Converting simple Ruby regex to Java

Question

I'm returning to Java after a several-year hiatus with Ruby. I'm looking for idiomatic and short Java code that accomplishes the following Ruby statement:

some_string.scan(/[\w|\']+/)

The above expression creates an array from a string. The elements in the array are all the sections of some_string that are composed of either alphanum chars (\w) or the apostrophe (\' so that "John's" is not split into two words.)

For example:

"(The farmer's daughter) went to the market".scan(/[\w|\']+/)

=>

["The", "farmer's", "daughter", ...]

Update

I know the solution will use something like this:

String[] words = sentence.split(" ");

I just need the regex part that goes in split().

I know java and regex in java - but I can't see what you're Ruby regex there is doing. Can you tell it in words? :) — Mads
– Mads, Commented Apr 18, 2012 at 23:07
You don't need | in a character class (surrounded by brackets [ ]), and you don't need to escape the '. The regular expression /[\w']+/ is correct, and yours is buggy. — Ken Bloom
– Ken Bloom, Commented Apr 19, 2012 at 1:02

bevanb · Accepted Answer · 2012-04-19 21:23:28Z

3

Java doesn't have a built-in scan method that can do this in a function call, so you need to roll the loop yourself. You can do this quite easily with Java's regex Matcher class.

import java.util.regex.*;

String yourString = "(The farmer's daughter) went to the supermarket";

/* The regex syntax is basically identical to Ruby, except that you need
 * to specify your regex as a normal string literal, and therefore you need to 
 * double up on your backslashes. The other differences between my regex and 
 * yours are all things that I think you need to change about the Ruby version
 * as well. */
Pattern p = Pattern.compile("[\\w']+");
Matcher m = p.matcher(yourString);
List<String> words = new Vector<String>();
while (m.find()) {
   words.add(m.group());
}

I'm not sure what the relative merits are of using Matcher versus using Scanner for this situation.

edited Apr 19, 2012 at 21:23

bevanb

8,59110 gold badges58 silver badges94 bronze badges

answered Apr 19, 2012 at 1:02

Ken Bloom

59.1k14 gold badges114 silver badges171 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Matheus Moreira · Accepted Answer · 2012-04-18 23:42:53Z

2

Regular expressions should behave more or less the same even across languages. In this case, the only difference is that you have to escape the backslashes and single quotes.

If in Ruby we write /[\w']+/, in Java we would write Pattern.compile("[\\w\']+").

Oh, Scanners can scan Strings as well!

final String s = "The farmer's daughter went to the market";
Scanner sc = new Scanner(s);
Pattern p = Pattern.compile("[\\w\\']+");
while (sc.hasNext(p)) { System.out.println(sc.next(p)); }

It is not exactly the same thing, but why not split the string on spaces, which are the word boundaries?

"The farmer's daughter went to the market".split("\s");

edited Apr 18, 2012 at 23:42

answered Apr 18, 2012 at 23:19

Matheus Moreira

17.1k3 gold badges73 silver badges112 bronze badges

4 Comments

bevanb Over a year ago

That's pretty close. I know I need to use the .split, I just need the regex to filter out non-alphanum chars except apostrophes.

Matheus Moreira Over a year ago

@bevanb, I just learned that Scanners work with Strings as well. See if it can solve your problem. Also, the | inside the square brackets is unnecessary.

Ken Bloom Over a year ago

The regex in Ruby should be /[\w']+/ and the equivalent regex in Java is "[\\w']+".

bevanb Over a year ago

@MatheusMoreira, thanks for pointing out that the | is unnecessary.

Rich Drummond · Accepted Answer · 2012-04-19 00:48:34Z

0

How about

String[] words = test.split("[^a-zA-Z0-9']+");

or

words = test.split("[^\\w']+");

The difference in these patterns from your Ruby example is because you were using Ruby's String#scan - where you supply the pattern which matches a word. Java's String#split is like's Ruby's method of the same name - you supply the pattern which matches your word delimiters.

answered Apr 19, 2012 at 0:48

Rich Drummond

3,5191 gold badge18 silver badges16 bronze badges

Collectives™ on Stack Overflow

Converting simple Ruby regex to Java

Update

3 Answers 3

Comments

4 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Update

3 Answers 3

Comments

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related