✅ Week 01 Lab - Solutions

2023/24 Autumn Term

Author

Published

20 Jan 2024

This solution file follows the format of the Quarto file .qmd you had to fill in during the lab session. If you want to render the document yourselves and play with the code, you can download the .qmd version of this solution file by clicking on the button below:

📋 Lab Tasks

⚙️ Setup

We are using the pipe operator in this lab so we need to load the magrittr library. As explained before, we load all the libraries we’ll need in our code in the very first code chunk in our file.

library(magrittr) #we're using pipes in this lab so this is the one import we'll need.

(Part 3) Reviewing fundamental concepts of R/tidyverse

Vectors

Try the basic vector manipulations from the examples and check that you understand the results.


# This is a code chunk. You can use it to write and run R code.
# Once you render the document, the code will be executed and the output will be displayed.

#Creating vectors (chunk 1)

x <- c(22L,30L,42L) #vector of integers
y <- c(6.02214,3.14159,6.674) #vector of numerics
z <- c(FALSE, TRUE, FALSE) #vector of logicals
s <- c("Darth Vader", "Luke Skywalker", "Han Solo") #vector of characters

You can try and print out the actual content of your vectors, once you’ve created them:

print(x)
print(y)
print(z)
print(s)

This outputs the following response:

> print(x)
[1] 22 30 42
> print(y)
[1] 6.02214 3.14159 6.67400
> print(z)
[1] FALSE  TRUE FALSE
> print(s)
[1] "Darth Vader"    "Luke Skywalker" "Han Solo"

If you now run the chunk below (chunk 2)

#(chunk 2)
#run this chunk after chunk 1 (creation of vectors)
y <- sort(y) # operation to sort the content of y
v <- x^2/2 - y*3 + 5 
w <- mean(y) # computing the mean of y

then try to output the values of the variables you defined in (chunk 2):

print(y)
print(v)
print(w)

then this is the output you get:

> print(y)
[1] 3.14159 6.02214 6.67400
> print(v)
[1] 237.5752 436.9336 866.9780
> print(w)
[1] 5.279243

Running (chunk 3)

#(chunk 3)
a <- c(1:10) #a sequence of numbers from 1 to 10 with an increment of 1
b <- seq(-3,3,by=0.5) #a sequence of numbers from -3 to 3 with an increment of -0.5

then printing out the variables defined in it

print(a)
print(b)

results in the following output:

> print(a)
 [1]  1  2  3  4  5  6  7  8  9 10
> print(b)
 [1] -3.0 -2.5 -2.0 -1.5 -1.0 -0.5  0.0  0.5  1.0  1.5  2.0  2.5  3.0

Running (chunk 4):

# (chunk 4)
# run this chunk after (chunk 3)
d <- rep(a, times=2)
e <- rep(a, each=2)

then printing out the variables defined in it:

print(d)
print(e)

results in the following output:

> print(d)
 [1]  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10
print(e)
> print(e)
 [1]  1  1  2  2  3  3  4  4  5  5  6  6  7  7  8  8  9  9 10 10

The difference between using the times keyword with the rep function ,as opposed to the each keyword, is that the full vector is repeated (consecutively) as many times specified by the times keyword whereas with each, it’s each vector element that is replicated consecutively as many times as specified by the keyword.

Finally, running (chunk 5)

# (chunk 5)
s[2]
s[c(1,3)]
s[-3]
s[3] <- "Princess Leia"
s <- append(s,c("Han Solo","Yoda"))

then printing out the result of the chunk:

print(s)

results in this output:

> print(s)
[1] "Darth Vader"    "Luke Skywalker" "Princess Leia"  "Han Solo"      
[5] "Yoda"

Could you create three vectors:

a vector k which is a sequence from 8 to 1 (i.e descending order) in increments of 2
a vector m which contains the content of k repeated three times
a vector n containing three copies of k, with a 0 separating each copy from the next one?

k <- seq(8,1,by=-2)

Here’s what you get if you print out k:

print(k)

m <- rep(k,times=3) #you make use of the `rep` function here with keyword `times`

A simple solution uses already defined vector m then modifies it.

m_onezero <- append(m,0,after=4)
n <- append(m_onezero,0,after=9)

We could also have written the previous solution as:

n <- append(append(m,0,after=4),0,after=9)

However, this nested function call is less legible.

You could also use the pipe operator (%>%) here (after loading the right library!):

n <- m %>%
     append(0,after=4) %>%
     append(0,after=9)

(As before, use print(n) to check the result and see that we get the same result each time)

Can you check whether m is equal to n?

m==n

Note that you should also obtain a warning message because the 2 vectors are not of the same length. How can you check the length of both vectors (i.e m and n)?

length(m)
length(n)

What will happen in the examples from the code chunk below? Try to guess and confirm your guess by running the code.

   num_char <- c(1, 2, 3, "a")
   num_logical <- c(1, 2, 3, TRUE)
   char_logical <- c("a", "b", "c", TRUE)
   tricky <- c(1, 2, 3, "4")

Vectors can be of only one data type. R tries to convert (coerce) the content of this vector to find a “common denominator” that doesn’t lose any information. Therefore, elements in num_char, char_logical and tricky are coerced to type character, while elements in num_logical are coerced to type double.

6.Suppose the following vector represents flu cases over a given number of weeks:

   flu_cases <- c(NA, 1, 0, 0, 3, NA, 3, 3, 61, 411, 2570,7208)

Could you remove the missing values from the vector?

flu_cases <- na.omit(flu_cases)

How many weeks were the number of cases over 10?

sum(flu_cases>10)

Could you compute the mean and standard deviation associated with the number of flu cases?

mean(flu_cases)
sd(flu_cases)

Was there a week where the number of cases was equal to 42? And if so, which was it? Same questions with value 3.

42 %in% flu_cases # returns FALSE, which means that no week had 42 cases, so no need to find which week had 42 cases
3 %in% flu_cases # returns TRUE, which means that at least one week had 3 cases, so we need to find which week(s) had 3 cases
which(flu_cases==3) # finds all indices in vector flu_cases that satisfy the condition, in this case, we find all the indices, i.e weeks, where the value of the vector, i.e the number of cases, is equal to 3

Chaining operations with the pipe operator (`%>%`)

The pipe operator, %>%, comes from the magrittr package by Stefan Milton Bache. Packages in tidyverse load %>% automatically, so you don’t usually load magrittr explicitly.

This means that, before we are able to use the pipe operator, we either have to load the magrittr library or the tidyverse library (we would usually load tidyverse if we are going to use other tidyverse functions) after having made sure we have magrittr and/or tidyverse installed.

Note on installing and loading packages in R

How do I install a package in R e.g tidyverse?

Just run install.packages("tidyverse") in your R console. This will install all the packages that are part of the tidyverse ecosystem.

DO NOT leave an install.packages() command in your .qmd file. Always do this in the R Console. Otherwise, you won’t be able to render your markdown file as an HTML later.

How do I load a package/library that I have already installed e.g magrittr?

To load magrittr, you simply need to execute the call library(magrittr) before you use any function that comes from this package (e.g pipe).

However, in your .qmd file, make it a habit to create and dedicate your first chunk of code to loading all the packages you’ll use using the library() function. While it’s true that you may not know all the packages you will need when starting a new file, making the first chunk a reserved space for the task of loading libraries creates a neat, centralised hub. In this lab, since we simply need the pipe operator for now, we would simply load magrittr with library(magrittr) in that first chunk of code: we can circle back to it if we need to add further libraries. You have to rerun the chunk for the packages to be loaded, every time you add a new library() call to the chunk.

It allows you to chain together sequences of operations and has four main advantages:

you structure the sequence of your data operations from left to right, as apposed to from inside and out;
you avoid nested function calls (that was one of the possible solutions to create vector n in question 2 in Section 1.2.1);
you minimize the need for local variables and function definitions
you make it easy to add steps anywhere in the sequence of operations

What does it mean in practice?

Let’s go back to the flu cases example.

Suppose we want to only look at the first 9 weeks of data, replace the missing values in this subset of data with value 0 and then count how many weeks within the period the number of cases was equal to 0.

We could write the sequence of operations as follows:


flu_cases <- c(NA, 1, 0, 0, 3, NA, 3, 3, 61, 411, 2570, 7208)
flu_cases <- flu_cases[1:9]
flu_cases <- replace(flu_cases,is.na(flu_cases),0)
zero_cases <- length(subset(flu_cases,flu_cases==0))

Notice the number of assignments and nested function calls.

If we print out, the variables from the previous block, here’s what we’d see:

print(flu_cases)
print(zero_cases)

Alternatively, we could rewrite the sequence with the pipe (%>%) operator:

zero_cases <- c(NA, 1, 0, 0, 3, NA, 3, 3, 61, 411, 2570, 7208) %>% 
   .[1:9] %>% #in this line and the following . replaces the vector c(NA, 1, 0, 0, 3, NA, 3, 3, 61, 411, 2570, 7208) i.e flu_cases
   replace(is.na(.),0) %>%
   subset(.==0) %>%
   length()

The result of this block is the same as previously, but we’ve rewritten the block in a more legible, modular and ste-by-step fashion.

In this case, each line performs an operation from the sequence : the pipe is essentially equivalent to an English language ‘then’; define this vector then take its first nine values then replace the missing values within it with the value 0 then take the subset of this vector where the value is equal to 0 then take the length of the resulting vector (i.e count the number of weeks within the first nine weeks where the number of cases is 0).

For details on pipes, have a look at this tutorial.

🎯 ACTION POINTS

Can you re-create vector n from question 2 of Section 1.2.1 using the pipe operator?

n <- m %>%
     append(0,after=4) %>%
     append(0,after=9)

Use the sample function to create two vectors (you can choose the vector length) whose values are in the range [7-42]. Append both vectors together, then scale the resulting vector before only selecting negative values and getting a count of negative values. Use the pipe operator to write your sequence of operations.

#write your answer here
vector_range <- 7:42 #specifying the value range of both vectors we'll create
l1 <- 10 #length of the first vector
l2 <- 23 #length of the second vector
v1 <- sample(vector_range,l1) #creating a first vector whose values are in the range [7-42] and of length l1, i.e 10 here
v2 <- sample(vector_range,l2) #creating a second vector whose values are in the range [7-42] and of length l2, i.e 23 here
neg_count <- append(v1,v2) %>% # line that appends both created vectors together
             scale() %>% #scaling the values of the resulting vector
             .[,1] %>%   # since the `scale` function also returns scaling attributes, this line is necessary to only extract scaled values
             subset(.<0) %>% #finding the subset of values that are negative
             length() #counting the number of negative values i.e taking the length of the vector obtained in the previous step

`For` loops

Suppose that, for some reasons, you want to print out sentences of the form: “The year is [year]”” where [year] is equal to 2019, 2020, up to 2024. You can do this as follows:

print(paste("The year is", 2019))
print(paste("The year is", 2020))
print(paste("The year is", 2021))
print(paste("The year is",2022))
print(paste("The year is", 2023))
print(paste("The year is", 2024))

Running this code block gives you a result that looks like this:

> print(paste("The year is", 2019))
[1] "The year is 2019"
print(paste("The year is", 2020))
> print(paste("The year is", 2020))
[1] "The year is 2020"
> print(paste("The year is", 2021))
[1] "The year is 2021"
> print(paste("The year is",2022))
[1] "The year is 2022"
> print(paste("The year is", 2023))
[1] "The year is 2023"
> print(paste("The year is", 2024))
[1] "The year is 2024"

As you quickly see, this is rather tedious since you copy the same code chunk over and over again. Rather than doing this, you could use a for loop to write repetitive parts of code.

Using a for loop, the code above transforms into:

for (year in 2019:2024){
  print(paste("The year is", year))
}

Running this code block gives the following result:

> for (year in 2019:2024){
+   print(paste("The year is", year))
+ }
[1] "The year is 2019"
[1] "The year is 2020"
[1] "The year is 2021"
[1] "The year is 2022"
[1] "The year is 2023"
[1] "The year is 2024"

The best way to understand this loop is as follows: “For each year that is in the sequence 2019:2024, you execute the code chunk print(paste("The year is", year))”. Once the for loop has executed the code chunk for every year in the vector (i.e sequence 2019:2024), the loop stops and goes to the first instruction after the loop block.

🎯 ACTION POINTS

Suppose you have a new vector of characters:

t <- c("R2-D2","Chewbacca","Obi-Wan Kenobi")

Could you write code that creates a vector that contains all the Star Wars character names, then, for each name, counts the number of characters, then finds the Star Wars character with longest name? Use a for loop in your code.

#write your answer here
#Let's start with vector `s` that was defined in @sec-vectors

s <- c("Darth Vader", "Luke Skywalker", "Han Solo")
# we modify `s` as shown in @sec-vectors to include all possible known Starwars characters

s[3] <- "Princess Leia"
s <- append(s,c("Han Solo","Yoda"))

#we append `s` and `t` in a single vector that includes all (known) Starwars character names
char_names <- append(s,t)

#we now write our `for` loop

vec_chars <- c() #we need a variable (vector) that stores the number of characters per Starwars character name

# now we start the loop
for (i in char_names){
    vec_chars[i]=nchar(i)
}

#finding the Starwars character with the longest name

which(vec_chars==max(vec_chars)) #this returns two values: "Luke Skywalker" (with its index in the vector i.e 2) and "Obi-Wan Kenobi" (with its index in the vector i.e 8)

Could you write code that creates a vector that contains all the Star Wars character names, then, for each name, counts the number of characters and prints out a line of the form “The character name [name of character] is composed of [x] characters” (e.g “The character name Yoda is composed of 4 characters”)?

 #write your answer here

 #We start as we did before with vectors `s` and `t`
 t <- c("R2-D2","Chewbacca","Obi-Wan Kenobi")

 s <- c("Darth Vader", "Luke Skywalker", "Han Solo")
 # we modify `s` as shown in @sec-vectors to include all possible known Starwars characters
 s[3] <- "Princess Leia"
 s <- append(s,c("Han Solo","Yoda"))

 #we append `s` and `t` in a single vector that includes all (known) Starwars character names
 char_names <- append(s,t)

 # we write our `for` loop

 for (i in char_names){
    n_chars <- nchar(i) 
    print(paste("The character name",i,"is composed of",n_chars,"characters"))
 }

Functions

In this lab, we’ve encountered and used quite a few different pre-made functions in (e.g c(), replace(), subset(), sample(), rep(), length()), but sometimes you just need to write your own function to tackle your data, i.e your set/succession of (reproducible) instructions.

A function is simply a code block that performs a specific task (which can be more or less complex), e.g as calculating a sum.

You should think of writing a function whenever you’ve copied and pasted a block of code more than twice.

In R, functions are of the form:

name_of_the_function <- function(arguments) {
function_content
}

You give your function a (meaningful) name (name_of_the_function ), define your function arguments (arguments ) i.e the parameters it needs to perform the task it supposed to perform, and put some content in the function. You define how the function should deal with the input/arguments to perform the task it needs to perform in function_content.

Here’s, as an example, a very simple function to sum two numbers:

sum_twonumbers <- function(number1, number2){
     result <- number1 + number2
     return(result)
     }

As expected, you can invoke this function multiple times with different parameters and get different sum results e.g:

sum_twonumbers(45,6)
sum_twonumbers(1007888,177999)
sum_twonumbers(42,1337)

The result of running the code block above looks like this:

> sum_twonumbers(45,6)
[1] 51
> sum_twonumbers(1007888,177999)
[1] 1185887
> sum_twonumbers(42,1337)
[1] 1379

🎯 ACTION POINTS

Could you write a function that takes a vector of characters as an input, counts the number of characters for each element of the vector and and prints out a line of the form “The character name [name of character] is composed of [x] characters” (e.g “The character name Yoda is composed of 4 characters”)?

#write your answer here
# It's basically the same code as in the second question of Question 9 (`for` loop section) but wrapped in a function call!
# We pass the function an argument `vec_characters` which is a vector of characters
# The rest of the code is unchanged from Question 9 :)
print_vecElements_charCount <- function(vec_characters) {
   for (i in vec_characters){
    n_chars <- nchar(i) 
    print(paste("The character name",i,"is composed of",n_chars,"characters"))
    }
}

Write both_na(), a function that takes two vectors of the same length and returns the number of positions that have an NA in both vectors (hint: create example vectors with NA in them to test your function).

#write your answer here
#One possible solution is the following code:
both_na <- function(x, y) {
  sum(is.na(x) & is.na(y))
}
# Note that we could still modify the function to include conditions to verify the vectors are of the same length and how to handle cases where they're not. But we leave that as extra exercise :)

We can now test our function:

both_na(
  c(NA, NA, 1, 2, NA, NA, 1),
  c(NA, 1, NA, 2, NA, NA, 1)
)

This should return this printout:

> both_na(
+   c(NA, NA, 1, 2, NA, NA, 1),
+   c(NA, 1, NA, 2, NA, NA, 1)
+ )
[1] 3

(i.e there are 3 positions where both vectors tested have the value NA at the same position)

Running the code block below:

both_na(
  c(NA, 1, 42, NA, 1, NA),
  c(0 , 1, 42, 2, NA, 6 )
)

returns this:

> both_na(
+   c(NA, 1, 42, NA, 1, NA),
+   c(0 , 1, 42, 2, NA, 6 )
+ )
[1] 0

i.e there are no positions where both vectors tested both have a value NA at the same time.

Transforming `for` loops with `sapply`

for loops are all well and good and they are rather convenient (and quite easy to grasp and write!). But they’re not exactly the most efficient solution (computationally) when it comes to executing repetitive pieces of code. R supports vectorization and vectorized solutions that make use of apply functions, such as lapply and sapply (it’s often better to use sapply as it outputs a vector and is slightly more efficient) are more efficient than solutions that use loops, in particular for loops.

🎯 ACTION POINTS

Take the for loops from question 9 and see if you can rewrite them using the sapply function. Consult the R documentation to help you with your task or have a look at this link.

The first for loop in Question 9 can simply be rewritten, as follows, when using sapply

  
  vec_chars <- sapply(append(s,t), function(i) nchar(i)) #sapply returns a vector so we can assign its result directly to a vector named vec_chars

If we print out the vector vec_chars from the previous code block:

print(vec_chars)

the result is the same as the for loop from Question 9:

> print(vec_chars)
   Darth Vader Luke Skywalker  Princess Leia       Han Solo           Yoda 
            11             14             13              8              4 
          R2-D2      Chewbacca Obi-Wan Kenobi 
             5              9             14

How about the second for loop? Remember how, in Question 10, we wrote a function that did the same thing as the second for loop of Question 9? We’ll use this function to do our rewritting.

sapply(append(s,t), function(i) print_vecElements_charCount(i))

which returns the following:

> sapply(append(s,t), function(i) print_vecElements_charCount(i))
[1] "The character name Darth Vader is composed of 11 characters"
[1] "The character name Luke Skywalker is composed of 14 characters"
[1] "The character name Princess Leia is composed of 13 characters"
[1] "The character name Han Solo is composed of 8 characters"
[1] "The character name Yoda is composed of 4 characters"
[1] "The character name R2-D2 is composed of 5 characters"
[1] "The character name Chewbacca is composed of 9 characters"
[1] "The character name Obi-Wan Kenobi is composed of 14 characters"
$`Darth Vader`
NULL

$`Luke Skywalker`
NULL

$`Princess Leia`
NULL

$`Han Solo`
NULL

$Yoda
NULL

$`R2-D2`
NULL

$Chewbacca
NULL

$`Obi-Wan Kenobi`
NULL

Note how sapply assigns a return value to each element of the vector it is applied on and since the function we specified does not have a return value (it only prints out a string but its return value is NULL), it assigns a value of NULL to each element.

📋 Lab Tasks

⚙️ Setup

(Part 3) Reviewing fundamental concepts of R/tidyverse

Vectors

Chaining operations with the pipe operator (%>%)

For loops

Functions

Transforming for loops with sapply

Chaining operations with the pipe operator (`%>%`)

`For` loops

Transforming `for` loops with `sapply`