select string after second space

user_14518614 0
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I have a column containning different names, I would like to get all the strings after the second space of characters.

My example.

df <- data.frame(col = c("Adenia macrophylla", "Adinobotrys atropurpureus (Wall.) Dunn", "Ardisia purpurea Reinw. ex Blume"))

My desired outcome like this

                                     col
1                                    
2                           (Wall.) Dunn
3                        Reinw. ex Blume

Any sugesstions for me? The way before I did is to separate them and unite, but I consider whether we have any fancy way or better to do it, since I already have many columns.

Update Just solve it

xx %>% 
  mutate(col = str_pad(col, 20,"right")) %>% 
  mutate(col = str_remove(col, '\\w+\\s\\w+\\s'))

Thanks @Ronak and @U12-Forward for providing me regex

tidyverse r regex
3 Answers

You may use sub -

sub('\\w+\\s\\w+\\s', '', df$col)
#[1] "(Wall.) Dunn"    "Reinw. ex Blume"
#Also
#sub('.*?\\s.*?\\s', '', df$col)

If you want a tidyverse answer.

library(dplyr)
library(stringr)

df %>% mutate(val = str_remove(col, '\\w+\\s\\w+\\s'))
user_39629140

In case you want to select string after n space's it might be good to use repetition in sub.

sub("([^ ]* ){2}(.*)|.*", "\\2", df$col)
#sub("([^ ]* ){2}|.*", "", df$col, perl=TRUE) #Alternative
#[1] ""                "(Wall.) Dunn"    "Reinw. ex Blume"

[^ ] get everything but not a space * 0 to n times, match a space, {2} match it two times, .* match everything.

user_104885040

Or use this regex:

df$col <- sub('^\\S+\\s+\\S+', '', df$col)

Output df:

> df
               col
1                 
2     (Wall.) Dunn
3  Reinw. ex Blume
>
user_87083640
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