| Verb | Frequency |
|---|---|
| start | 418 |
| enjoy | 139 |
| begin | 337 |
| help | 281 |
2.4 Data frames
Preparation
You can find the full R script associated with this unit here.
Recommended reading
Winter (2020): Chapter 1.10-1.16
Suggested video tutorial:
Using the Data Frame in R (DataCamp, 5min)
Learn How to Subset, Extend & Sort Data Frames in R (DataCamp, 7min)
Word frequencies II
Recall our simple linguistic dataset from the previous unit:
We thought of the columns as one-dimensional, indexed lists of elements:
lemma <- c("start", "enjoy", "begin", "help")
frequency <- c(418, 139, 337, 281)In fact, R allows us to combine these two vectors into an actual spreadsheet. To this end, we need to apply the data.frame() function to two vectors of our choice. Note that they need to have the same length:
data <- data.frame(lemma, frequency)
print(data) lemma frequency
1 start 418
2 enjoy 139
3 begin 337
4 help 281
Essential R concepts
The variable data is no longer a vector, but a data frame (often abbreviated as ‘df’). Once again, each element carries its own label and can be, therefore, accessed or manipulated.
Since data frames are two-dimensional objects, the subsetting notation in square brackets [ ] needs to reflect that. This is the general pattern:
\[ \text{df[row, column]} \tag{1}\]
Say, we’re looking for the element at the intersection of the first row and first column. Applying the pattern above, we can access it like so:
data[1,1][1] "start"
But what if we needed the entire first row? We’d simply omit the column part. Note, however, that the comma , needs to remain:
data[1,] lemma frequency
1 start 418
Subsetting by columns is interesting. We can either use the square bracket notation [ ] or the column operator $:
data[,1][1] "start" "enjoy" "begin" "help"
data$lemma[1] "start" "enjoy" "begin" "help"
Filtering
Not all the observations contained in a data frame are necessarily relevant for our research. In such cases, it may be important to subset the rows and columns according to certain criteria.
Assume we only need those observations where the lemma frequencies are greater than 300. We can filter the dataset accordingly by specifying
- the data frame,
- the column of interest, and
- the condition to apply to the rows.
You can read the code below as
‘Take the data frame
dataand subset it according to the columndata$frequency. Show me those rows where the values ofdata$frequencyare greater than 300.’
data[data$frequency > 300, ] lemma frequency
1 start 418
3 begin 337
What if we wanted to filter by lemma instead? To make it more concrete, assume we’re looking for frequency data on the verbs start and help ( but not on begin and help).
We can start by accessing the rows with data on start first:
data[data$lemma == "start", ] lemma frequency
1 start 418
Next, we add a second, analogous condition. Combining multiple statements requires a logical operator. In this code chunk, we’re using | , which corresponds to a logical ‘or’ (also known as a “disjunction”).
data[data$lemma == "start" | data$lemma == "help", ] lemma frequency
1 start 418
4 help 281
The idea of combining statements somewhat naturally suggests a conjunction, which could be achieved via &. How come R doesn’t return anything if we do it that way?
data[data$lemma == "start" & data$lemma == "help", ][1] lemma frequency
<0 rows> (or 0-length row.names)
This looks unintuitive – is there another way to filter in R?
Yes, absolutely. The callouts below demonstrate a few popular alternatives. In the end, the exact way you filter doesn’t really matter, so long as you (as well as the people who have to work with your script) can understand what you’re trying to achieve with your code. Always make sure to add comments to your filtering operations!
subset()
Almost every subsetting operation we perform with square brackets can also be performed using the subset() function. Here are some expressions that are synonymous to the ones above:
subset(data, frequency > 300) lemma frequency
1 start 418
3 begin 337
subset(data, lemma == "start" | lemma == "help") lemma frequency
1 start 418
4 help 281
tidyverse
The tidyverse-ecosystem is a collection of packages specifically designed for handling typical data science tasks as comfortably and elegantly as possible, supplying countless helper functions for data manipulation, transformation and visualisation. Installation instructions are provided in 2.3 Libraries.
A extensive guide to the main tidyverse functions is provided in Chapter 3 of the free eBook R For Data Science (2nd edition). Due to its clarity, most of the more advanced code in this reader will draw on tidyverse syntax.
Let’s generate a tidyverse-style data frame, the tibble:
library(tidyverse)
data2 <- tibble(
lemma = c("start", "enjoy", "begin", "help"),
frequency = c(418, 139, 337, 281)
)
print(data2)# A tibble: 4 × 2
lemma frequency
<chr> <dbl>
1 start 418
2 enjoy 139
3 begin 337
4 help 281
We can single out certain columns with select():
select(data2, lemma)# A tibble: 4 × 1
lemma
<chr>
1 start
2 enjoy
3 begin
4 help
It is very easy to filter the data frame according to certain criteria:
filter(data2, frequency > 300)# A tibble: 2 × 2
lemma frequency
<chr> <dbl>
1 start 418
2 begin 337
filter(data2, lemma == "start" | lemma == "help")# A tibble: 2 × 2
lemma frequency
<chr> <dbl>
1 start 418
2 help 281
The tidyverse features a special pipe operator %>% that can be used to pass the output of one function on to the next one. It is conceptually similar to the coordinating conjunction and. The code can be rewritten in pipe notation as follows:
# Read as: "Take data2 and select the column with the name 'lemma'."
data2 %>%
select(lemma)
# Read as: "Take data2 and show me those rows where frequency is greater than 300."
data2 %>%
filter(frequency > 300)
# Read as: "Take data2 and show me those rows that correspond to the lemma 'start' or 'help' or both."
data %>%
filter(lemma == "start" | lemma == "help")Exercises
You can find the solutions to the exercises here.
Tier 1
Exercise 1 Recreate the barplot from the previous unit by subsetting the data variable accordingly.
Exercise 2 Print the following elements by subsetting the data frame data accordingly.
337
begin
enjoy
enjoy 139
the entire frequency column
Tier 2
Exercise 3 (Extension of Ex. 3 from Vectors) Verify that the following verbs are represented in the lemma column: enjoy, hit, find, begin. If they are in the data frame, print their frequency information.
Exercise 4 (Extension of Ex. 4 from Vectors) Use which() to find the rows where the frequency is greater than 200, and then print the lemma and frequency of those rows only.
Tier 3
Exercise 5 Diachronic corpora comprise data on language use across different time periods. This data frame indicates the frequencies of certain modal verbs across three time periods:
modals_df <- data.frame(
modal = c("can", "could", "may", "might", "must", "shall", "should", "will", "would"),
period1 = c(128, 68, 55, 21, 44, 19, 35, 85, 97),
period2 = c(142, 83, 41, 30, 39, 12, 52, 94, 119)
)Find the most and least frequent modal verb in each time period.
Calculate the percentage change in frequency for each modal verb between
period1andperiod2.Create a new column
trendwith the values"increasing"and"decreasing"based on whether the frequency increased or decreased across periods.
Exercise 6 (Extension of Ex. 8 in Vectors.) Write a function that performs part-of-speech (POS) annotation on the sentence The quick brown fox jumps over the lazy dog. Here are a few code snippets to help you get started:
- You can split up sentences into tokens using
tokenize_words()from thetokenizerslibrary.
library(tokenizers)
library(tidyverse)
text <- "Colorless green ideas sleep furiously."
text_tokenized <- tokenize_words(text)
# To lowercase the tokens
tokens_lower <- tolower(text_tokenized[[1]])- Vectors can have name attributes:
word <- "read"
# Give it a name
names(word) <- "verb"
# Get rid of its name
word <- unname(word)- There are several ways to apply conditional logic:
things <- c("apple", "cherry", "pear", "cucumber", "coconut")
fruits <- c("apple", "cherry", "pear")
vegetables <- c("cabbage", "carrot", "cucumber")
# Base R
food_analysis <- ifelse(things %in% fruits, "fruit", "not_fruit")
# Tidyverse
food_analysis2 <- case_when(
things %in% fruits ~ "yes", # if elements from "things" are in "fruits", print "yes", else
TRUE ~ "no" # print "no" (default)
)- If multiple conditions should be checked, the statements/cases have to be nested appropriately:
# Base R
complex_food_analysis <- ifelse(things %in% fruits, "fruit",
ifelse(things %in% vegetables, "vegatable",
"unknown"))
# Tidyverse
complex_food_analysis2 <- case_when(
things %in% fruits ~ "fruit", # if elements from "things" are in "fruits", print "yes", else
things %in% vegetables ~ "vegetable", # if they're in "vegetables", print "yes", else
TRUE ~ "unknown" # print "unknown"
)