Splitting a string using streams and returning a string with the same order of words

Question

I have a string containing words separated by one or more blank characters (space, tab, etc.). I'm trying to write the most optimized procedure possible that outputs the string with the same words in the same order, but separated by only one space.

I'm trying this but I still have a problem:

public class Test {  
    public static void main(String args[]) {  
        String str = "word1, word2 word3@+word4?.word5.word6";  
        Stream<String> stream = Arrays.stream(input.split( "[, ?.@]+_" ));  
            .stream().collect(Collectors.joining(" "));
    }  
}

If you want computationally fast, and you know the separators are always just a single character as in your example, use a char array and manually replace — Jems
– Jems, Commented Feb 2, 2021 at 23:11
We can have one or more blank characters. Thanks for your comment Jems , I've edited my example. — Amamra
– Amamra, Commented Feb 2, 2021 at 23:15
What's the problem? Is it the stray _? I'm guessing [, ?.@]+_ should be [, ?.@_]+. — John Kugelman
– John Kugelman, Commented Feb 2, 2021 at 23:20
Why use Stream for this? A simple s = s.replaceAll("\\P{Alnum}+", " ").trim() can do it. — Andreas
– Andreas, Commented Feb 3, 2021 at 0:04

Nowhere Man · Accepted Answer · 2021-02-02 23:19:54Z

1

Split by one or more non-word characters \W+ and collect using Collectors.joining

String input = "word1, word2 word3@word4?word5.word6";  

String str = Arrays.stream(input.split( "\\W+" ))
                    .collect(Collectors.joining(" "));
System.out.println(str);

answered Feb 2, 2021 at 23:19

Nowhere Man

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2 Comments

Andreas Over a year ago

Why not just str = input.replaceAll("\\W+", " ") and lose that Stream overhead?

Nowhere Man Over a year ago

This is possible too. As for "why Stream?" - should have asked myself this question before answering

Thiyanesh · Accepted Answer · 2021-02-02 23:24:10Z

1

Is this your requirement?

String s = "abc  a    b";
System.out.println(s.replaceAll("\\s+", " ")); // abc a b

Or if you need to remove any non-alpha numeric character as word boundary too and remove them, then please use this

  String s = "abc  a+.?b    c";
  System.out.println(s.replaceAll("\\W+", " ").replaceAll("\\s+", " ")); // abc a b c

Thank you @ohn Kugelman for fixing

edited Feb 2, 2021 at 23:24

answered Feb 2, 2021 at 23:15

Thiyanesh

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Comments

XO56 · Accepted Answer · 2021-02-02 23:45:12Z

In this case to get fast computation I think we can use String.charAt(i) method. We have to use the ASCII values of each characters,

Digits 0 to 9 has ASCII values of 48 to 57
Capital letters A to Z has ASCII values of 65 to 90
Small letters a to z has ASCII values of 97 to 122
Space or " " has ASCII value of 32

Except this ASCII values we will eliminate all other ASCII values.

String str = "word1,     word2 word3@+word4?.word5.word6";  
String newStr = "";
boolean flag = false;
for(int i = 0 ; i < str.length(); i++){
    int charAsciiValue = (int)str.charAt(i);
    if((charAsciiValue >= 48 && charAsciiValue <= 57) || (charAsciiValue >= 65 && charAsciiValue <= 90) || charAsciiValue >= 97 && charAsciiValue <= 122){
        newStr += str.charAt(i);
        flag = true;
    }
    else if(charAsciiValue == 32 && flag){
        newStr += " ";
        flag = false;
    }else if(charAsciiValue != 32 && flag){
        newStr += " ";
        flag = false;
    }
}
System.out.println(newStr);

Output:

word1 word2 word3 word4 word5 word6

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