取出r中字符串空格中间的字符

Question

I have several strings like

"AAA BBB CCC 1X2L BOT BR, DDD EEE FFF 3X4L BOT BR, GGG 5X6L BOT BR"

And I just want to take out the characters before the last last spaces, ie, I want

"1X2L, 3X4L, 5X6L"

only.

How can I reach this in R?

Answer 1

You can try using sub after splitting the string on comma ( , ).

x <- "AAA BBB CCC 1X2L BOT BR, DDD EEE FFF 3X4L BOT BR, GGG 5X6L BOT BR"
sub('.*?(\\w+)\\s\\w+\\s\\w+$', '\\1', strsplit(x, ',\\s')[[1]])
#[1] "1X2L" "3X4L" "5X6L"

.*? - matches as few characters as possible until

( (\\w+) - is a capture group to capture the word that we want

\\s - a whitespace followed by

\\w+ - a word followed by

\\s - another whitespace and a word ( \\w+ ) is encountered.)

Answer 2

Another regex you can use in this case

library(stringr)
str_extract_all(x, "\\d{1}\\w{1}\\d{1}\\w{1}")
#[1] "1X2L" "3X4L" "5X6L"

• \\d{1} : Matches one digit only
• \\w{1} : Matches one letter only