用java计算句子中每个单词的频率

Calculating frequency of each word in a sentence in java

提问by Sigma

I am writing a very basic java program that calculates frequency of each word in a sentence so far i managed to do this much

我正在编写一个非常基本的 java 程序，它计算一个句子中每个单词的频率，到目前为止我设法做到了这么多

import java.io.*;

class Linked {

    public static void main(String args[]) throws IOException {

        BufferedReader br = new BufferedReader(
            new InputStreamReader(System.in));
        System.out.println("Enter the sentence");
        String st = br.readLine();
        st = st + " ";
        int a = lengthx(st);
        String arr[] = new String[a];
        int p = 0;
        int c = 0;

        for (int j = 0; j < st.length(); j++) {
            if (st.charAt(j) == ' ') {
                arr[p++] = st.substring(c,j);
                c = j + 1;
            }
        }
    }

    static int lengthx(String a) {
        int p = 0;
        for (int j = 0; j < a.length(); j++) {
            if (a.charAt(j) == ' ') {
                p++;
            }
        }
        return p;
    }
}

I have extracted each string and stored it in a array , now problem is actually how to count the no of instances where each 'word' is repeated and how to display so that repeated words not get displayed multiple times , can you help me in this one ?

我已经提取了每个字符串并将其存储在一个数组中，现在的问题实际上是如何计算每个“单词”重复的实例数以及如何显示以便重复的单词不会被多次显示，你能帮我吗？一 ？

采纳答案by Evgeniy Dorofeev

Use a map with word as a key and count as value, somthing like this

使用以单词为键的映射并计为值，像这样

    Map<String, Integer> map = new HashMap<>();
    for (String w : words) {
        Integer n = map.get(w);
        n = (n == null) ? 1 : ++n;
        map.put(w, n);
    }

if you are not allowed to use java.util then you can sort arr using some sorting algoritm and do this

如果不允许使用 java.util，则可以使用某种排序算法对 arr 进行排序并执行此操作

    String[] words = new String[arr.length];
    int[] counts = new int[arr.length];
    words[0] = words[0];
    counts[0] = 1;
    for (int i = 1, j = 0; i < arr.length; i++) {
        if (words[j].equals(arr[i])) {
            counts[j]++;
        } else {
            j++;
            words[j] = arr[i];
            counts[j] = 1;
        }
    }

An interesting solution with ConcurrentHashMap since Java 8

自 Java 8 以来 ConcurrentHashMap 的一个有趣解决方案

    ConcurrentMap<String, Integer> m = new ConcurrentHashMap<>();
    m.compute("x", (k, v) -> v == null ? 1 : v + 1);

回答by Zeeshan

Try this

尝试这个

public class Main
{

    public static void main(String[] args)
    {       
        String text = "the quick brown fox jumps fox fox over the lazy dog brown";
        String[] keys = text.split(" ");
        String[] uniqueKeys;
        int count = 0;
        System.out.println(text);
        uniqueKeys = getUniqueKeys(keys);

        for(String key: uniqueKeys)
        {
            if(null == key)
            {
                break;
            }           
            for(String s : keys)
            {
                if(key.equals(s))
                {
                    count++;
                }               
            }
            System.out.println("Count of ["+key+"] is : "+count);
            count=0;
        }
    }

    private static String[] getUniqueKeys(String[] keys)
    {
        String[] uniqueKeys = new String[keys.length];

        uniqueKeys[0] = keys[0];
        int uniqueKeyIndex = 1;
        boolean keyAlreadyExists = false;

        for(int i=1; i<keys.length ; i++)
        {
            for(int j=0; j<=uniqueKeyIndex; j++)
            {
                if(keys[i].equals(uniqueKeys[j]))
                {
                    keyAlreadyExists = true;
                }
            }           

            if(!keyAlreadyExists)
            {
                uniqueKeys[uniqueKeyIndex] = keys[i];
                uniqueKeyIndex++;               
            }
            keyAlreadyExists = false;
        }       
        return uniqueKeys;
    }
}

Output:

输出：

the quick brown fox jumps fox fox over the lazy dog brown
Count of [the] is : 2
Count of [quick] is : 1
Count of [brown] is : 2
Count of [fox] is : 3
Count of [jumps] is : 1
Count of [over] is : 1
Count of [lazy] is : 1
Count of [dog] is : 1

回答by Bahul Jain

In Java 8, you can write this in two simple lines! In addition you can take advantage of parallel computing.

在 Java 8 中，您可以用两行简单的代码来编写它！此外，您还可以利用并行计算。

Here's the most beautiful way to do this:

这是执行此操作的最美丽的方法：

Stream<String> stream = Stream.of(text.toLowerCase().split("\W+")).parallel();

Map<String, Long> wordFreq = stream
     .collect(Collectors.groupingBy(String::toString,Collectors.counting()));

回答by user6009053

public class WordFrequencyProblem {

    public static void main(String args[]){
        String s="the quick brown fox jumps fox fox over the lazy dog brown";
        String alreadyProcessedWords="";
        boolean isCount=false;
        String[] splitWord = s.split("\s|\.");
        for(int i=0;i<splitWord.length;i++){
            String word = splitWord[i];
            int count = 0;
            isCount=false;
            if(!alreadyProcessedWords.contains(word)){
                for(int j=0;j<splitWord.length;j++){
                        if(word.equals(splitWord[j])){
                            count++;
                            isCount = true;
                            alreadyProcessedWords=alreadyProcessedWords+word+" ";
                        }
                    }
            }
            if(isCount)
            System.out.println(word +"Present "+ count);
        }
    }

}

回答by Nhan

You could try this

你可以试试这个

public static void frequency(String s) {
    String trimmed = s.trim().replaceAll(" +", " ");
    String[] a = trimmed.split(" ");
    ArrayList<Integer> p = new ArrayList<>();
    for (int i = 0; i < a.length; i++) {
        if (p.contains(i)) {
            continue;
        }
        int d = 1;
        for (int j = i+1; j < a.length; j++) {
            if (a[i].equals(a[j])) {
                d += 1;
                p.add(j);
            }
        }
        System.out.println("Count of "+a[i]+" is:"+d);
    }
}

回答by mohd naeem khan

class find
{
    public static void main(String nm,String w)
    {
        int l,i;
        int c=0;


        l=nm.length();String b="";

        for(i=0;i<l;i++)
        {
            char d=nm.charAt(i);
            if(d!=' ')
            {
                b=b+d;
            }
            if(d==' ')
            {
                if(b.compareTo(w)==0)
                {
                    c++;

                } 
               b="";           
            }        
        }       
        System.out.println(c);
    }
}

回答by Nishant Srivastava

public class TestSplit {

    public static void main(String[] args) {
            String input="Find the repeated word which is repeated in this string";
            List<String> output= (List) Arrays.asList(input.split(" "));

            for(String str: output) {
                    int occurrences = Collections.frequency(output, str);
                    System.out.println("Occurence of " + str+ " is "+occurrences);
            }

            System.out.println(output);
    }

}

回答by Rajesh NJ

public class wordFrequency {
    private static Scanner scn;

    public static void countwords(String sent) {
        sent = sent.toLowerCase().replaceAll("[^a-z ]", "");
        ArrayList<String> arr = new ArrayList<String>();
        String[] sentarr = sent.split(" ");
        Map<String, Integer> a = new HashMap<String, Integer>();
        for (String word : sentarr) {
            arr.add(word);
        }
        for (String word : arr) {
            int count = Collections.frequency(arr, word);
            a.put(word, count);
        }
        for (String key : a.keySet()) {
            System.out.println(key + " = " + a.get(key));
        }
    }

    public static void main(String[] args) {
        scn = new Scanner(System.in);
        System.out.println("Enter sentence:");
        String inp = scn.nextLine();
        countwords(inp);
    }

}

回答by Jitendra

Determine the frequency of words in a file.

确定文件中单词的频率。

File f = new File(fileName);
Scanner s = new Scanner(f);
Map<String, Integer> counts =
 new Map<String, Integer>(); 
while( s.hasNext() ){
 String word = s.next();
if( !counts.containsKey( word ) )
 counts.put( word, 1 );
else
 counts.put( word, 
  counts.get(word) + 1 );

}

}

回答by Kumar Ayush

The following program finds the frequency, sorts it accordingly, and prints it.

下面的程序找到频率，相应地对其进行排序，然后打印出来。

Below is the output grouped by frequency:

以下是按频率分组的输出：

0-10:
       The   2
       Is    4
11-20:
       Have 13
       Done 15

Here is my program:

这是我的程序：

package com.company;
import java.io.*;
import java.util.*;
import java.lang.*;

/**
 * Created by ayush on 12/3/17.
 */

public class Linked {

    public static void main(String args[]) throws IOException {

        BufferedReader br = new BufferedReader(
                new InputStreamReader(System.in));
        System.out.println("Enter the sentence");
        String st = br.readLine();
        st=st.trim();
        st = st + " ";
        int count = lengthx(st);
        System.out.println(count);
        String arr[] = new String[count];
        int p = 0;
        int c = 0;

        for (int i = 0; i < st.length(); i++) {
            if (st.charAt(i) == ' ') {
                arr[p] = st.substring(c,i);
                System.out.println(arr[p]);
                c = i + 1;
                p++;
            }
        }
        Map<String, Integer> map = new HashMap<>();

        for (String w : arr) {
            Integer n = map.get(w);
            n = (n == null) ? 1 : ++n;
            map.put(w, n);
        }
        for (String key : map.keySet()) {
            System.out.println(key + " = " + map.get(key));
        }

        Set<Map.Entry<String, Integer>> entries = map.entrySet();

        Comparator<Map.Entry<String, Integer>> valueComparator = new Comparator<Map.Entry<String,Integer>>() {

            @Override
            public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2) {
                Integer v1 = e1.getValue();
                Integer v2 = e2.getValue();
                return v1.compareTo(v2); }
        };

        List<Map.Entry<String, Integer>> listOfEntries = new ArrayList<Map.Entry<String, Integer>>(entries);
        Collections.sort(listOfEntries, valueComparator);

        LinkedHashMap<String, Integer> sortedByValue = new LinkedHashMap<String, Integer>(listOfEntries.size());

        for(Map.Entry<String, Integer> entry : listOfEntries){

            sortedByValue.put(entry.getKey(), entry.getValue());
        }

        for(Map.Entry<String, Integer> entry : listOfEntries){

            sortedByValue.put(entry.getKey(), entry.getValue());
        }

        System.out.println("HashMap after sorting entries by values ");
        Set<Map.Entry<String, Integer>> entrySetSortedByValue = sortedByValue.entrySet();
        for(Map.Entry<String, Integer> mapping : entrySetSortedByValue){
            System.out.println(mapping.getKey() + " ==> " + mapping.getValue());
        }


    }

    static int lengthx(String a) {
        int count = 0;
        for (int j = 0; j < a.length(); j++) {
            if (a.charAt(j) == ' ') {
                count++;
            }
        }
        return count;
    }
}