CODEWITHSUNDEEP
javaregex
DarK
DarK

Reputation: 55

Regex extract numbers of all lengths

I am trying to extract numbers of length 3 to 5 from a long string of text. Let me explain

Say there is a string like this 123456 and I want to extract all the number that are between length 3 and 5 output would be

123
234
345
456
1234
2345
3456
12345
23456

I can run multiple regex that individually find the lengths but there might be a better way to do that than what I am doing.

import java.io.IOException;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexTestHarness {

    public static void main(String[] args) throws IOException {

        String data = "123456";

        Matcher m = Pattern.compile("\\d{3}").matcher(data);

        // m = Pattern.compile("\\d{4}").matcher(data);
        // m = Pattern.compile("\\d{5}").matcher(data);

        int position = 0;

        while (m.find(position++)) {
            System.out.println(m.group());
        }
    }
}

Premature Optimization Idea - I can match everything to 5 and then run smaller length matchers on the result of those. That way I cut down on reading the data over and over which in my case is an external source.

Upvotes: 1

Views: 981

Answers (3)

user557597
user557597

Reputation:

You can do this with a single regex. Just globally find.
Print capture groups 1,2,3 if thier lengths are greater than 0 

 #  "(?=(\\d{3}))(?=(\\d{4}))?(?=(\\d{5}))?"

 (?=
      ( \d{3} )         # (1)
 )
 (?=
      ( \d{4} )         # (2)
 )?
 (?=
      ( \d{5} )         # (3)
 )?

Perl test case 

while ( '123456' =~ /(?=(\d{3}))(?=(\d{4}))?(?=(\d{5}))?/g )
{
     print "$1\n";
     if ( length ($2) ) {
         print "$2\n";
     }
     if ( length ($3) ) {
         print "$3\n";
     }
}

Output >> 

123
1234
12345
234
2345
23456
345
3456
456

Upvotes: 2

Justin
Justin

Reputation: 25287

This looks a lot harder with regex. If you don't need to use it, loop through each one of the starting positions and extract the numbers:

// If this string is just plain numbers, skip the dataArray and the
// for (String s: dataArray) and replace the s's in the loops with data's

String data = "123456 some other datas 654321";
String[] dataArray = data.split("\\D+");

for (String s: dataArray){
    for (int length = 3; length <= 5; length++){
        for (int index = 0; index <= s.length() - length; index++) {
            int maxIndex = index + length;
            System.out.println(s.substring(index, maxIndex));
        }
    }
}

Output:

123
234
345
456
1234
2345
3456
12345
23456
654
543
432
321
6543
5432
4321
65432
54321

Upvotes: 2

Sabuj Hassan
Sabuj Hassan

Reputation: 39355

Try this solution for 3-5 match only. I am using lookahead to find the overlapping matching from the string. I have used three regex here practically.

String text = "123456";
Pattern pattern = Pattern.compile("(?=(\\d\\d\\d))(?=(\\d\\d\\d\\d?))(?=(\\d\\d\\d\\d?\\d?))");
Matcher m = pattern.matcher(text);

// taking out all the captures from the Matcher
List<String> list = new ArrayList<String>();
while (m.find()) {
    list.add(m.group(1));
    list.add(m.group(2));
    list.add(m.group(3));
}

// making the list unique using HashSet
list = new ArrayList<String>(new HashSet<String>(list));

// printing
for(String s : list){
    System.out.println(s);
}

// output is not sorted, if you want you can sort the List<>

Upvotes: 0

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