Java: make a valid REGEX

Boombom 0
Viewed 60

i am stuck. I have a text with “And how about you, Count Peter Kirílych? If they call up the militia, you too will have to mount a horse,” remarked the old count, addressing Pierre.

so i am trying to split it and Count how often the word occurs in the text and return the results with the Max repeated to less ones. So the question how can i return results with Max repeated to less one And Some of results are wrong ,

Unfortunately my regex is not working for dash(-), (`) and quote ("). Would be glad for any help Here is what i did :

StringBuilder sb= new StringBuilder();
        Map<String, Integer> counterMap = new HashMap<>();
        for(Object str:lines){
            sb.append(str.toString()+" ");
        }
        String boom[] = sb.toString().split("['@?-` \\p{Punct}]+\\s*");
        for (String word : boom) {
            if(word.length()>=4){
            if(!word.isEmpty()) {
                word = word.trim();
                Integer count = counterMap.get(word);
                if(count == null) {
                    count = 0;
                }
                counterMap.put(word, ++count);
            }}
        }
StringBuilder bb = new StringBuilder();
        Map.Entry<String,Integer> maxEntry = null;

        for(String word : counterMap.keySet()) {
            System.out.println(word + ": " + counterMap.get(word));

            bb.append(word + " - "+counterMap.get(word)+"\n");
        }

        return bb.toString();
    }
java regex
2 Answers

Instead of trying to split your input string at word boundaries, it is easier to actually look for words - i.e. a sequence of letters. This way you can even have your regex check the length of the word.

I tried this:

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

// ...

Matcher matcher = Pattern.compile("\\p{IsLatin}{4,}").matcher(sb);
while (matcher.find()) {
    String word = matcher.group();
    Integer count = counterMap.getOrDefault(word, 0);
    counterMap.put(word, ++count);
}
Jochen Reinhardt0

Here is another way to accomplish this task. Instead of splitting on all those punctuation characters...get rid of them. Clean the string so that it is nothing more than a string of words separated by a whitespaces. Split the string then find occurrences. Read the comments in code:

public static List<String> getWordOccurrences(String inputString, boolean... ignoreLetterCase) {
    boolean ignoreCase = true;              // Ignore letter case by default.
    if (ignoreLetterCase.length > 0) {      // Has otherwise been optionally supplied?
        ignoreCase = ignoreLetterCase[0];   // Let it be what was supplied.
    }
    
    // Clean the String...
    String strg = inputString.replaceAll("[\\\\@:;.-/]"," ")      // Separate joined words...
                             .replaceAll("\\s+", " ")             // Remove any extra spacing...
                             .replaceAll("[“”\\p{Punct}]+", "");  // Remove all unwanted characters...
    
    // If ignore Letter case then convert string to lower case.
    if (ignoreCase) {
        strg = strg.toLowerCase();
    }

    // Split String into a words[] String array...
    String[] words = strg.split("\\s+");
    
    Arrays.sort(words);     // Sort the Array. 

    // Creat a List<String> to store ALL word Occurrances
    List<String> occurrences = new ArrayList<>();
    
    // Determine word occurrences... 
    /* Counter to count the number of times the current 
       word in play is found within the supplied string. */
    int count;
    // Iterate through the `words[]` array...
    for (int i = 0; i < words.length; i++) {
        count = 1;  // Reset count to 1 for new word.
        /* Iterate through the `words[]` array again
           ignoring the same index as previous loop. */
        for (int j = 0; j < words.length; j++) {
            if (i != j && words[i].equals(words[j])) {
                count++; // increment counter if there is a match.
            }
        }
        
        /* Format the word string to add to the List.
           This would contain the word and the number 
           of times it was detected within the string
           separated by a single whitespace.       */
        String toAdd = words[i] + " " + count;
        
        /* Before we add the word to the list, let's 
           make sure the list doesn't already have it. */
        if (!occurrences.contains(toAdd)) {
            occurrences.add(toAdd);
        }
    }

    // In order to meet requirements:
    /* Sort the list of word occurrences in ascending order based on the 
       highest word counts first as outlined as a requirement.  A custom 
       comparator with the `Collections.sort()` method is used for this:  */
    java.util.Collections.sort(occurrences,
            (String s1, String s2) -> s2.split("\\s+")[1].compareTo(s1.split("\\s+")[1]));
    
    return occurrences;   // Return the created List.
}

To use the above method you might do something like this:

// The Input String to process...
String strg = "“And how about you, Count Peter Kirílych? If they call up the "
        + "militia, you too will have to mount a horse,” remarked the old "
        + "count, addressing Pierre.";

System.out.println(strg); // Print the input string to console:
    
// List to hold determined occurrences.
List<String> occurrences = getWordOccurrences(strg);
    
// Display the List of occurrences...
System.out.println();
System.out.println("--------------------");
System.out.println("Words (" + occurrences.size() + "):  No. of:");
System.out.println("--------------------");
for (String str : occurrences) {
    System.out.println(String.format("%-15s%-5s", 
            str.split("\\s")[0], str.split("\\s")[1]));
}
System.out.println();

And the Console Window should display something like:

“And how about you, Count Peter Kirílych? If they call up the militia, you too will have to mount a horse,” remarked the old count, addressing Pierre.

--------------------
Words (24):  No. of:
--------------------
count          2    
the            2    
you            2    
a              1    
about          1    
addressing     1    
and            1    
call           1    
have           1    
horse          1    
how            1    
if             1    
kirílych       1    
militia        1    
mount          1    
old            1    
peter          1    
pierre         1    
remarked       1    
they           1    
to             1    
too            1    
up             1    
will           1
DevilsHnd0
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