Convert string vector to dataframe in R - r

Question

I am working on a quick scraping project that involves grabbing historical NFL football data. Here is a quick glance of what my data looks like:
allgames_thisweek = c("Chicago Bears 21, Tampa Bay Buccaneers 9 -- Box Score", "Cleveland Browns 28, Cincinnati Bengals 20 -- Box Score",
"Dallas Cowboys 26, Pittsburgh Steelers 9 -- Box Score", "Detroit Lions 31, Atlanta Falcons 28 (OT) -- Box Score",
"Green Bay Packers 16, Minnesota Vikings 10 -- Box Score", "Indianapolis Colts 45, Houston Oilers 21 -- Box Score",
"Kansas City Chiefs 30, New Orleans Saints 17 -- Box Score",
"Los Angeles Rams 14, Arizona Cardinals 12 -- Box Score", "Miami Dolphins 39, New England Patriots 35 -- Box Score",
"New York Giants 28, Philadelphia Eagles 23 -- Box Score", "New York Jets 23, Buffalo Bills 3 -- Box Score",
"San Diego Chargers 37, Denver Broncos 34 -- Box Score", "San Francisco 49ers 44, Los Angeles Raiders 14 -- Box Score",
"Seattle Seahawks 28, Washington Redskins 7 -- Box Score")
allgames_thisweek[1]
"Chicago Bears 21, Tampa Bay Buccaneers 9 -- Box Score"
Each row has the following data [team1, team1score, team2, team2score, --, Box Score]
My data is all formatted the exact same way, meaning there's always a comma after the first team's score, and there's always a -- after the 2nd team's score. I'd like to create a dataframe that has 4 columns (team1, team1score, team2, team2score), so an output might look like this:
output_df
team1 team1score team2 team2score
1. Chicago Bears 21 Tampba Bay Buccaneers 9
Any thoughts on how I could achieve this? Any help is appreciated! Thanks

Answer 1

You can do this with dplyr + stringr:
library(dplyr)
library(stringr)
string %>%
str_replace("(?<=\\d)\\s.*--.+$", "") %>%
str_replace_all("\\s(?=\\d+\\b)", ",") %>%
strsplit(",") %>%
do.call(rbind, .) %>%
data.frame() %>%
setNames(c("team1", "team1score", "team2", "team2score"))
Result:
team1 team1score team2 team2score
1 Chicago Bears 21 Tampa Bay Buccaneers 9
2 Cleveland Browns 28 Cincinnati Bengals 20
3 Dallas Cowboys 26 Pittsburgh Steelers 9
4 Detroit Lions 31 Atlanta Falcons 28
5 Green Bay Packers 16 Minnesota Vikings 10
6 Indianapolis Colts 45 Houston Oilers 21
7 Kansas City Chiefs 30 New Orleans Saints 17
8 Los Angeles Rams 14 Arizona Cardinals 12
9 Miami Dolphins 39 New England Patriots 35
10 New York Giants 28 Philadelphia Eagles 23
11 New York Jets 23 Buffalo Bills 3
12 San Diego Chargers 37 Denver Broncos 34
13 San Francisco 49ers 44 Los Angeles Raiders 14
14 Seattle Seahawks 28 Washington Redskins 7
Notes:
(?<=\\d)\\s.*--.+$ matches a space (\\s) followed by any character zero or more times (.*), the literal --, any character one or more times (.+), and which ends the string ($). This pattern has an extra condition that it has to be following a digit (?<=\\d).
(?<=...) is called a positive lookbehind, which checks whether what comes after is immediately following the pattern in ....
\\s(?=\\d+\\b) matches a space that immediately follows ((?=...)) a digit one or more times and a word boundary (\\b). So this matches the space between the team names and the team scores.
(?=...) is a positive lookahead, which checks whether what comes before immediately follows the pattern in ....
Data:
string = c("Chicago Bears 21, Tampa Bay Buccaneers 9 -- Box Score", "Cleveland Browns 28, Cincinnati Bengals 20 -- Box Score",
"Dallas Cowboys 26, Pittsburgh Steelers 9 -- Box Score", "Detroit Lions 31, Atlanta Falcons 28 (OT) -- Box Score",
"Green Bay Packers 16, Minnesota Vikings 10 -- Box Score", "Indianapolis Colts 45, Houston Oilers 21 -- Box Score",
"Kansas City Chiefs 30, New Orleans Saints 17 -- Box Score",
"Los Angeles Rams 14, Arizona Cardinals 12 -- Box Score", "Miami Dolphins 39, New England Patriots 35 -- Box Score",
"New York Giants 28, Philadelphia Eagles 23 -- Box Score", "New York Jets 23, Buffalo Bills 3 -- Box Score",
"San Diego Chargers 37, Denver Broncos 34 -- Box Score", "San Francisco 49ers 44, Los Angeles Raiders 14 -- Box Score",
"Seattle Seahawks 28, Washington Redskins 7 -- Box Score")