Calcul de la fréquence de chaque mot dans une phrase en java
J'écris un programme Java très basique qui calcule la fréquence de chaque mot dans une phrase jusqu'à présent, j'ai réussi à le faire beaucoup
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;
}
}
J'ai extrait chaque chaîne et l'ai stockée dans un tableau. Le problème est maintenant de savoir comment compter le nombre d'occurrences où chaque 'mot' est répété et comment l'afficher pour que les mots répétés ne s'affichent pas plusieurs fois. Pouvez-vous m'aider dans cette tâche? un ?
Utiliser une carte avec Word comme clé et compter comme valeur, quelque chose comme ça
Map<String, Integer> map = new HashMap<>();
for (String w : words) {
Integer n = map.get(w);
n = (n == null) ? 1 : ++n;
map.put(w, n);
}
si vous n'êtes pas autorisé à utiliser Java.util, vous pouvez alors trier arr en utilisant un algorithme de tri
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;
}
}
Une solution intéressante avec ConcurrentHashMap depuis Java 8
ConcurrentMap<String, Integer> m = new ConcurrentHashMap<>();
m.compute("x", (k, v) -> v == null ? 1 : v + 1);
En Java 8, vous pouvez écrire cela en deux lignes simples! De plus, vous pouvez tirer parti du calcul parallèle.
Voici la plus belle façon de faire ceci:
Stream<String> stream = Stream.of(text.toLowerCase().split("\\W+")).parallel();
Map<String, Long> wordFreq = stream
.collect(Collectors.groupingBy(String::toString,Collectors.counting()));
Essaye ça
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;
}
}
Sortie:
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
import Java.util.*;
public class WordCounter {
public static void main(String[] args) {
String s = "this is a this is this a this yes this is a this what it may be i do not care about this";
String a[] = s.split(" ");
Map<String, Integer> words = new HashMap<>();
for (String str : a) {
if (words.containsKey(str)) {
words.put(str, 1 + words.get(str));
} else {
words.put(str, 1);
}
}
System.out.println(words);
}
}
Sortie: quoi = 1, soin = 1}
À partir de Java 10, vous pouvez utiliser les éléments suivants:
import Java.util.Arrays;
import Java.util.stream.Collectors;
public class StringFrequencyMap {
public static void main(String... args){
String[] wordArray = {"One", "One", "Two","Three", "Two", "two"};
var freq = Arrays.stream(wordArray)
.collect(Collectors.groupingBy(x -> x, Collectors.counting()));
System.out.println(freq);
}
}
Sortie:
{One=2, two=1, Two=2, Three=1}
package naresh.Java;
import Java.util.HashMap;
import Java.util.HashSet;
import Java.util.Set;
public class StringWordDuplicates {
static void duplicate(String inputString){
HashMap<String, Integer> wordCount = new HashMap<String,Integer>();
String[] words = inputString.split(" ");
for(String Word : words){
if(wordCount.containsKey(Word)){
wordCount.put(Word, wordCount.get(Word)+1);
}
else{
wordCount.put(Word, 1);
}
}
//Extracting of all keys of Word count
Set<String> wordsInString = wordCount.keySet();
for(String Word : wordsInString){
if(wordCount.get(Word)>1){
System.out.println(Word+":"+wordCount.get(Word));
}
}
}
public static void main(String args[]){
duplicate("I am Java Programmer and IT Server Programmer with Java as Best Java lover");
}
}
Tu pourrais essayer ça
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);
}
}
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);
}
}
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);
}
}
Déterminez la fréquence des mots dans un fichier.
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 );
}
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;
}
}
Le programme suivant trouve la fréquence, la trie en conséquence et l’imprime.
Ci-dessous, la sortie regroupée par fréquence:
0-10:
The 2
Is 4
11-20:
Have 13
Done 15
Voici mon programme:
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;
}
}
Utilisez simplement la fonction groupby de collecteurs Java 8 Stream:
import Java.util.function.Function;
import Java.util.stream.Collectors;
static String[] COUNTRY_NAMES
= { "China", "Australia", "India", "USA", "USSR", "UK", "China",
"France", "Poland", "Austria", "India", "USA", "Egypt", "China" };
Map<String, Long> result = Stream.of(COUNTRY_NAMES).collect(
Collectors.groupingBy(Function.identity(), Collectors.counting()));