String.split() not on regular expression?

Question

Since String.split() works with regular expressions, this snippet:

String s = "str?str?argh";
s.split("r?");

... yields: [, s, t, , ?, s, t, , ?, a, , g, h]

What's the most elegant way to split this String on the r? sequence so that it produces [st, st, argh]?

EDIT: I know that I can escape the problematic ?. The trouble is I don't know the delimiter offhand and I don't feel like working this around by writing an escapeGenericRegex() function.

This is mentioned in the accepted answer of [How to split a string in Java - Stack Overflow](https://stackoverflow.com/q/3481828). Related questions: How to split string by [(space)](https://stackoverflow.com/q/7899525)/[(backslash)](https://stackoverflow.com/q/23751618)/[(newline)](https://stackoverflow.com/q/454908)/[(pipe)](https://stackoverflow.com/q/10796160)? ; [How to escape text for regular expression in Java](https://stackoverflow.com/questions/60160/how-to-escape-text-for-regular-expression-in-java) — user202729, Feb 02 '21 at 02:35

Stephen C · Accepted Answer · 2018-03-05T12:20:47.283

88

A general solution using just Java SE APIs is:

String separator = ...
s.split(Pattern.quote(separator));

The quote method returns a regex that will match the argument string as a literal.

edited Mar 05 '18 at 12:20

answered Jun 16 '11 at 15:10

Stephen C

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score 11 · Answer 2 · edited Apr 08 '12 at 18:37

11

You can use

StringUtils.split("?r")

from commons-lang.

edited Apr 08 '12 at 18:37

Tomasz Nurkiewicz

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answered Jun 16 '11 at 15:07

BastiS

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3

StringUtils.split() should be much faster than String.split() since StringUtils.split is using linear scanning for the separator, whereas String.split() is using regex, which is really slow – Michael P Feb 15 '17 at 19:11
3

Something to be aware of - according to the JavaDoc this treats adjacent separators as one separator. In my situation this was not desired – Tarmo Jun 29 '18 at 10:56
be aware that this accepts a list of _characters_ to split on, not a string. so this would split the string on instances of `?` or `r`, not instances of `r?` – Starwarswii May 13 '21 at 20:14

score 5 · Answer 3 · answered Jun 16 '11 at 15:06

5

Escape the ?:

s.split("r\\?");

answered Jun 16 '11 at 15:06

Etienne de Martel

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1

a more generic solution was asked – dvhh Jun 17 '11 at 07:59

score 5 · Answer 4 · edited Oct 18 '15 at 13:22

5

This works perfect as well:

public static List<String> splitNonRegex(String input, String delim)
{
    List<String> l = new ArrayList<String>();
    int offset = 0;

    while (true)
    {
        int index = input.indexOf(delim, offset);
        if (index == -1)
        {
            l.add(input.substring(offset));
            return l;
        } else
        {
            l.add(input.substring(offset, index));
            offset = (index + delim.length());
        }
    }
}

edited Oct 18 '15 at 13:22

Community

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answered Jun 16 '11 at 15:21

Martijn Courteaux

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The performance of this solution is not ideal since it creates temporary substrings. – BladeCoder May 20 '14 at 09:30
1

@BladeCoder: You're right. I fixed it :) (When I wrote this, I must have been 16, I guess) – Martijn Courteaux May 20 '14 at 10:44
Much better indeed :) – BladeCoder May 20 '14 at 21:25
I have an app (and tests) where I split frequently, and I do not need a single split on a regular expression. And Android-Studio keeps kvetching about my regular expressions (which I do not need) are not efficiently pre-compiled patterns. I will use this, and not use it in the production code inside a loop. Thanks! – Phlip Feb 13 '21 at 00:22

score 3 · Answer 5 · edited Jun 20 '20 at 09:12

3

Use Guava Splitter:

Extracts non-overlapping substrings from an input string, typically by recognizing appearances of a separator sequence. This separator can be specified as a single character, fixed string, regular expression or CharMatcher instance. Or, instead of using a separator at all, a splitter can extract adjacent substrings of a given fixed length.

edited Jun 20 '20 at 09:12

Community

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answered Jun 16 '11 at 15:16

mindas

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贼小气 · Answer 6 · 2011-06-17T13:20:31.080

3

String[] strs = str.split(Pattern.quote("r?"));

edited Jun 17 '11 at 13:20

answered Jun 16 '11 at 16:20

贼小气

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score 1 · Answer 7 · edited Aug 18 '20 at 05:44

1

Using directly the Pattern class, is possible to define the expression as LITERAL, and in that case, the expression will be evaluated as is (not regex expression).

Pattern.compile(<literalExpression>, Pattern.LITERAL).split(<stringToBeSplitted>);

example:

String[] result = Pattern.compile("r?", Pattern.LITERAL).split("str?str?argh");

will result:

[st, st, argh]

edited Aug 18 '20 at 05:44

Stephen C

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answered Jul 30 '18 at 10:27

Manuel Romeiro

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Your answer would be best if you explained your code. It will also be more useful to new users who search something similar in the future. – Nic3500 Jul 30 '18 at 11:46
I think that `Pattern.quote(...)` is a better solution. Certainly it is fewer characters :-) – Stephen C Sep 13 '18 at 00:03
There should be no difference in performance. They will do the same thing under the hood. – Stephen C Sep 17 '18 at 00:00
I have to agree with you. In theory, LITERAL should be more performant than evaluate the regex expression, but I done a little test with java 8, and for some inputs, LITERAL was best than QUOTE, but for others was the reverse. Conclusion: for now there is no relevant difference on performance. – Manuel Romeiro Sep 18 '18 at 01:25

score -4 · Answer 8 · edited Jun 16 '11 at 15:22

-4

try

String s = "str?str?argh";
s.split("r\?");

edited Jun 16 '11 at 15:22

Martijn Courteaux

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answered Jun 16 '11 at 15:08

Akash Agrawal

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