Recap before the exam

From objects to dataframes

Objects

Creating and naming objects

name <- value: <- creates objects

age <- 25

age <- age + 1          # overwrite

age_months <- age * 12  # use it to create a new object

age_months

[1] 312

• names start with a letter, no spaces: my_data, x1 (not 1x, my data)
• R is case-sensitive: Name and name are two different objects
• rm(x) removes one object, rm(list = ls()) removes all

Data types

class(3.14)      # numeric

[1] "numeric"

class(5L)        # integer

[1] "integer"

class("hello")   # character: always in quotes

[1] "character"

class(TRUE)      # logical

[1] "logical"

class("5")       # in quotes: character

[1] "character"

Operators

Mathematical operators and functions

operator	meaning	example	result
+ - * /	arithmetic	6 * 2 / 4	3
^	power	2 ^ 3	8
%%	remainder of the division	9 %% 2	1

(1 + 2) * 3                  # parentheses first

[1] 9

sqrt(16); abs(-7); round(3.14159, 2)

[1] 4

[1] 7

[1] 3.14

Relational and logical operators

x <- 30

x == 30; x != 30; x >= 18     # compare: TRUE or FALSE

[1] TRUE

[1] FALSE

[1] TRUE

x %in% c(10, 20, 30)          # is x one of these values?

[1] TRUE

x > 25 & x < 60               # AND: both TRUE

[1] TRUE

x < 18 | x > 60               # OR: at least one TRUE

[1] FALSE

!(x > 25)                     # NOT

[1] FALSE

TRUE and FALSE are numbers

TRUE counts as 1, FALSE as 0: sum() counts, mean() gives the proportion

scores <- c(18, 25, 30, 22, 27)

scores >= 25

[1] FALSE  TRUE  TRUE FALSE  TRUE

sum(scores >= 25)       # how many?

[1] 3

mean(scores >= 25)      # which proportion?

[1] 0.6

Vectors

Creating vectors

One dimension, all elements of the same type

num1 <- c(4, 8, 15); num1

[1]  4  8 15

num2 <- 1:5; num2

[1] 1 2 3 4 5

num3 <- seq(from = 0, to = 10, by = 2); num3

[1]  0  2  4  6  8 10

ch1 <- rep(c("x", "y"), times = 2); ch1   # whole vector

[1] "x" "y" "x" "y"

ch2 <- rep(c("x", "y"), each = 2); ch2    # each element

[1] "x" "x" "y" "y"

ch3 <- rep(c("x", "y"), each = 2, times = 2); ch3

[1] "x" "x" "y" "y" "x" "x" "y" "y"

Types: is. and as.

Mixing types converts everything to one type:

c(1, 2, "three")                     # all character

[1] "1"     "2"     "three"

c(1, 2, NA, 4)                       # NA keeps the type

[1]  1  2 NA  4

is. tests the type, as. converts it:

is.numeric(c(1, 0, 3))

[1] TRUE

as.logical(c(1, 0, 3))               # 0 is FALSE, others TRUE

[1]  TRUE FALSE  TRUE

as.numeric(c("1", "2", "three"))     # cannot convert: NA

Warning: NAs introduced by coercion

[1]  1  2 NA

Operations and missing values

Operations work element by element, summary functions return one value

temps <- c(12, 18, NA, 31, 7)

temps * 2

[1] 24 36 NA 62 14

mean(temps)                  # one NA makes the result NA

[1] NA

mean(temps, na.rm = TRUE)    # remove the NAs first

[1] 17

is.na(temps)

[1] FALSE FALSE  TRUE FALSE FALSE

sum(is.na(temps))            # how many NAs?

[1] 1

Indexing by position

temps <- c(12, 18, 25, 31, 7)

temps[2]                  # the second element

[1] 18

temps[c(1, 3)]            # the first and the third

[1] 12 25

temps[2:4]                # from the second to the fourth

[1] 18 25 31

temps[length(temps)]      # the last one

[1] 7

temps[-1]                 # all except the first

[1] 18 25 31  7

Indexing: logical conditions and which()

temps

[1] 12 18 25 31  7

temps > 20                        # the condition

[1] FALSE FALSE  TRUE  TRUE FALSE

temps[temps > 20]                 # keep where TRUE

[1] 25 31

temps[temps > 10 & temps < 30]

[1] 12 18 25

which(temps > 20)                 # the positions

[1] 3 4

Modifying with indexing

Indexing + <- modifies only the selected elements:

scores <- c(18, NA, 30, 22, NA)

scores[1] <- 20                   # by position

scores[is.na(scores)] <- 0        # by condition

scores[scores < 21] <- 21

scores

[1] 21 21 30 22 21

Factors

Factors: codes + levels

group <- factor(x = c("placebo", "caffeine", "nicotine",
                  "placebo", "caffeine"))

group

[1] placebo  caffeine nicotine placebo  caffeine
Levels: caffeine nicotine placebo

levels(group)

[1] "caffeine" "nicotine" "placebo" 

as.integer(group)  # caffeine=1, nicotine=2, placebo=3

[1] 3 1 2 3 1

table(group)

group
caffeine nicotine  placebo 
       2        1        2 

Levels stay, even when unused

no_placebo <- group[group != "placebo"]
no_placebo # "placebo" is still a level

[1] caffeine nicotine caffeine
Levels: caffeine nicotine placebo

no_placebo <- droplevels(no_placebo)

levels(no_placebo)

[1] "caffeine" "nicotine"

Adding a new level

To use a new value, first add it to the levels

group

[1] placebo  caffeine nicotine placebo  caffeine
Levels: caffeine nicotine placebo

levels(group) <- c(levels(group), "water")

group[1] <- "water"
group

[1] water    caffeine nicotine placebo  caffeine
Levels: caffeine nicotine placebo water

Reorder vs rename

group <- factor(x = c("placebo", "caffeine", "nicotine",
                  "placebo", "caffeine"))

# REORDER: placebo first (reference category), same data
group_ord <- factor( x = c("placebo", "caffeine", "nicotine",
                  "placebo", "caffeine"),
  levels = c("placebo", "caffeine", "nicotine"))

levels(group_ord)

[1] "placebo"  "caffeine" "nicotine"

# RENAME: by position, check levels() first!
levels(group_ord) <- c("PLA", "CAF", "NIC")

group_ord

[1] PLA CAF NIC PLA CAF
Levels: PLA CAF NIC

Reorder and rename in one step

factor() with both arguments:

• levels: the existing values, in the order we want
• labels: their new names, in the same order

group_new <- factor(x = c("placebo", "caffeine", "nicotine",
                  "placebo", "caffeine"),
  levels = c("placebo", "caffeine", "nicotine"),
  labels = c("PLA", "CAF", "NIC"))

group_new

[1] PLA CAF NIC PLA CAF
Levels: PLA CAF NIC

levels(group_new)

[1] "PLA" "CAF" "NIC"

Ordered factors and numbers

For ordinal categories (< and > make sense), set the order and ordered = TRUE

freq <- factor(x = c("often", "never", "sometimes"),
  levels = c("never", "sometimes", "often"), ordered = TRUE)

freq > "never"

[1]  TRUE FALSE  TRUE

To get back numbers saved as a factor: first to text, then to numbers

points_f <- factor(c(10, 30, 10, 20))

as.numeric(as.character(points_f))

[1] 10 30 10 20

Lists

Lists: different types and sizes

participant <- list(name = "Anna",
                    rts = c(450, 520, 380),
                    passed = TRUE,
                    other = list(mom_age = c(27, 34, 40, 18),
                                 degree = c("bachelor")
                    ))

str(participant)

List of 4
 $ name  : chr "Anna"
 $ rts   : num [1:3] 450 520 380
 $ passed: logi TRUE
 $ other :List of 2
  ..$ mom_age: num [1:4] 27 34 40 18
  ..$ degree : chr "bachelor"

Looking at a list

length(participant)

[1] 4

names(participant)

[1] "name"   "rts"    "passed" "other" 

[ ], [[ ]] and $

participant[2]          # [ ]: a smaller LIST

$rts
[1] 450 520 380

participant[[2]]        # [[ ]]: the element itself

[1] 450 520 380

participant$rts         # $: the element, by name

[1] 450 520 380

participant$rts[3]      # then index inside it

[1] 380

class(participant[2]); class(participant[[2]])

[1] "list"

[1] "numeric"

Modifying lists

participant$age <- 24              # add an element

participant$passed <- FALSE        # change an element

participant$rts[1] <- 455          # change inside an element

str(participant)

List of 5
 $ name  : chr "Anna"
 $ rts   : num [1:3] 455 520 380
 $ passed: logi FALSE
 $ other :List of 2
  ..$ mom_age: num [1:4] 27 34 40 18
  ..$ degree : chr "bachelor"
 $ age   : num 24

Matrices and arrays

Matrices: two dimensions, one type

m <- matrix(data = 1:6, nrow = 2, ncol = 3)   # by column

m

     [,1] [,2] [,3]
[1,]    1    3    5
[2,]    2    4    6

# by row
matrix(data = 1:6, nrow = 2, ncol = 3, byrow = TRUE)

     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    4    5    6

dim(m); nrow(m); ncol(m)

[1] 2 3

[1] 2

[1] 3

Indexing matrices: [row, column]

m[2, 3]          # one element

[1] 6

m[1, ]           # the whole first row

[1] 1 3 5

m[, 2]           # the whole second column

[1] 3 4

m[m > 3]         # logical condition

[1] 4 5 6

m[m > 3] <- 0    # modify

m

     [,1] [,2] [,3]
[1,]    1    3    0
[2,]    2    0    0

Joining matrices and vectors

a <- 1:3

b <- c(10, 20, 30)

cbind(a, b)   # as columns

     a  b
[1,] 1 10
[2,] 2 20
[3,] 3 30

rbind(a, b)   # as rows

  [,1] [,2] [,3]
a    1    2    3
b   10   20   30

Still one type only: adding text turns everything into text

cbind(a, c("x", "y", "z"))

     a      
[1,] "1" "x"
[2,] "2" "y"
[3,] "3" "z"

Dataframes

Dataframes: columns of different types

Each column is a vector (all the same length): a list with a rectangular shape

df <- data.frame(id = c("P1", "P2", "P3", "P4", "P5"),
                 group = c("ctrl", "trt", "ctrl", "trt", "trt"),
                 age = c(22, 27, 31, NA, 24),
                 score = c(18, 25, 30, 22, 27))

df

  id group age score
1 P1  ctrl  22    18
2 P2   trt  27    25
3 P3  ctrl  31    30
4 P4   trt  NA    22
5 P5   trt  24    27

Looking at a dataframe

str(df)

'data.frame':   5 obs. of  4 variables:
 $ id   : chr  "P1" "P2" "P3" "P4" ...
 $ group: chr  "ctrl" "trt" "ctrl" "trt" ...
 $ age  : num  22 27 31 NA 24
 $ score: num  18 25 30 22 27

dim(df); names(df)

[1] 5 4

[1] "id"    "group" "age"   "score"

head(df, n = 2)

  id group age score
1 P1  ctrl  22    18
2 P2   trt  27    25

summary(df) and table(df$group) are also useful checks

Extracting a column as a vector

df$age               # a column: a vector

[1] 22 27 31 NA 24

df[["age"]]          # the same

[1] 22 27 31 NA 24

Extracting a column as a dataframe

df["age"]            # a dataframe with one column

  age
1  22
2  27
3  31
4  NA
5  24

df[3]                # the same, by position

  age
1  22
2  27
3  31
4  NA
5  24

Extracting rows

df[rows, columns]: leave a position empty to keep everything

df

  id group age score
1 P1  ctrl  22    18
2 P2   trt  27    25
3 P3  ctrl  31    30
4 P4   trt  NA    22
5 P5   trt  24    27

df[2, ]              # the second row

  id group age score
2 P2   trt  27    25

Extracting cells

df[2, "score"]       # one cell: row 2, column score

[1] 25

df[1:2, c("id", "score")]

  id score
1 P1    18
2 P2    25

Selecting rows with conditions

df[rows, columns]: the condition goes in the rows position

df

  id group age score
1 P1  ctrl  22    18
2 P2   trt  27    25
3 P3  ctrl  31    30
4 P4   trt  NA    22
5 P5   trt  24    27

df[df$score > 20, ]

  id group age score
2 P2   trt  27    25
3 P3  ctrl  31    30
4 P4   trt  NA    22
5 P5   trt  24    27

Combining conditions

df[df$group == "trt" & df$score > 22, c("id", "score")]

  id score
2 P2    25
5 P5    27

Selecting rows with subset()

Same result, without repeating df$:

subset(df, subset = group == "trt" & score > 22,
       select = c(id, score))

  id score
2 P2    25
5 P5    27

Creating a new column

df$new_column <- value: one value for each row

df$passed <- df$score >= 25
df

  id group age score passed
1 P1  ctrl  22    18  FALSE
2 P2   trt  27    25   TRUE
3 P3  ctrl  31    30   TRUE
4 P4   trt  NA    22  FALSE
5 P5   trt  24    27   TRUE

A new column from another column

df$score_100 <- df$score / 30 * 100
df

  id group age score passed score_100
1 P1  ctrl  22    18  FALSE  60.00000
2 P2   trt  27    25   TRUE  83.33333
3 P3  ctrl  31    30   TRUE 100.00000
4 P4   trt  NA    22  FALSE  73.33333
5 P5   trt  24    27   TRUE  90.00000

Creating a column with ifelse()

A new column that depends on a condition

df$level <- ifelse(df$score >= 25, "high", "low")
df

  id group age score passed score_100 level
1 P1  ctrl  22    18  FALSE  60.00000   low
2 P2   trt  27    25   TRUE  83.33333  high
3 P3  ctrl  31    30   TRUE 100.00000  high
4 P4   trt  NA    22  FALSE  73.33333   low
5 P5   trt  24    27   TRUE  90.00000  high

Changing the type of a column

df$group <- as.factor(df$group)
str(df)

'data.frame':   5 obs. of  7 variables:
 $ id       : chr  "P1" "P2" "P3" "P4" ...
 $ group    : Factor w/ 2 levels "ctrl","trt": 1 2 1 2 2
 $ age      : num  22 27 31 NA 24
 $ score    : num  18 25 30 22 27
 $ passed   : logi  FALSE TRUE TRUE FALSE TRUE
 $ score_100: num  60 83.3 100 73.3 90
 $ level    : chr  "low" "high" "high" "low" ...

Changing one cell

Select the cell with a condition, then assign the new value

df$age[df$id == "P4"] <- 29
df

  id group age score passed score_100 level
1 P1  ctrl  22    18  FALSE  60.00000   low
2 P2   trt  27    25   TRUE  83.33333  high
3 P3  ctrl  31    30   TRUE 100.00000  high
4 P4   trt  29    22  FALSE  73.33333   low
5 P5   trt  24    27   TRUE  90.00000  high

Removing a column

Assign NULL to the column

df$score_100 <- NULL
df

  id group age score passed level
1 P1  ctrl  22    18  FALSE   low
2 P2   trt  27    25   TRUE  high
3 P3  ctrl  31    30   TRUE  high
4 P4   trt  29    22  FALSE   low
5 P5   trt  24    27   TRUE  high

Counting

table(df$group)                     # frequencies

ctrl  trt 
   2    3 

table(df$group, df$level)           # contingency table

      
       high low
  ctrl    1   1
  trt     2   1

Sorting rows: order()

order() gives the positions of the rows, from the lowest to the highest value

df$score

[1] 18 25 30 22 27

order(df$score)

[1] 1 4 2 5 3

We use these positions to select the rows: the rows are now sorted by score

df_sorted <- df[order(df$score), ]
df_sorted

  id group age score passed level
1 P1  ctrl  22    18  FALSE   low
4 P4   trt  29    22  FALSE   low
2 P2   trt  27    25   TRUE  high
5 P5   trt  24    27   TRUE  high
3 P3  ctrl  31    30   TRUE  high

Sorting from the highest

decreasing = TRUE

df_sorted <- df[order(df$score, decreasing = TRUE), ]
df_sorted

  id group age score passed level
3 P3  ctrl  31    30   TRUE  high
5 P5   trt  24    27   TRUE  high
2 P2   trt  27    25   TRUE  high
4 P4   trt  29    22  FALSE   low
1 P1  ctrl  22    18  FALSE   low

Missing values in a dataframe

df$age[2] <- NA

df$age

[1] 22 NA 31 29 24

sum(is.na(df$age))                  # how many NAs?

[1] 1

mean(df$age, na.rm = TRUE)

[1] 26.5

df[complete.cases(df), "id"]        # rows without any NA

[1] "P1" "P3" "P4" "P5"

Adding rows: rbind()

The new rows need the same number of columns and the same column names

new <- data.frame(id = "P6", group = "ctrl", age = 25,
                  score = 21, passed = FALSE, level = "low")
new

  id group age score passed level
1 P6  ctrl  25    21  FALSE   low

Adding rows: the result

df2 <- rbind(df, new)
df2

  id group age score passed level
1 P1  ctrl  22    18  FALSE   low
2 P2   trt  NA    25   TRUE  high
3 P3  ctrl  31    30   TRUE  high
4 P4   trt  29    22  FALSE   low
5 P5   trt  24    27   TRUE  high
6 P6  ctrl  25    21  FALSE   low

nrow(df2)

[1] 6

Adding columns: merge()

Add the columns of another dataframe, matching the rows by a key (here id)

sleep <- data.frame(id = c("P3", "P1", "P6", "P2", "P5", "P4"),
                    hours = c(8, 7, 6, 6, 9, 7))
sleep

  id hours
1 P3     8
2 P1     7
3 P6     6
4 P2     6
5 P5     9
6 P4     7

merge(): the result

Each row of df2 gets the hours of its own id, even if the order is different

df3 <- merge(df2, sleep, by = "id")
df3

  id group age score passed level hours
1 P1  ctrl  22    18  FALSE   low     7
2 P2   trt  NA    25   TRUE  high     6
3 P3  ctrl  31    30   TRUE  high     8
4 P4   trt  29    22  FALSE   low     7
5 P5   trt  24    27   TRUE  high     9
6 P6  ctrl  25    21  FALSE   low     6

merge() when some ids are missing

Here sleep2 has only some of the ids of df2, and one more (P7)

sleep2 <- data.frame(id = c("P1", "P2", "P7"),
                     hours = c(7, 6, 8))
sleep2

  id hours
1 P1     7
2 P2     6
3 P7     8

# only the ids in both dataframes
df4 <- merge(df2, sleep2, by = "id")
df4

  id group age score passed level hours
1 P1  ctrl  22    18  FALSE   low     7
2 P2   trt  NA    25   TRUE  high     6

Keeping all the rows: all.x = TRUE

The rows of df2 without a match get NA

df5 <- merge(df2, sleep2, by = "id", all.x = TRUE)
df5

  id group age score passed level hours
1 P1  ctrl  22    18  FALSE   low     7
2 P2   trt  NA    25   TRUE  high     6
3 P3  ctrl  31    30   TRUE  high    NA
4 P4   trt  29    22  FALSE   low    NA
5 P5   trt  24    27   TRUE  high    NA
6 P6  ctrl  25    21  FALSE   low    NA

Before the exam

Remember

• <- creates objects, == compares
• text always goes in quotes: "hello", "5"
• with missing values: mean(x, na.rm = TRUE)
• vectors and matrices hold one type of data
• before renaming levels, look at levels(x); to reorder and rename use factor(x, levels, labels)
• to use a new value in a factor, add its level first
• lists: [[ ]] or $ to get an element
• dataframes: df[rows, columns], e.g. df[df$score > 20, ]
• rbind(): same column names; merge(): a common key

Function summary

structure	dimensions	types	create	index
vector	1	one	c(), seq(), rep()	x[i]
factor	1	categories	factor()	x[i], levels()
list	1 (nested)	any	list()	x[i], x[[i]], x$name
matrix	2	one	matrix(), cbind()	x[row, col]
dataframe	2	one per column	data.frame()	x$col, x[row, col]

Good luck!

During the exam you may use the material provided during the lessons.

Other online material (or ChatGPT) is not allowed.