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Basics of R Programming



Thiyanga S. Talagala, University of Sri Jayewardenepura

IASSL - Feb 21/25, 2022

1

Data structures

Way to store and organize data so that it can be used efficiently.

marks <- c(100, 40, 34, 97, 98)
marks
[1] 100  40  34  97  98
2

Data structures

Way to store and organize data so that it can be used efficiently.

marks <- c(100, 40, 34, 97, 98)
marks
[1] 100  40  34  97  98

Functions

Tell R to do something

mean(marks)
[1] 73.8
summary(marks)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   34.0    40.0    97.0    73.8    98.0   100.0
3

Data structures

Source: Ceballos and Cardiel, 2013

4

Creating vectors

Syntax

vector_name <- c(element1, element2, element3)

Example

x <- c(5, 6, 3, 1, 100)
x
[1]   5   6   3   1 100
5

Combine two vectors

p <- c(1, 2, 3)
p
[1] 1 2 3
q <- c(10, 20, 30)
q
[1] 10 20 30
r <- c(p, q)
r
[1]  1  2  3 10 20 30
6

Vector with charactor elements

names <- c("USJ", "UM", "UC", "UJ")
names
[1] "USJ" "UM"  "UC"  "UJ"

Logical vector

result <- c(TRUE, FALSE, FALSE, TRUE, FALSE)
result
[1]  TRUE FALSE FALSE  TRUE FALSE
7

Simplifying vector creation

id <- 1:10
id
 [1]  1  2  3  4  5  6  7  8  9 10
treatment <- rep(1:3, each=2)
treatment
[1] 1 1 2 2 3 3

Additional resources: https://hellor.netlify.app/2021/week1/l12021.html#62

8

Vector operations

x <- c(1, 2, 3)
y <- c(10, 20, 30)
x+y
[1] 11 22 33
p <- c(100, 1000)
x+p
[1]  101 1002  103
9

Your turn

10

  1. Generate a sequence using the code seq(from=1, to=10, by=1).

  2. What other ways can you generate the same sequence?

  3. Using the function rep , create the below sequence 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4

03:00
11

Vectors: Subsetting

myvec <- 1:20; myvec
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
12

Vectors: Subsetting

myvec <- 1:20; myvec
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
myvec[1]
[1] 1
13

Vectors: Subsetting

myvec <- 1:20; myvec
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
myvec[1]
[1] 1
myvec[5:10]
[1]  5  6  7  8  9 10
14

Vectors: Subsetting (cont.)

15

Vectors: Subsetting (cont.)

myvec[-1]
 [1]  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
16

Vectors: Subsetting (cont.)

myvec[-1]
 [1]  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
myvec[myvec > 3]
 [1]  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
17

Changing values of a vector

covid <- c(100, 30, 40, 50, -1, 100)
covid
[1] 100  30  40  50  -1 100
covid[1] <- 50000
covid
[1] 50000    30    40    50    -1   100
18

Changing values of a vector (cont.)

covid[covid < 0] <- 0
covid
[1] 50000    30    40    50     0   100
covid[c(1, 2)] <- c(1000, 10000)
covid
[1]  1000 10000    40    50     0   100
19

factor

20

Required R package

library(tidyverse)
── Attaching packages ─────────────────────────────────────── tidyverse 1.3.1 ──
✓ ggplot2 3.3.5          ✓ purrr   0.3.4     
✓ tibble  3.1.6          ✓ dplyr   1.0.8.9000
✓ tidyr   1.2.0          ✓ stringr 1.4.0     
✓ readr   2.1.2          ✓ forcats 0.5.1
Warning: package 'tidyr' was built under R version 4.1.2
Warning: package 'readr' was built under R version 4.1.2
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
x dplyr::filter() masks stats::filter()
x dplyr::lag()    masks stats::lag()
21

Character vector vs Factor

  • Factor prints all possible levels of the variable.

Character vector

grade_character_vctr <- c("A", "D", "A", "C", "B")
grade_character_vctr
[1] "A" "D" "A" "C" "B"

Factor vector

grade_factor_vctr <- factor(c("A", "D", "A", "C", "B"), levels = c("A", "B", "C", "D", "E"))
grade_factor_vctr
[1] A D A C B
Levels: A B C D E
22

Character vector vs Factor (cont.)

  • Let's create a contingency table with table function.

Character vector output with table function

grade_character_vctr <- c("A", "D", "A", "C", "B")
table(grade_character_vctr)
grade_character_vctr
A B C D 
2 1 1 1
23

Factor vector (with levels) output with table function

grade_factor_vctr <- 
  factor(c("A", "D", "A", "C", "B"), 
         levels = c("A", "B", "C", "D", "E"))
table(grade_factor_vctr)
grade_factor_vctr
A B C D E 
2 1 1 1 0
  • Output corresponds to factor prints counts for all possible levels of the variable. Hence, with factors it is obvious when some levels contain no observations.
24

Character vector vs Factor (cont.)

  • With factors you can't use values that are not listed in the levels, but with character vectors there is no such restrictions.

Character vector

grade_character_vctr[2] <- "A+"
grade_character_vctr
[1] "A"  "A+" "A"  "C"  "B"
25

Factor vector

grade_factor_vctr[2] <- "A+"
grade_factor_vctr
[1] A    <NA> A    C    B   
Levels: A B C D E
26

Factor: order levels

fv2 <- factor(c("1T","2T","3A","4A", "5A", "6B", "3A"))
fv2
[1] 1T 2T 3A 4A 5A 6B 3A
Levels: 1T 2T 3A 4A 5A 6B
27

Factor: order levels

fv2 <- factor(c("1T","2T","3A","4A", "5A", "6B", "3A"))
fv2
[1] 1T 2T 3A 4A 5A 6B 3A
Levels: 1T 2T 3A 4A 5A 6B
library(ggplot2)
qplot(fv2, geom = "bar")

28

You can change the order of levels

fv2 <- factor(c("1T","2T","3A","4A", "5A", "6B", "3A"), 
              levels = c("3A", "4A", "5A", "6B", "1T", "2T"))
fv2
[1] 1T 2T 3A 4A 5A 6B 3A
Levels: 3A 4A 5A 6B 1T 2T
qplot(fv2, geom = "bar")

29

Data set

30

Required R package

library(tidyverse)
31

Create a tibble

marks <- c(90, 50, 20, 60)
grade <- factor(c("A+", "C", "E", "B"))
final <- tibble(Marks = marks, Grade = grade)
final
# A tibble: 4 × 2
  Marks Grade
  <dbl> <fct>
1    90 A+   
2    50 C    
3    20 E    
4    60 B
32

Create a tibble

marks <- c(90, 50, 20, 60)
grade <- factor(c("A+", "C", "E", "B"),
                 level = c("A+", "A", "B+", "B", "C", "D", "E"))
final <- tibble(Marks = marks, Grade = grade)
final
# A tibble: 4 × 2
  Marks Grade
  <dbl> <fct>
1    90 A+   
2    50 C    
3    20 E    
4    60 B
33

Functions in R

34

Data set: tibble

final
# A tibble: 4 × 2
  Marks Grade
  <dbl> <fct>
1    90 A+   
2    50 C    
3    20 E    
4    60 B

Functions

summary(final)
     Marks      Grade 
 Min.   :20.0   A+:1  
 1st Qu.:42.5   A :0  
 Median :55.0   B+:0  
 Mean   :55.0   B :1  
 3rd Qu.:67.5   C :1  
 Max.   :90.0   D :0  
                E :1
35

Your Turn

01:00
36

h <- c(100, 101, 102, 150, NA)
w <- c(50, 60, 80, 43, 50)
hwdata <- tibble(Height=h, Weight=w)
hwdata
# A tibble: 5 × 2
  Height Weight
   <dbl>  <dbl>
1    100     50
2    101     60
3    102     80
4    150     43
5     NA     50
37
hwdata
# A tibble: 5 × 2
  Height Weight
   <dbl>  <dbl>
1    100     50
2    101     60
3    102     80
4    150     43
5     NA     50
summary(hwdata)
     Height          Weight    
 Min.   :100.0   Min.   :43.0  
 1st Qu.:100.8   1st Qu.:50.0  
 Median :101.5   Median :50.0  
 Mean   :113.2   Mean   :56.6  
 3rd Qu.:114.0   3rd Qu.:60.0  
 Max.   :150.0   Max.   :80.0  
 NA's   :1
38

Subsetting

hwdata
# A tibble: 5 × 2
  Height Weight
   <dbl>  <dbl>
1    100     50
2    101     60
3    102     80
4    150     43
5     NA     50
hwdata[1, 1]
# A tibble: 1 × 1
  Height
   <dbl>
1    100
hwdata[, 1]
# A tibble: 5 × 1
  Height
   <dbl>
1    100
2    101
3    102
4    150
5     NA
hwdata[1, ]
# A tibble: 1 × 2
  Height Weight
   <dbl>  <dbl>
1    100     50
hwdata$Height
[1] 100 101 102 150  NA
39

Help file

hwdata$Weight
[1] 50 60 80 43 50
mean(hwdata$Weight)
[1] 56.6
hwdata$Height
[1] 100 101 102 150  NA
mean(hwdata$Height)
[1] NA
40

Help file

hwdata$Weight
[1] 50 60 80 43 50
mean(hwdata$Weight)
[1] 56.6
hwdata$Height
[1] 100 101 102 150  NA
mean(hwdata$Height)
[1] NA
mean(hwdata$Height, na.rm=TRUE)
[1] 113.25
41

Help file

?mean
help(mean)

42

Commenting

mean(hwdata$Height, na.rm=TRUE) # compute mean of height
[1] 113.25
43

Some useful functions

mean(hwdata$Weight)
[1] 56.6
median(hwdata$Weight)
[1] 50
sd(hwdata$Weight)
[1] 14.41527
sum(hwdata$Weight)
[1] 283
length(hwdata$Weight)
[1] 5
44

Pipe operator (%>%)

mean(hwdata$Weight)
[1] 56.6
mean(hwdata$Height, na.rm=TRUE)
[1] 113.25
library(magrittr)
hwdata$Weight %>% mean()
[1] 56.6
hwdata$Height %>% mean(na.rm=TRUE)
[1] 113.25
45

Pipe operator (%>%)

46

Built-in dataset

library(palmerpenguins)
data(penguins)
head(penguins)
# A tibble: 6 × 8
  species island bill_length_mm bill_depth_mm flipper_length_… body_mass_g sex  
  <fct>   <fct>           <dbl>         <dbl>            <int>       <int> <fct>
1 Adelie  Torge…           39.1          18.7              181        3750 male 
2 Adelie  Torge…           39.5          17.4              186        3800 fema…
3 Adelie  Torge…           40.3          18                195        3250 fema…
4 Adelie  Torge…           NA            NA                 NA          NA <NA> 
5 Adelie  Torge…           36.7          19.3              193        3450 fema…
6 Adelie  Torge…           39.3          20.6              190        3650 male 
# … with 1 more variable: year <int>
47

Skim data

library(skimr)
skim(penguins)
48

iris dataset

49

50

Use the R dataset “iris” to answer the following questions:

  1. How many rows and columns does iris have?

  2. Select the first 4 rows.

  3. Select the last 6 rows.

  4. Select rows 10 to 20, with all columns in the iris dataset.

  5. Select rows 10 to 20 with only the Species, Petal.Width and Petal.Length.

  6. Create a single vector (a new object) called ‘width’ that is the Sepal.Width column of iris.

  7. What are the column names and data types of the different columns in iris?

  8. How many rows in the iris dataset have Petal.Length larger than 5 and Sepal.Width smaller than 3?

05:00
51

Recap

✅ Data structures and functions

✅ Factors

✅ Working with packages

✅ Create a tibble

✅ Help file

✅ Commenting

52

Data structures

Way to store and organize data so that it can be used efficiently.

marks <- c(100, 40, 34, 97, 98)
marks
[1] 100  40  34  97  98
2
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