Remove everything after space in string

Question

I would like to remove everything after a space in a string.

For example:

"my string is sad"

should return

"my"

I've been trying to figure out how to do this using sub/gsub but have been unsuccessful so far.

Solution 1

strsplit("my string is sad"," ")[[1]][1]

Solution 2

You may use a regex like

sub(" .*", "", x)

See the regex demo.

Here, sub will only perform a single search and replace operation, the .* pattern will find the first space (since the regex engine is searching strings from left to right) and .* matches any zero or more characters (in TRE regex flavor, even including line break chars, beware when using perl=TRUE, then it is not the case) as many as possible, up to the string end.

Some variations:

sub("[[:space:]].*", "", x) # \s or [[:space:]] will match more whitespace chars
sub("(*UCP)(?s)\\s.*", "", x, perl=TRUE) # PCRE Unicode-aware regex
stringr::str_replace(x, "(?s) .*", "")   # (?s) will force . to match any chars

See the online R demo.

Solution 3

or, substitute everything behind the first space to nothing:

gsub(' [A-z ]*', '' , 'my string is sad')

And with numbers:

gsub('([0-9]+) .*', '\\1', c('c123123123 0320.1'))

Solution 4

If you want to do it with a regex:

gsub('([A-z]+) .*', '\\1', 'my string is sad')

Solution 5

Stringr is your friend.

library(stringr)
word("my string is sad", 1)

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Comments

[Monica Heddneck]

a vectorized version to apply across a column of a dataframe would be even cooler

[Monica Heddneck]

Watch out, the top example can't remove a period!