Remove everything after space in string

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.

If you want to do it with a regex:

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

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

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'))