javaで文中の各単語の頻度を計算する
私は非常に基本的なJavaプログラムを書いています。これは、文中の各単語の頻度を計算します。
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;
}
}
私は各文字列を抽出して配列に格納しましたが、実際には問題は各「単語」が繰り返されるインスタンスの数をカウントする方法と、繰り返される単語が複数回表示されないように表示する方法です、あなたはこれで私を助けることができます1 ？
Wordをキーとしてマップを使用し、値としてカウントします。
Map<String, Integer> map = new HashMap<>();
for (String w : words) {
Integer n = map.get(w);
n = (n == null) ? 1 : ++n;
map.put(w, n);
}
java.utilの使用が許可されていない場合は、いくつかのソートアルゴリズムを使用してarrをソートし、これを行うことができます
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;
}
}
Java 8以降のConcurrentHashMapを使用した興味深いソリューション
ConcurrentMap<String, Integer> m = new ConcurrentHashMap<>();
m.compute("x", (k, v) -> v == null ? 1 : v + 1);
Java 8では、これを2つの簡単な行で記述できます！さらに、並列計算を利用できます。
これを行う最も美しい方法は次のとおりです。
Stream<String> stream = Stream.of(text.toLowerCase().split("\\W+")).parallel();
Map<String, Long> wordFreq = stream
.collect(Collectors.groupingBy(String::toString,Collectors.counting()));
これを試して
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;
}
}
出力：
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);
}
}
出力：{a = 3、be = 1、may = 1、yes = 1、this = 7、about = 1、i = 1、is = 3、it = 1、do = 1、not = 1、what = 1 、care = 1}
Java 10から、次を使用できます。
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);
}
}
出力：
{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");
}
}
これを試すことができます
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);
}
}
ファイル内の単語の頻度を決定します。
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;
}
}
次のプログラムは、頻度を見つけ、それに応じてソートし、印刷します。
以下は、頻度別にグループ化された出力です。
0-10:
The 2
Is 4
11-20:
Have 13
Done 15
私のプログラムは次のとおりです。
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;
}
}
String s[]=st.split(" ");
String sf[]=new String[s.length];
int count[]=new int[s.length];
sf[0]=s[0];
int j=1;
count[0]=1;
for(int i=1;i<s.length;i++)
{
int t=j-1;
while(t>=0)
{
if(s[i].equals(sf[t]))
{
count[t]++;
break;
}
t--;
}
if(t<0)
{
sf[j]=s[i];
count[j]++;
j++;
}
}
Java 8 Stream collectors groupby function：
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()));
Java 8のリストの要素の頻度を数える
List<Integer> list = new ArrayList<Integer>();
Collections.addAll(list,3,6,3,8,4,9,3,6,9,4,8,3,7,2);
Map<Integer, Long> frequencyMap = list.stream().collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
System.out.println(frequencyMap);
注：文字列の頻度カウントでは、文字列を分割してリストに変換し、カウント頻度のストリームを使用します=>（Map frequencyMap）*