Java字数统计程序

Question

I am trying to make a program on word count which I have partially made and it is giving the correct result but the moment I enter space or more than one space in the string, the result of word count show wrong results because I am counting words on the basis of spaces used. I need help if there is a solution in a way that no matter how many spaces are I still get the correct result. I am mentioning the code below.

public class CountWords 
{
    public static void main (String[] args)
    {

            System.out.println("Simple Java Word Count Program");

            String str1 = "Today is Holdiay Day";

            int wordCount = 1;

            for (int i = 0; i < str1.length(); i++) 
            {
                if (str1.charAt(i) == ' ') 
                {
                    wordCount++;
                } 
            }

            System.out.println("Word count is = " + wordCount);
    }
}

Answer 1

public static void main (String[] args) {

     System.out.println("Simple Java Word Count Program");

     String str1 = "Today is Holdiay Day";

     String[] wordArray = str1.trim().split("\\s+");
     int wordCount = wordArray.length;

     System.out.println("Word count is = " + wordCount);
}

The ideas is to split the string into words on any whitespace character occurring any number of times. The split function of the String class returns an array containing the words as its elements. Printing the length of the array would yield the number of words in the string.

Answer 2

Two routes for this. One way would be to use regular expressions. You can find out more about regular expressions here . A good regular expression for this would be something like "\w+" Then count the number of matches.

If you don't want to go that route, you could have a boolean flag that remembers if the last character you've seen is a space. If it is, don't count it. So the center of the loop looks like this:

boolean prevCharWasSpace=true;
for (int i = 0; i < str1.length(); i++) 
{
    if (str1.charAt(i) == ' ') {
        prevCharWasSpace=true;
    }
else{
        if(prevCharWasSpace) wordChar++;
        prevCharWasSpace = false;

    }
}

Update
Using the split technique is exactly equivalent to what's happening here, but it doesn't really explain why it works. If we go back to our CS theory, we want to construct a Finite State Automa (FSA) that counts words. That FSA may appear as:

If you look at the code, it implements this FSA exactly. The prevCharWasSpace keeps track of which state we're in, and the str1.charAt('i') is decideds which edge (or arrow) is being followed. If you use the split method, a regular expression equivalent of this FSA is constructed internally, and is used to split the string into an array.

Answer 3

Java does have StringTokenizer API and can be used for this purpose as below.

String test = "This is a test app";
int countOfTokens = new StringTokenizer(test).countTokens();
System.out.println(countOfTokens);

OR

in a single line as below

System.out.println(new StringTokenizer("This is a test app").countTokens());

StringTokenizer supports multiple spaces in the input string, counting only the words trimming unnecessary spaces.

System.out.println(new StringTokenizer("This    is    a test    app").countTokens());

Above line also prints 5

Answer 4

You can use String.split ( read more here ) instead of charAt, you will get good results. If you want to use charAt for some reason then try trimming the string before you count the words that way you won't have the extra space and an extra word

Answer 5

My implementation, not using StringTokenizer:

Map<String, Long> getWordCounts(List<String> sentences, int maxLength) {
    Map<String, Long> commonWordsInEventDescriptions = sentences
        .parallelStream()
        .map(sentence -> sentence.replace(".", ""))
        .map(string -> string.split(" "))
        .flatMap(Arrays::stream)
        .map(s -> s.toLowerCase())
        .filter(word -> word.length() >= 2 && word.length() <= maxLength)
        .collect(groupingBy(Function.identity(), counting()));
    }

Then, you could call it like this, as an example:

getWordCounts(list, 9).entrySet().stream()
                .filter(pair -> pair.getValue() <= 3 && pair.getValue() >= 1)
                .findFirst()
                .orElseThrow(() -> 
    new RuntimeException("No matching word found.")).getKey();

Perhaps flipping the method to return Map<Long, String> might be better.

Answer 6

public class wordCOunt
{
public static void main(String ar[])
{
System.out.println("Simple Java Word Count Program");

    String str1 = "Today is Holdiay Day";

    int wordCount = 1;

    for (int i = 0; i < str1.length(); i++) 
    {
        if (str1.charAt(i) == ' '&& str1.charAt(i+1)!=' ') 
        {
            wordCount++;
        } 
    }

    System.out.println("Word count is = " +(str1.length()- wordCount));
}

}

Answer 7

public class wordCount
{
public static void main(String ar[]) throws Exception
{
System.out.println("Simple Java Word Count Program");


    int wordCount = 1,count=1;
 BufferedReader br = new BufferedReader(new FileReader("C:/file.txt"));
            String str2 = "", str1 = "";

            while ((str1 = br.readLine()) != null) {

                    str2 += str1;

            }


    for (int i = 0; i < str2.length(); i++) 
    {
        if (str2.charAt(i) == ' ' && str2.charAt(i+1)!=' ') 
        {
            wordCount++;
        } 


        }

    System.out.println("Word count is = " +(wordCount));
}

}

Answer 8

you should make your code more generic by considering other word separators as well.. such as "," ";" etc.

public class WordCounter{
    public int count(String input){
        int count =0;
        boolean incrementCounter = false;
        for (int i=0; i<input.length(); i++){
            if (isValidWordCharacter(input.charAt(i))){
                incrementCounter = true;
            }else if (incrementCounter){
                count++;
                incrementCounter = false;
            }
        }
        if (incrementCounter) count ++;//if string ends with a valid word
        return count;
    }
    private boolean isValidWordCharacter(char c){
        //any logic that will help you identify a valid character in a word
        // you could also have a method which identifies word separators instead of this
        return (c >= 'A' && c<='Z') || (c >= 'a' && c<='z'); 
    }
}

Answer 9

import com.google.common.base.Optional;
import com.google.common.base.Splitter;
import com.google.common.collect.HashMultiset;
import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Multiset;

String str="Simple Java Word Count count Count Program";
Iterable<String> words = Splitter.on(" ").trimResults().split(str);


//google word counter       
Multiset<String> wordsMultiset = HashMultiset.create();
for (String string : words) {   
    wordsMultiset.add(string.toLowerCase());
}

Set<String> result = wordsMultiset.elementSet();
for (String string : result) {
    System.out.println(string+" X "+wordsMultiset.count(string));
}

Answer 10

public static int CountWords(String str){

   if(str.length() == 0)
          return 0;

   int count =0;
   for(int i=0;i< str.length();i++){


      if(str(i) == ' ')
          continue;

      if(i > 0 && str.charAt(i-1) == ' '){
        count++;
      } 

      else if(i==0 && str.charAt(i) != ' '){
       count++;
      }


   }
   return count;

}

Answer 11

    String data = "This world is mine";
    System.out.print(data.split("\\s+").length);

Answer 12

try this

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class wordcount {
    public static void main(String[] args) {
        String s = "India is my country. I love India";
        List<String> qw = new ArrayList<String>();
        Map<String, Integer> mmm = new HashMap<String, Integer>();
        for (String sp : s.split(" ")) {
            qw.add(sp);
        }
        for (String num : qw) {
            mmm.put(num, Collections.frequency(qw, num));
        }
        System.out.println(mmm);

    }

}

Answer 13

To count total words Or to count total words without repeat word count

public static void main(String[] args) {
    // TODO Auto-generated method stub
    String test = "I am trying to make make make";
    Pattern p = Pattern.compile("\\w+");
    Matcher m = p.matcher(test);
    HashSet<String> hs =  new HashSet<>();
    int i=0;
    while (m.find()) {
        i++;
        hs.add(m.group());
    }
    System.out.println("Total words Count==" + i);
    System.out.println("Count without Repetation ==" + hs.size());
}

}

Output :

Total words Count==7

Count without Repeatation ==5

Answer 14

Not sure if there is a drawback, but this worked for me...

    Scanner input = new Scanner(System.in);
    String userInput = input.nextLine();
    String trimmed = userInput.trim();
    int count = 1;

    for (int i = 0; i < trimmed.length(); i++) {
      if ((trimmed.charAt(i) == ' ') && (trimmed.charAt(i-1) != ' ')) {
        count++;
      }
    }

Answer 15

This could be as simple as using split and count variable.

public class SplitString {

    public static void main(String[] args) {
        int count=0;        
        String s1="Hi i love to code";

        for(String s:s1.split(" "))
        {
            count++;
        }
        System.out.println(count);
    }
}

Answer 16

    public class TotalWordsInSentence {
    public static void main(String[] args) {

        String str = "This is sample sentence";
        int NoOfWOrds = 1;

        for (int i = 0; i<str.length();i++){
            if ((str.charAt(i) == ' ') && (i!=0) && (str.charAt(i-1) != ' ')){
                NoOfWOrds++;
            }
        }
         System.out.println("Number of Words in Sentence: " + NoOfWOrds);
    }
}

In this code, There wont be any problem regarding white-space in it.
just the simple for loop. Hope this helps...

Answer 17

To count specified words only like John, John99, John_John and John's only. Change regex according to yourself and count the specified words only.

    public static int wordCount(String content) {
        int count = 0;
        String regex = "([a-zA-Z_’][0-9]*)+[\\s]*";     
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(content);
        while(matcher.find()) {
            count++;
            System.out.println(matcher.group().trim()); //If want to display the matched words
        }
        return count;
    }

Answer 18

class HelloWorld {

public static void main(String[] args) {
String str = "User is in for an interview";
int counter=0;
String arrStr[] = str.split(" ");
    for (int i = 0; i< arrStr.length; i++){
        String charStr = arrStr[i];
        for(int j=0; j<charStr.length(); j++) {
            if(charStr.charAt(j) =='i') {
              counter++;
            }
       }
    }
    System.out.println("i " + counter);
}

}

Answer 19

Use split(regex) method. The result is an array of strings that was splited by regex .

String s = "Today is Holdiay Day";
System.out.println("Word count is = " + s.split(" ").length);

Answer 20

You can use this code.It may help you:

public static void main (String[] args)
{

   System.out.println("Simple Java Word Count Program");

   String str1 = "Today is Holdiay Day";
   int count=0;
   String[] wCount=str1.split(" ");

   for(int i=0;i<wCount.length;i++){
        if(!wCount[i].isEmpty())
        {
            count++;
        }
   }
   System.out.println(count);
}

Answer 21

You need to read the file line by line and reduce the multiple occurences of the whitespaces appearing in your line to a single occurence and then count for the words. Following is a sample:

public static void main(String... args) throws IOException {   

    FileInputStream fstream = new FileInputStream("c:\\test.txt");
    DataInputStream in = new DataInputStream(fstream);
    BufferedReader br = new BufferedReader(new InputStreamReader(in));
    String strLine;
    int wordcount = 0;
    while ((strLine = br.readLine()) != null)   {
        strLine = strLine.replaceAll("[\t\b]", "");
        strLine = strLine.replaceAll(" {2,}", " ");
        if (!strLine.isEmpty()){
            wordcount = wordcount + strLine.split(" ").length;
        }
    }

    System.out.println(wordcount);
    in.close();
}

Answer 22

 public class CountWords 
    {
        public static void main (String[] args)
        {
            System.out.println("Simple Java Word Count Program");
            String str1 = "Today is Holdiay Day";
            int wordCount = 1;
            for (int i = 0; i < str1.length(); i++) 
            {
                if (str1.charAt(i) == ' ' && str1.charAt(i+1)!=' ') 
                {
                    wordCount++;
                } 
            }
            System.out.println("Word count is = " + wordCount));
        }
    }   

This gives the correct result because if space comes twice or more then it can't increase wordcount. Enjoy.

Answer 23

The full program working is:

public class main {

    public static void main(String[] args) {

        logicCounter counter1 = new logicCounter();
        counter1.counter("I am trying to make a program on word count which I have partially made and it is giving the correct result but the moment I enter space or more than one space in the string, the result of word count show wrong results because I am counting words on the basis of spaces used. I need help if there is a solution in a way that no matter how many spaces are I still get the correct result. I am mentioning the code below.");
    }
}

public class logicCounter {

    public void counter (String str) {

        String str1 = str;
        boolean space= true;
        int i;

        for ( i = 0; i < str1.length(); i++) {

            if (str1.charAt(i) == ' ') {
                space=true;
            } else {
                i++;
            }
        }

        System.out.println("there are " + i + " letters");
    }
}