Java String split with regex

Question

I'm trying to split an arithmetic equation represented as a String and I want to retain the multi-character delimiters: {==, !=, >=, <=, >, <}

This is what I have:

String expression = "2*(5 +1)- 3 * 2 >= 6^3.1 + 5";
expression = expression.replaceAll("\\s", "");
String[] parsedExpression = expression.split("((?<===)|(?===))|"
            + "((?<=>=)|(?=>=))|"
            + "((?<=!=)|(?=!=))|"
            + "((?<=<=)|(?=<=))|"
            + "((?<=>)|(?=>))|"
            + "((?<=<)|(?=<))");

However it splits it like this:

[2*(5+1)-3*2, >, =, 6^3.1+5]

When the desired split would be this:

[2*(5+1)-3*2, >=, 6^3.1+5]

I'm guessing the problem is that it's my rule for using > and < as a delimiter that's causing the problem but I don't know how to fix it.

I don't know about regex, but would you be ok with doing it manually? You could just loop until you get to a token or part of a token (">") and then just check the next one. If it's "=" your splitting token is ">=", else it's ">" — pushkin
– pushkin, Commented Oct 21, 2015 at 21:51

Wiktor Stribiżew · Accepted Answer · 2015-10-21 22:02:51Z

1

Append negative look-aheads (?!=) to the < and > look-arounds to make sure they do not match when = is part of the operator:

String[] parsedExpression = expression.split("((?<===)|(?===)|"
        + "(?<=>=)|(?=>=)|"
        + "(?<=!=)|(?=!=)|"
        + "(?<=<=)|(?=<=)|"
        + "(?<=>(?!=))|(?=>(?!=))|"   // See here
        + "(?<=<(?!=))|(?=<(?!=)))"); // and here

See IDEONE demo

The System.out.println(Arrays.toString(parsedExpression)); prints [2*(5+1)-3*2, >=, 6^3.1+5].

answered Oct 21, 2015 at 22:02

Wiktor Stribiżew

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1 Comment

chmza Over a year ago

Thank you, this clears up how to solve it using my method. Thanks for the answer!

pushkin · Accepted Answer · 2015-10-21 22:04:41Z

0

Can there be only one token (">", ">=", "==", etc.) in the expression? Are you allowing something like "5 < 6 < 7"?

Though it doesn't use regex, you could try something like this.

String[] parsedExpression = new String[3]; // assuming form "3 < 4". may need to modify a little
String[] tokens = {"==", "!=", "<", ">", "<=", ">="};

int idxOfToken = expression.indexOf(try every token until one is present);
String comparOp = ""; // set to operator you found
int additional = comparOp.length() == 2 ? 2 : 1;

parsedExpression[0] = expression.substring(0, idxOfToken);
parsedExpression[1] = comparOp;
parsedExpression[2] = expression.substring(idxOfToken + additional);

answered Oct 21, 2015 at 22:04

pushkin

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Comments

user1919238 · Accepted Answer · 2015-10-21 22:09:00Z

0

Because you are splitting on lookarounds, which have zero width, even if you match a two character pattern initially, the match position doesn't move past the whole pattern. Instead, you can match a further part of the pattern.

So, even if you initially match: >=, the = within that will match a second time.

While split with lookarounds can be made to work for your problem, it makes a real mess of a regex that is hard to understand. It would be better, and simpler, to take a different approach.

For example, you could match either a delimiter or a not delimiter:

/[^><=]+|[><=]+/

The list of all matches to such a pattern would split the string as you want it. This makes certain assumptions about your input data, but it could be easily tweaked if necessary. For example, it could be expanded to only match on valid delimiters.

See it work here.

answered Oct 21, 2015 at 22:09

user1919238

1 Comment

chmza Over a year ago

Thanks for the answer, I like the idea of approaching it differently. I will look it over as well!

Andreas · Accepted Answer · 2015-10-21 22:38:55Z

Update 1: The 4th pattern wasn't working right.

You want to split before and after only these: ==, !=, >=, <=, >, <, so (spaces for clarity, e.g. using COMMENTS/(?x)):

(?= [=!]= ): Before ==, !=
(?= [><] ): Before >, < (incl. >=, <=)
(?<= [=!><]= ): After ==, !=, >=, <=
(?<= [><](?!=) ): After >, <, not followed by =

The first two can be combined using | as (?= [=!]= | [><] ).
The last two can be combined using | as (?<= [=!><]= | [><](?!=) ).

So, all combined it means (?= [=!]= | [><] ) | (?<= [=!><]= | [><](?!=) ) using COMMENTS flag, or just:

(?=[=!]=|[><])|(?<=[=!><]=|[><](?!=))

Test

String regex = "(?=[=!]=|[><])|(?<=[=!><]=|[><](?!=))";
String[] split = "2*(5+1)-3*2 >= 6^3.1+5".split(regex);
System.out.println(Arrays.toString(split));

split = "a == b != c >= d <= e > f < g = h ! i".split(regex);
System.out.println(Arrays.toString(split));

Output

[2*(5+1)-3*2 , >=,  6^3.1+5]
[a , ==,  b , !=,  c , >=,  d , <=,  e , >,  f , <,  g = h ! i]

Update 2

For a complete answer, I'd like to showcase a solution based on ideas by both Pushkin and dan1111, which is to simply find the operator.

The pattern is much simpler and easier to understand. Likely performs better too.

String text = "2*(5+1)-3*2 >= 6^3.1+5";
Matcher m = Pattern.compile("[=!]=|[><]=?").matcher(text);
if (m.find()) {
    String left  = text.substring(0, m.start());
    String oper  = m.group(); // or text.substring(m.start(), m.end());
    String right = text.substring(m.end());
    System.out.printf("[%s, %s, %s]%n", left, oper, right);
}

Output

[2*(5+1)-3*2 , >=,  6^3.1+5]

Thank you so much, this simplifies what I was trying to do immensely.

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