¿Contar palabras en un método de cadena?

Question

Me preguntaba cómo escribiría un método para contar el número de palabras en una cadena java solo mediante el uso de métodos de cadena como charAt, longitud o subcadena.

¡Loops y si las declaraciones son correctas!

¡Realmente agradezco cualquier ayuda que pueda obtener! ¡Gracias!

Answer 1

public static int countWords(String s){ 

    int wordCount = 0; 

    boolean word = false; 
    int endOfLine = s.length() - 1; 

    for (int i = 0; i < s.length(); i++) { 
     // if the char is a letter, word = true. 
     if (Character.isLetter(s.charAt(i)) && i != endOfLine) { 
      word = true; 
      // if char isn't a letter and there have been letters before, 
      // counter goes up. 
     } else if (!Character.isLetter(s.charAt(i)) && word) { 
      wordCount++; 
      word = false; 
      // last word of String; if it doesn't end with a non letter, it 
      // wouldn't count without this. 
     } else if (Character.isLetter(s.charAt(i)) && i == endOfLine) { 
      wordCount++; 
     } 
    } 
    return wordCount; 
} 

Answer 2

Esto funcionaría incluso con múltiples espacios y principales y/o finales espacios y líneas en blanco:

String trim = s.trim(); 
if (trim.isEmpty()) 
    return 0; 
return trim.split("\\s+").length; // separate string around spaces 

Espero que ayude. Más información sobre dividida here.

Answer 3

private static int countWordsInSentence(String input) { 
    int wordCount = 0; 

    if (input.trim().equals("")) { 
     return wordCount; 
    } 
    else { 
     wordCount = 1; 
    } 

    for (int i = 0; i < input.length(); i++) { 
     char ch = input.charAt(i); 
     String str = new String("" + ch); 
     if (i+1 != input.length() && str.equals(" ") && !(""+ input.charAt(i+1)).equals(" ")) { 
      wordCount++; 
     } 
    } 

    return wordCount; 
} 

Answer 4

Uso

myString.split("\\s+"); 

Esto funcionará.

Answer 5

Algo en O (N)

count : 0; 

if(str[0] == validChar) : 
     count++; 
else : 
     for i = 1 ; i < sizeOf(str) ; i++ : 

      if(str[i] == validChar AND str[i-1] != validChar) 

      count++; 

      end if; 

     end for; 

end if; 

return count; 

Answer 6

public static int countWords(String str){ 
     if(str == null || str.isEmpty()) 
      return 0; 

     int count = 0; 
     for(int e = 0; e < str.length(); e++){ 
      if(str.charAt(e) != ' '){ 
       count++; 
       while(str.charAt(e) != ' ' && e < str.length()-1){ 
        e++; 
       } 
      } 
     } 
     return count; 
    } 

Answer 7

public class TestStringCount { 

    public static void main(String[] args) { 
    int count=0; 
    boolean word= false; 
    String str = "how ma ny wo rds are th ere in th is sente nce"; 
    char[] ch = str.toCharArray(); 
    for(int i =0;i<ch.length;i++){ 
     if(!(ch[i]==' ')){ 
      for(int j=i;j<ch.length;j++,i++){ 
       if(!(ch[j]==' ')){ 
        word= true; 
        if(j==ch.length-1){ 
         count++; 
        } 
        continue; 
       } 
       else{ 
        if(word){ 
         count++; 
        } 
        word = false; 
       } 
      } 
     } 
     else{ 
      continue; 
     } 
    } 
    System.out.println("there are "+(count)+" words");  
    } 
} 

Answer 8

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)); 
    } 


add at the pom.xml 
<dependency> 
    <groupId>com.google.guava</groupId> 
    <artifactId>guava</artifactId> 
    <version>r09</version> 
</dependency> 

Answer 9

Contar palabras de una cadena:
Esto también podría ayudar ->

package data.structure.test; 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
public class CountWords { 

    public static void main(String[] args) throws IOException { 
// Couting number of words in a string 
     BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
     System.out.println("enter Your String"); 
     String input = br.readLine(); 

     char[] arr = input.toCharArray(); 
     int i = 0; 
    boolean notCounted = true; 
    int counter = 0; 
    while (i < arr.length) { 
     if (arr[i] != ' ') { 
      if (notCounted) { 
       notCounted = false; 
       counter++; 
      } 
     } else { 
      notCounted = true; 
     } 
     i++; 
    } 
    System.out.println("words in the string are : " + counter); 
} 

} 

Answer 10

simplemente usar ,

str.split("\\w+").length ; 

Answer 11

if(str.isEmpty() || str.trim().length() == 0){ 
    return 0; 
} 
return (str.trim().split("\\s+").length); 

Answer 12

Hola acabo de descubrir con StringTokenizer así: java.util

String words = "word word2 word3 word4"; 
StringTokenizer st = new Tokenizer(words); 
st.countTokens(); 

Answer 13

importación. ; import java.io.;

public class Principal {

public static void main(String[] args) { 

    File f=new File("src/MyFrame.java"); 
    String value=null; 
    int i=0; 
    int j=0; 
    int k=0; 
try { 
    Scanner in =new Scanner(f); 
    while(in.hasNextLine()) 
    { 
    String a=in.nextLine(); 
    k++; 
    char chars[]=a.toCharArray(); 
    i +=chars.length; 
    } 
    in.close(); 
    Scanner in2=new Scanner(f); 
    while(in2.hasNext()) 
      { 

     String b=in2.next(); 
     System.out.println(b); 
     j++; 
      } 
    in2.close(); 

    System.out.println("the number of chars is :"+i); 
    System.out.println("the number of words is :"+j); 
    System.out.println("the number of lines is :"+k); 





} 
catch (Exception e) { 
    e.printStackTrace(); 

} 


} 

}

Answer 14

Hay una solución simple que usted puede probar este código

String s = "hju vg jhdgsf dh gg g g g "; 

    String[] words = s.trim().split("\\s+"); 

    System.out.println("count is = "+(words.length)); 

Answer 15

public static int countWords(String input) { 
     int wordCount = 0; 
     boolean isBlankSet = false; 
     input = input.trim(); 

     for (int j = 0; j < input.length(); j++) { 
      if (input.charAt(j) == ' ') 
       isBlankSet = true; 
      else { 
       if (isBlankSet) { 
        wordCount++; 
        isBlankSet = false; 
       } 
      } 

     } 

     return wordCount + 1; 
    } 

Answer 16

Mi idea de este programa es que:

package text; 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 

public class CoutingWords { 

    public static void main(String[] args) throws IOException { 
     String str; 
     int cWords = 1; 
     char ch; 

     BufferedReader buffor = new BufferedReader(new InputStreamReader(System.in)); 

     System.out.println("Enter text: "); 
     str = buffor.readLine(); 

     for(int i =0; i<str.length(); i++){ 
      ch = str.charAt(i); 
      if(Character.isWhitespace(ch)){ cWords++; } 
     } 
     System.out.println("There are " + (int)cWords +" words."); 
    } 
} 

Answer 17

Soy nuevo a stackoverflow pero espero que mi código de ayuda:

private int numOfWordsInLineCounter(String line){ 

    int words = 0; 

     for(int i = 1 ; i<line.length();i++){ 
     Character ch = line.charAt(i-1); 
     Character bch = line.charAt(i); 
      if(Character.isLetterOrDigit(ch) == true && Character.isLetterOrDigit(bch)== false) words++; 
      if(i == line.length()-1 && Character.isLetterOrDigit(bch))words++; 
     } 
    return words; 
} 

Answer 18

Una frase cadena normaly tiene palabras separadas por un espacio. Bueno, puedes dividir la frase usando los espacios como personajes separadores y contarlos de la siguiente manera.

import java.util.HashMap; 

import java.util.Map; 

public class WordCountMethod { 

    public static void main (String [] args){ 

     Map<String, Integer>m = new HashMap<String, Integer>(); 
     String phrase = "hello my name is John I repeat John"; 
     String [] array = phrase.split(" "); 

     for(int i =0; i < array.length; i++){ 
      String word_i = array[i]; 
      Integer ci = m.get(word_i); 
      if(ci == null){ 
       m.put(word_i, 1); 
      } 
      else m.put(word_i, ci+1); 
     } 

     for(String s : m.keySet()){ 
      System.out.println(s+" repeats "+m.get(s)); 
     } 
    } 

} 

Answer 19

Tomando la respuesta elegida como punto de partida los siguientes acuerdos con algunos problemas de idioma Inglés, incluyendo palabras con guiones, apóstrofes para los posesivos y mantecas, números y también los caracteres fuera de UTF-16:

public static int countWords(final String s) { 
    int wordCount = 0; 
    boolean word = false; 
    final int endOfLine = s.length() - 1; 

    for (int i = 0; i < s.length(); i++) { 
     // if the char is a letter, word = true. 
     if (isWordCharacter(s, i) && i != endOfLine) { 
      word = true; 
      // if char isn't a letter and there have been letters before, 
      // counter goes up. 
     } else if (!isWordCharacter(s, i) && word) { 
      wordCount++; 
      word = false; 
      // last word of String; if it doesn't end with a non letter, it 
      // wouldn't count without this. 
     } else if (isWordCharacter(s, i) && i == endOfLine) { 
      wordCount++; 
     } 
    } 
    return wordCount; 
} 

private static boolean isWordCharacter(final String s, final int i) { 
    final char ch = s.charAt(i); 
    return Character.isLetterOrDigit(ch) 
      || ch == '\'' 
      || Character.getType(ch) == Character.DASH_PUNCTUATION 
      || Character.isSurrogate(ch); 
}