CODEWITHSUNDEEP
javaregex
anta40
anta40

Reputation: 6743

How to display only some characters from a string with multiple consecutive characters?

Let's say I have this string: fffooooobbbbaarrr. Given a number N, for each duplicated characters, I want to display N of them.

If N=2, the output is ffoobbaarr

If N=3, the output is fffooobbbaarrr

If N=1, the output is fobar

And if N=0, the output is (empty)

As I'm learning regex, after some experimentation, I found that this works for N=2:

Pattern pattern = Pattern.compile("(\\w)\\1{2,}");
System.out.println(pattern.matcher(input.replaceAll("$1$1"));

Of course, won't work for N=3, 4, etc. How to fix this?

Upvotes: 4

Views: 91

Answers (3)

anubhava
anubhava

Reputation: 785721

You can use this regex replacement:

int n = 3 // or whatever number;
String repl = "";

if (n > 0) {
   repl = str.replaceAll("((\\S)\\2{" + (n-1) + "})\\2*", "$1");
}

Example: (for N=3)

RegEx Demo 1

Example: (for N=2)

RegEx Demo 2

Explanation:

  • (: Start capture group #1
  • (\S): Match 1+ non-whitespace char and capture as group #2
  • \2{2}: Match 2 instances of same char
  • ): End capture group #1
  • \2*: Match 0+ instances of same character outside capture group

Code Demo

Upvotes: 3

revo
revo

Reputation: 48751

Use below regex as looker:

(\\w)(\\1{N})\\1*

Breakdown:

  • (\w) Match and capture a letter to capturing group 1
  • (\1{N}) Match previous captured letter N times (capturing group 2)
  • \1* Match any number of following repetitions

N is the number of letters you need to retain (you could use it as a variable. 0 results an empty output) and for replacement use:

$2

Regex live demo

Java code (demo):

String str = "fffooooobbbbaarrr";
int N = 3;
str = str.replaceAll("(\\w)(\\1{" + N + "})\\1*", "$2");
System.out.println(str); // fffooobbbaarrr

Upvotes: 1

Youcef LAIDANI
Youcef LAIDANI

Reputation: 60046

You can Pattern and matcher like this :

    String text = "fffooooobbbbaarrr";
    Pattern pattern = Pattern.compile("(.)\\1*");
    Matcher matcher = pattern.matcher(text);
    String result = "";
    int len = 3;
    while (matcher.find()) {
        if(matcher.group().length() >= len) {
            result += matcher.group().substring(0, len);
        }else {
            result += matcher.group();
        }

    }
    System.out.println(result);

Result :

3 --> fffooobbbaarrr
2 --> ffoobbaarr
1 --> fobar
0 --> empty

The idea is :

  • match any repetitive character (.)\1* zero or more time
  • then check if the length of that matches is great or equal to your length, if so use substring to get the length you want.
  • else use the matched characters as it is.

Upvotes: 1

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