Data Wrangling: Subsetting vectors and factors Answer Key

Author

Will Gammerdinger

Published

July 1, 2025

Exercise 1

Create a vector called alphabets with the following letters, C, D, X, L, F.

# Create a vector with the letters C, D, X, L, F
alphabets <- c("C", "D", "X", "L", "F")

Use the associated indices along with [ ] to do the following:

Only display C, D and F

# Subset C, D and F from the alphabets vector
alphabets[c(1,2,5)]

[1] "C" "D" "F"

Display all except X

# Subset everything but X from the alphabets vector
alphabets[-3]

[1] "C" "D" "L" "F"

Display the letters in the opposite order (F, L, X, D, C)

# Reverse the order of the alphabets vector
alphabets[5:1]

[1] "F" "L" "X" "D" "C"

Exercise 2

Extract only those elements in samplegroup that are not KO (nesting the logical operation is optional).

# Create an boolean index vector for elements in samplegroup that are not "KO" 
idx <- samplegroup != "KO"

# Subset the samplegroup vector by the boolean index
samplegroup[idx]

[1] CTL CTL CTL OE  OE  OE 
Levels: CTL KO OE

Alternatively, you can use a nested approach:

# Subset the samplegroup vector by elements in samplegroup that are not "KO" 
samplegroup[samplegroup != "KO"]

[1] CTL CTL CTL OE  OE  OE 
Levels: CTL KO OE

Exercise 3

Use the samplegroup factor we created in a previous lesson, and relevel it such that KO is the first level followed by CTL and OE.

# Re-level the samplegroup factor vector with KO being the first level followed by CTL and OE
factor(samplegroup, levels = c("KO", "CTL", "OE"))

[1] CTL CTL CTL KO  KO  KO  OE  OE  OE 
Levels: KO CTL OE

---
title: "Data Wrangling: Subsetting vectors and factors Answer Key"
author:
  - Will Gammerdinger
date: "2025-07-01"
---

```{r}
#| label: load_data
#| echo: false
# Create samplegroup
samplegroup <- c("CTL", "CTL", "CTL", "KO", "KO", "KO", "OE", "OE", "OE")
# Turn samplegroup into a factor
samplegroup <- factor(samplegroup)
```

# Exercise 1

1. Create a vector called `alphabets` with the following letters, C, D, X, L, F.

```{r}
#| label: create_alphabets
# Create a vector with the letters C, D, X, L, F
alphabets <- c("C", "D", "X", "L", "F")
```

2. Use the associated indices along with `[ ]` to do the following:

* Only display C, D and F

```{r}
#| label: subset_c_d_f
# Subset C, D and F from the alphabets vector
alphabets[c(1,2,5)]
```

* Display all except X

```{r}
#| label: subset_all_but_x
# Subset everything but X from the alphabets vector
alphabets[-3]
```

* Display the letters in the opposite order (F, L, X, D, C)
    
```{r}
#| label: alphabets_reverse_sequence
# Reverse the order of the alphabets vector
alphabets[5:1]
```

# Exercise 2

Extract only those elements in `samplegroup` that are not KO (*nesting the logical operation is optional*).

```{r}
#| label: subset_all_but_ko
# Create an boolean index vector for elements in samplegroup that are not "KO" 
idx <- samplegroup != "KO"

# Subset the samplegroup vector by the boolean index
samplegroup[idx]
```

Alternatively, you can use a nested approach:

```{r}
#| label: nested_subset_all_but_ko
# Subset the samplegroup vector by elements in samplegroup that are not "KO" 
samplegroup[samplegroup != "KO"]
```

# Exercise 3

Use the `samplegroup` factor we created in a previous lesson, and relevel it such that KO is the first level followed by CTL and OE. 

```{r}
#| label: relevel_samplegroup
# Re-level the samplegroup factor vector with KO being the first level followed by CTL and OE
factor(samplegroup, levels = c("KO", "CTL", "OE"))
```