Split string by all spaces except those in brackets

Question

Possible Duplicate:
Split a String based on regex

I've never been a regular expression guru, so I need your help! I have a string like this:

String s = "a [b c] d [e f g]";

I want to split this string using spaces as delimiters -- but I don't want to split on spaces that appear within the [] brackets. So, from the example above, I would like this array:

{"a", "[b c]", "d", "[e f g]"}

Any advice on what regex could be used in conjunction with split in order to achieve this?

Here's another example:

"[a b] c [[d e] f g]"

becomes

{"[a b]", "c", "[[d e] f g]"}

[Lesson: Regular Expressions](http://docs.oracle.com/javase/tutorial/essential/regex/) — user1329572, Oct 14 '12 at 17:17
@artbristol Yes they can, I'd like no splitting to happen within any set of brackets. I edited to include another example. — arshajii, Oct 14 '12 at 17:20
@A.R.S, then you can't do it with regular expressions. Time to write a parser. — Carl Norum, Oct 14 '12 at 17:21
this is the third exact duplicate question..[this](http://stackoverflow.com/questions/12756651/split-a-string-based-on-regex/12756722#12756722) and [this](http://stackoverflow.com/questions/12305182/parse-commas-not-surrounded-by-brackets/12305883#12305883) — Anirudha, Oct 14 '12 at 17:32
@CarlNorum you can...check out the above similar question i answered — Anirudha, Oct 14 '12 at 18:08

Bergi · Accepted Answer · 2012-10-14T17:40:23.327

10

I think this should work, using negative lookahead - it matches no whitespace that comes before closing bracket without an opening bracket:

"a [b c] d [e f g]".split("\\s+(?![^\\[]*\\])");

For nested brackets you will need to write a parser, regexes can't afford an infinite level and get too complicated for more than one or two levels. My expression for example fails for

"[a b [c d] e] f g"

edited Oct 14 '12 at 17:40

answered Oct 14 '12 at 17:22

Bergi

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It doesn't compile - I get "malformed regular expression" error – Nir Alfasi Oct 14 '12 at 17:25
Hm, does java need `[` to be escaped inside a character class? Thanks for the hint – Bergi Oct 14 '12 at 17:28
@Jimmy: I just edited in the escaping, before it was `"…[^[]…"` – Bergi Oct 14 '12 at 17:31
@Bergi yes- the corrected version works! +1 – Nir Alfasi Oct 14 '12 at 17:34
@Bergi Any way to have this work for both `[]` and `{}` brackets (non-nested)? e.g. `a b {c d} [e f]` would become `a, b, {c d}, [e f]`. – arshajii Oct 16 '12 at 20:06
@A.R.S.: Then use `"\\s+(?![^\\[]*\\])(?![^\\{]*\\})"` – Bergi Oct 16 '12 at 22:46
@Bergi how would be the express to split by white space but to ignore spaces in `()` instead of `{}` – Code Drop Nov 25 '22 at 13:34
@DanielNascimento Just replace the `\[` and `\]` in the regex by `\(` and `\)` respectively. – Bergi Nov 25 '22 at 23:55

score 3 · Answer 2 · answered Oct 14 '12 at 17:35

3

You can not do that with single regex, simply because it can not match open/close braces and handle nested braces.

Regexes are not turing-complete, so even if it might look as working, there will be the case where it fails to.

So I'd rather suggest to program your own few lines of code which will definitely handle all cases.

You may create very simple grammar for JavaCC or AntLR or use simple stack-based parser.

answered Oct 14 '12 at 17:35

jdevelop

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you can do it in regex..nd it works fine.. – Anirudha Oct 14 '12 at 18:12
[[xx], [y]] z t [z [x] y] go ahead. – jdevelop Oct 14 '12 at 18:15
did it..2 times [here](http://stackoverflow.com/questions/12756651/split-a-string-based-on-regex/12756722#12756722) and above.. – Anirudha Oct 14 '12 at 18:17
works in c# but not in java...thxx to java's limited support of lookbehind – Anirudha Oct 14 '12 at 18:18
Regexes with look-back/look forward and other perl-inspired stuff has nothing to do with regular expressions themselves. This question related to Java, I don't care much what works with C# or Perl. – jdevelop Oct 14 '12 at 18:26
java BUG? Come on. Not implementing strange crutch over regex like is done in Perl is not a reason to call it "bug". – jdevelop Oct 14 '12 at 19:03
for me its a bug..c# does it.y not java.. – Anirudha Oct 14 '12 at 19:05
why do I or TS care about C#? It's Java. C# lacks JDBC/JMS support, its a bug of C#. – jdevelop Oct 14 '12 at 19:06
u r getting it all wrong..java supports lookbehind but cant do it properly..thts y i called it a bug..no hard feelings for java.. – Anirudha Oct 14 '12 at 19:08
please share the BUGID in Oracle issue tracking. – jdevelop Oct 14 '12 at 19:10

Marco Martinelli · Answer 3 · 2012-10-14T17:56:07.487

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As said in other answers you need a parser for that. Here a string that fail with previous regex solutions.

"[a b] c [a [d e] f g]"

EDIT:

public static List<String> split(String s){
    List<String> l = new LinkedList<String>();
    int depth=0;
    StringBuilder sb = new StringBuilder();
    for(int i=0; i<s.length(); i++){
        char c = s.charAt(i);
        if(c=='['){
            depth++;
        }else if(c==']'){
            depth--;
        }else if(c==' ' && depth==0){
            l.add(sb.toString());
            sb = new StringBuilder();
            continue;
        }
        sb.append(c);
    }
    l.add(sb.toString());

    return l;
}

edited Oct 14 '12 at 17:56

answered Oct 14 '12 at 17:39

Marco Martinelli

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you can do it in regex..u dont need a parser – Anirudha Oct 14 '12 at 18:11
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How can you manage multiple nested [] with a regex ? – Marco Martinelli Oct 14 '12 at 19:13

taufique · Answer 4 · 2012-10-14T17:32:23.493

0

If I understood your question correctly then may be the answer is following rule4.

rule1 -> ((a-z).(\w))*.(a-z)

rule2 -> ([).rule1.(])

rule3 -> ([).(rule1.(\w))*.rule2.((\w).rule1)*.(])

rule4 -> rule1 | rule3

edited Oct 14 '12 at 17:32

answered Oct 14 '12 at 17:21

taufique

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Anirudha · Answer 5 · 2012-10-14T17:54:15.003

-1

FOR NON NESTED

\\s+(?![^\\[]*\\])

FOR NESTED([] inside [])

(?<!\\[[^\\]]*)\\s+(?![^\\[]*\\])

edited Oct 14 '12 at 17:54

answered Oct 14 '12 at 17:27

Anirudha

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The compiler complains that there is an invalid escape sequence in the second regex. – arshajii Oct 14 '12 at 17:48
Same thing for the first regex. – arshajii Oct 14 '12 at 17:50
@A.R.S. try removing the escaped character \ – Anirudha Oct 14 '12 at 17:51
@Anirudha I managed to get the first regex to work, still having trouble with the second one. – arshajii Oct 14 '12 at 17:53
@Anirudha For the second regex: `java.util.regex.PatternSyntaxException: Look-behind group does not have an obvious maximum length near index 11` – arshajii Oct 14 '12 at 17:55
@A.R.S. it should work..its working in c# regex..i dnt knw y its givin such error – Anirudha Oct 14 '12 at 18:08
I guess there are subtle differences between C# and Java in terms of regex. – arshajii Oct 14 '12 at 18:10
@Anirudha: Java lookbehind is very limited – Bergi Oct 14 '12 at 18:14
@Bergi Does that mean a Java equivalent of this regex does not exist? – arshajii Oct 14 '12 at 18:16
@Anirudha Any way to have this work for *both* `[]` and `{}` brackets (non-nested)? – arshajii Oct 16 '12 at 19:08

Split string by all spaces except those in brackets

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