Regex replace all words with given character numbers

Question

I have text, for exp: Test numbers test count gggg aaaaaa

I need replace all words with count of characters 4(or other number) to "SUPER". What is the easiest way to do it?

Now I tried to do something this, but it not work properly:

String pattern = "[aA-zZ]+";
    Pattern p = Pattern.compile(pattern);
    Matcher m = p.matcher(myText);
    while (m.find()) {
        String word = myText.substring(m.start(), m.end());
        System.out.println("one word |" + word + "|");
        if (m.end() - m.start() == myWord.length) {
            m.replaceAll("SUPER");
        }
    }

See my answer below to make sure you only replace 4-*letter* words, and not numbers. Please feel free to drop a comment if you need more help with this. — Wiktor Stribiżew, Sep 28 '17 at 13:51

score 3 · Answer 1 · edited Sep 28 '17 at 13:29

3

str.replaceAll("\\b\\w{4}\\b", "SUPER"); should work, \\w means word character; \\b means word boundary

edited Sep 28 '17 at 13:29

Procrastinator

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answered Sep 28 '17 at 13:06

Evgeniy Dorofeev

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This is the best that can be achieved. – Taras Velykyy Sep 28 '17 at 13:07
`\w` matches also `[0-9]` and `_`. They want to match letters only. – Toto Sep 28 '17 at 13:09
@WiktorStribiżew It works with unicode strings of exact 4 characters. I tried with `"\u0048\u0065\u006C\u006C".replaceAll("\\b\\w{4}\\b", "SUPER");` and it worked. – Procrastinator Sep 28 '17 at 13:25
1

@procrastinator If you do not believe me, [here is a demo](https://ideone.com/Fthn0m). And `"\\b\\w{4}\\b"` does not replace neither `Маша` nor `кашу`. – Wiktor Stribiżew Sep 28 '17 at 13:27

score 2 · Accepted Answer · answered Sep 28 '17 at 13:07

You can use this pattern : \b\w{4}\b a for-letter group with a word boundary at start and at end

public static String rplcWordWithSize(int size, String sentence) {
    return sentence.replaceAll("\\b\\w{" + size + "}\\b", "SUPER");
}

Example of use :

public static void main(String argv[]) {
    String str = "Test numbers test count gggg aaaaaa";
    System.out.println(rplcWordWithSize(3, str));  //Test numbers test count gggg aaaaaa
    System.out.println(rplcWordWithSize(4, str));  //SUPER numbers SUPER count SUPER aaaaaa
    System.out.println(rplcWordWithSize(5, str));  //Test numbers test SUPER gggg aaaaaa
}

[Same comments](https://stackoverflow.com/a/46469954/3832970) apply here. — Wiktor Stribiżew, Sep 28 '17 at 13:12

score 1 · Answer 3 · answered Sep 28 '17 at 13:48

Note that [aA-zZ]+ matches more than just letters, as the A-z range matches [, \, ], ^, _, ` beside the English letters.

If you do not expect to replace "words" like 1234 or wrd5, and just want to replace natural language non-compound words, use either of the two solutions below.

This one is Unicode-aware, \p{L} matches any Unicode letters and \b (a word boundary) "supports" Unicode word boundaries thanks to the Pattern.UNICODE_CHARACTER_CLASS modifier embedded flag, (?U):

s = s.replaceAll("(?U)\\b\\p{L}{4}\\b", "SUPER");

Or, if you only plan to work with ASCII:

s = s.replaceAll("\\b[a-zA-Z]{4}\\b", "SUPER");

See the online Java demo:

System.out.println("Test numbers test count gggg aaaaaa".replaceAll("\\b[a-zA-Z]{4}\\b", "SUPER"));
// => SUPER numbers SUPER count SUPER aaaaaa
System.out.println("Маша ела кашу".replaceAll("(?U)\\b\\p{L}{4}\\b", "SUPER")); 
// => SUPER ела SUPER

score 0 · Answer 4 · answered Sep 28 '17 at 13:03

Try this:

Matcher m = p.matcher(myText);
String word = myText.substring(m.start(), m.end());

String[] words = word.Split(" ");
String newword = "";
for(String w : words){
if(w.length == myWord.length){
newword += "SUPER ";
}
else{
newword += w + " ";
}
}
Console.println(newword);

Did this straight out of the texteditor so there could be some minor errors.

Regex replace all words with given character numbers

4 Answers4