 -2 -1 0 1 2 3`

still another one:

`"Darth Vader" "Luke Skywalker" "Han Solo"`

and our final example:

`389.3491`

These examples show that a vector is, simply speaking, just a collection of one (fourth example) or more (first and second example) numbers or character strings (third example). In R, a vector is considered a data structure (it’s a very simple data structure, and we’ll cover more complex structures in another tutorial).

We can construct a vector from a series of individual elements, using the `c()`

function, as follows:

c(10, 150, 30, 45, 20.3)

## [1] 10.0 150.0 30.0 45.0 20.3

(In examples like these, lines starting with `##`

show the output from R on the screen).

As you’ll see, once you have entered the vector, R will respond by displaying its elements. In many cases it will be convenient to refer to this vector using a name, instead of having to enter it over and over again. We can accomplish this using the `assign()`

function, which is equivalent to the `<-`

and `=`

operators:

assign('a', c(10, 150, 30, 45, 20.3)) a <- c(10, 150, 30, 45, 20.3) a = c(10, 150, 30, 45, 20.3)

The second statement (using the <`-`

operator) is the most common way of assigning in R, and we’ll therefore use this form rather than the `=`

operator or the `assign()`

function.

Once we have assigned a vector to a name, we can refer to the vector using this name. For example, if we type a, R will now show the elements of vector a.

```
a
```

## [1] 10.0 150.0 30.0 45.0 20.3

Instead of a, we could have chosen any other name, e.g.:

aVeryLongNameWhichIsCaseSensitive_AndDoesNotContainSpaces <- c(10, 150, 30, 45, 20.3)

Strictly speaking, we call this “name” an object.

To familiarize yourself with the vector data structure, now try to construct a couple of vectors in R and assign them to a named object, as in the example above.

**To summarize: A vector is a data structure, which can be constructed using the c() function, and assigned to a named object using the <- operator.**

Now, let’s move on to the first set of real exercises on vectors!

]]>[For this exercise, first write down your answer, without using R. Then, check your answer using R.]

Answers to the exercises are available here.

**Exercise 1**

Consider two vectors, `x, y`

`x=c(4,6,5,7,10,9,4,15)`

`y=c(0,10,1,8,2,3,4,1)`

What is the value of: `x*y`

**Exercise 2**

Consider two vectors, `a, b`

`a=c(1,2,4,5,6)`

`b=c(3,2,4,1,9)`

What is the value of: `cbind(a,b)`

**Exercise 3**

Consider two vectors, `a, b`

`a=c(1,5,4,3,6)`

`b=c(3,5,2,1,9)`

What is the value of: `a<=b`

**Exercise 4**

Consider two vectors, `a, b`

`a=c(10,2,4,15)`

`b=c(3,12,4,11)`

What is the value of: `rbind(a,b)`

**Exercise 5**

If `x=c(1:12)`

What is the value of: `dim(x)`

What is the value of: `length(x)`

**Exercise 6**

If `a=c(12:5)`

What is the value of: `is.numeric(a)`

**Exercise 7**

Consider two vectors, `x, y`

`x=c(12:4)`

`y=c(0,1,2,0,1,2,0,1,2)`

What is the value of: `which(!is.finite(x/y))`

**Exercise 8**

Consider two vectors, `x, y`

` x=letters[1:10]`

`y=letters[15:24]`

What is the value of: `x<y`

**Exercise 9**

If `x=c('blue','red','green','yellow')`

What is the value of: `is.character(x)`

**Exercise 10**

If `x=c('blue',10,'green',20)`

What is the value of: `is.character(x)`

**Want to practice vectors a bit more? We have more exercise sets on this topic here.**

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[For this exercise, first write down your answer, without using R. Then, check your answer using R.]

Answers to the exercises are available here.

**Exercise 1**

if `x <- c(a = 1, b = 2,c=3,d=4)`

What is the output for the code:

`seq(5,11,along.with =x)`

**Exercise 2**

If `x= seq(4,12,4)`

,

what is the output for the code:

`rep(x,each=2)`

**Exercise 3**

What is the output for the code:

`seq(5,11,by=2,length.out=3)`

**Exercise 4**

What is the output for the code:

`rep(letters[1:10],3)`

**Exercise 5**

Create a sequence with values:

`100 95 90 85 80 75 70 65 60 55 50`

**Exercise 6**

What is the output for the code:

`seq(10,0,by=5)`

**Exercise 7**

What is the output for the code:

`seq(2,10,by=4)==c(2,6,10)`

**Exercise 8**

What is the output for the code:

` rep(c('seq','rep'),each=4)`

**Exercise 9**

Consider two variables, A and B,

`A= as.Date("2016-11-01")`

`B = as.Date("2016-11-15")`

What is the output for the code:

`seq.Date(A,B, by = "1 day")`

**Exercise 10**

Consider two variables, C and D,

`C= as.Date("2016-02-01")`

`D = as.Date("2016-06-15")`

What is the output for the code:

`seq.Date(D,C, by = "-1 month")`

**Want to practice regular sequences a bit more? We have more exercise sets on this topic here.**

`sum`

, and `which`

functions.
Answers to the exercises are available here.

**Exercise 1**

If `x <- c("ww", "ee", "ff", "uu", "kk")`

, what will be the output for x[c(2,3)]?

a. "ee", "ff"

b. "ee"

c. "ff"

**Exercise 2**

If `x <- c("ss", "aa", "ff", "kk", "bb")`

, what will be the third value in the index vector operation `x[c(2, 4, 4)]`

?

a. "uu"

b. NA

c. "kk"

**Exercise 3**

If ` x <- c("pp", "aa", "gg", "kk", "bb")`

, what will be the fourth value in the index vector operation `x[-2]`

?

a. "aa"

b. "gg"

c. "bb"

**Exercise 4**

Let `a <- c(2, 4, 6, 8)`

and `b <- c(TRUE, FALSE, TRUE, FALSE)`

, what will be the output for the R expression `max(a[b])`

?

**Exercise 5**

Let `a <- c (3, 4, 7, 8)`

and `b <- c(TRUE, TRUE, FALSE, FALSE)`

, what will be the output for the R expression `sum(a[b])`

?

**Exercise 6**

Write an R expression that will return the sum value of 10 for the vector `x <- c(2, 1, 4, 2, 1, NA)`

**Exercise 7**

If `x <- c(1, 3, 5, 7, NA)`

write an r expression that will return the output 1, 3, 5, 7.

**Exercise 8**

Consider the data frame `s <- data.frame(first= as.factor(c("x", "y", "a", "b", "x", "z")), second=c(2, 4, 6, 8, 10, 12))`

. Write an R statement that will return the output 2, 4, 10, by using the variable `first`

as an index vector.

**Exercise 9**

What will be the output for the R expression `(c(FALSE, TRUE)) || (c(TRUE, TRUE))`

?

**Exercise 10**

Write an R expression that will return the positions of 3 and 7 in the vector `x <- c(1, 3, 6, 7, 3, 7, 8, 9, 3, 7, 2)`

.

`nchar`

, `substr`

and `sub`

functions.
Answers to the exercises are available here.

**Exercise 1**

If `x <- “Good Morning! “`

, find out the number of characters in X

a. 1

b. 14

c. 13

**Exercise 2**

Consider the character vector ` x <- c (“Nature’s”, “Best “)`

, how many characters are there in x?

a. 12

b. 13

c. 8,5

**Exercise 3**

If ` x <- c("Nature’s"," At its best ") `

, how many characters are there in x?

a. 19

b. 8, 13

c. 8, 9

**Exercise 4**

If ` fname <- “James“ `

and ` lname <- “Bond”`

, write some R code that will produce the output "James Bond".

**Exercise 5**

If `m <- “Capital of America is Washington” `

then extract the string “Capital of America” from the character vector m.

**Exercise 6**

Write some R code to replace the first occurrence of the word “failed” with “failure” in the string “Success is not final, failed is not fatal”.

**Exercise 7**

Consider two character vectors:

`Names <- c("John", "Andrew", "Thomas") `

and

`Designation <- c("Manager", "Project Head", "Marketing Head")`

.

Write some R code to obtain the following output.

Names Designation

1 John Manager

2 Andrew Project Head

3 Thomas Marketing Head

**Exercise 8**

Write some R code that will initialise a character vector with fixed length of 10.

**Exercise 9**

Write some R code that will generate a vector with the following elements, without using loops.

"aa" "ba" "ca" "da" "ea" "ab" "bb" "cb" "db" "eb" "ac" "bc" "cc" "dc" "ec"

"ad" "bd" "cd" "dd" "ed" "ae" "be" "ce" "de" "ee"

**Exercise 10**

Let ` df <- data.frame(Date = c("12/12/2000 12:11:10")) `

. Write some R code that will convert the given date to character values and gives the following output:

"2000-12-12 12:11:10 GMT"

Solutions are available here.

**Exercise 1**

If ` X <- c (22,3,7,NA,NA,67) `

what will be the output for the R statement ` length(X) `

**Exercise 2**

If ` X = c(NA,3,14,NA,33,17,NA,41) `

write some R code that will remove all occurrences of NA in X.

a. `X[!is.na(X)]`

b. `X[is.na(X)]`

c. `X[X==NA]= 0`

**Exercise 3**

If ` Y = c(1,3,12,NA,33,7,NA,21) `

what R statement will replace all occurrences of NA with 11?

a. `Y[Y==NA]= 11`

b. `Y[is.na(Y)]= 11`

c. `Y[Y==11] = NA`

**Exercise 4**

If ` X = c(34,33,65,37,89,NA,43,NA,11,NA,23,NA) `

then what will count the number of occurrences of NA in X?

a. `sum(X==NA)`

b. `sum(X == NA, is.na(X))`

c. `sum(is.na(X))`

**Exercise 5**

Consider the following vector ` W <- c (11, 3, 5, NA, 6) `

Write some R code that will return `TRUE`

for value of `W`

missing in the vector.

**Exercise 6**

Load ‘Orange’ dataset from R using the command ` data(Orange) `

. Replace all values of `age=118`

to NA.

**Exercise 7**

Consider the following vector ` A <- c (33, 21, 12, NA, 7, 8) `

.

Write some R code that will calculate the mean of A without the missing value.

**Exercise 8**

Let:

` c1 <- c(1,2,3,NA) `

;

` c2 <- c(2,4,6,89) `

;

` c3 <- c(45,NA,66,101) `

.

If ` X <- rbind (c1,c2,c3, deparse.level=1) `

, write a code that will display all rows with missing values.

**Exercise 9**

Consider the following data obtained from ` df <- data.frame (Name = c(NA, “Joseph”, “Martin”, NA, “Andrea”), Sales = c(15, 18, 21, 56, 60), Price = c(34, 52, 21, 44, 20), stringsAsFactors = FALSE) `

Write some R code that will return a data frame which removes all rows with NA values in `Name`

column

**Exercise 10**

Consider the following data obtained from ` df <- data.frame(Name = c(NA, “Joseph”, “Martin”, NA, “Andrea”), Sales = c(15, 18, 21, NA, 60), Price = c(34, 52, 33, 44, NA), stringsAsFactors = FALSE) `

Write some R code that will remove all rows with NA values and give the following output

Name Sales Price

2 Joseph 18 52

3 Martin 21 33

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`data <- mtcars`

Solutions are available here.

**Exercise 1**

Use logical operators to output only those rows of `data`

where column `mpg`

is between 15 and 20 (excluding 15 and 20).

**Exercise 2**

Use logical operators to output only those rows of `data`

where column `cyl`

is equal to 6 and column `am`

is not 0.

**Exercise 3**

Use logical operators to output only those rows of `data`

where column `gear`

or `carb`

has the value 4.

**Exercise 4**

Use logical operators to output only the even rows of `data`

.

**Exercise 5**

Use logical operators and change every fourth element in column `mpg`

to 0.

**Exercise 6**

Output only those rows of `data`

where columns `vs`

and `am`

have the same value 1, solve this without using `==`

operator.

**Exercise 7**

`(TRUE + TRUE) * FALSE`

, what does this expression evaluate to and why?

**Exercise 8**

Output only those rows of `data`

where at least `vs`

or `am`

have the value 1, solve this without using `==`

or `!=`

.

**Exercise 9**

Explain the difference between `|`

, `||`

, `&`

and `&&`

.

**Exercise 10**

Change all values that are 0 in the column `am`

in `data`

to 2.

**Exercise 11**

Add 2 to every element in the column `vs`

without using numbers.

**Exercise 12**

Output only those rows of `data`

where `vs`

and `am`

have different values, solve this without using `==`

or `!=`

.