Separating a column using big spaces in strings in R - r

Question

This is my data frame, composed only of the 1 observation. This is a long string where 4 different parts are identifiable:
example <- "4.6 (19 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. 151 students enrolled "
df <- data.frame(example)
As you can see, the first observation is composed of a string with 4 different parts: rating (4.6), number of ratings (19 ratings), a sentence (Course...accurately), and students enrolled (151).
I employed the separate() function to divide that column in 4 one:
df1 <- separate(df, example, c("Rating", "Number of rating", "Sentence", "Students"), sep = " ")
Thus, this does not behave as expected.
Any idea.
UPDATE:
This is what I get with your comment #nicola
> df1 <- separate(df, example, c("Rating", "Number of rating", "Sentence", "Students"), sep=" {4,}")
Warning message:
Expected 4 pieces. Additional pieces discarded in 1 rows [1].

Answer 1

How about this:
x <- str_split(example, " ") %>%
unlist()
x <- x[x != ""]
df <- tibble("a", "b", "c", "d")
df[1, ] <- x
colnames(df) <- c("Rating", "Number of rating", "Sentence", "Students")
> str(df)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 1 obs. of 4 variables:
$ Rating : chr "4.6"
$ Number of rating: chr " (19 ratings)"
$ Sentence : chr " Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of ra"| __truncated__
$ Students : chr "151 students enrolled"

Answer 2

There are two keys to the answer. The first is to the correct regex used as separator sep = "[[:space:]]{2,}" which means two or more whitespace (\\s{2,} would be a more common alterantive). The second one is that your example actually has a lot a trailing whitespace which separate() tries to put into another column. It can simply be removed using trimws(). The solution therefore looks like this:
library(tidyr)
library(dplyr)
example <- "4.6 (19 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. 151 students enrolled "
df <- data.frame(example)
df_new <- df %>%
mutate(example = trimws(example)) %>%
separate(col = "example",
into = c("rating", "number_of_ratings", "sentence", "students_enrolled"),
sep = "[[:space:]]{2,}")
as_tibble(df_new)
# A tibble: 1 x 4
rating number_of_ratings sentence students_enrolled
<chr> <chr> <chr> <chr>
1 4.6 (19 ratings) Course Ratings are calculated from individual students’ ratings and a vari~ 151 students enr~
tibble is only used to formatting the output.

Answer 3

Certainly possible with the stringr package and a bit of regular expressions:
rating_mean n_ratings n_students descr
1 4.65 19 151 "Course (...) accurately."
Code
library(stringr)
# create result data frame
result <- data.frame(cbind(rating_mean = 0, n_ratings = 0, n_students = 0, descr = 0))
# loop through rows of example data frame
for (i in 1:nrow(df)){
# replace spaces
example[i, 1] <- gsub("\\s+", " ", example[i, 1])
# match and extract mean rating
result[i, 1] <- as.numeric(str_match(example[i], "^[0-9]+\\.[0-9]+"))
# match and extract number of ratings
result[i, 2] <- as.numeric(str_match(str_match(example[i, 1], "\\(.+\\)"), "[0-9]+"))
# match and extract number of enrolled students
result[i, 3] <- as.numeric(str_match(str_match(example[i, 1], "\\s[0-9].+$"), "[0-9]+"))
# match and extract sentence
result[i, 4] <- str_match(example[i, 1], "[A-Z].+\\.")
}
Data
example <- "4.65 (19 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. 151 students enrolled "
example <- data.frame(example, stringsAsFactors = FALSE)