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单词词频统计代码java 单词词频统计软件
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java计算一篇英文文档词频 并按出现次数从高到低输出（以下基础上补充）谢谢！
String result = sb.toString();
String[] Str = result.split("[^A-Za-z0-9]"); //quanbu
for(String string:Str){
singleSet.add(string);
if("".equals(string)){ //这里是我加的，去除空格次数的处理
singleSet.remove("");
}
}
MapString, Integer map=new HashMapString, Integer();
for (String childString : singleSet){
int count=0;
for(String fatherString : Str){
if(fatherString.equals(childString)){
count++;
}
}
map.put(childString, count); //存储在hashmap中
}
ArrayListEntryString,Integer l = new ArrayListEntryString,Integer(map.entrySet());
Collections.sort(l, new ComparatorObject(){
public int compare(Object e1, Object e2){
int v1 = Integer.parseInt(((EntryString,Integer)e1).getValue().toString());
int v2 = Integer.parseInt(((Entry)e2).getValue().toString());
return v2-v1; //改为v1-v2就是从小到大了
}
});
for (EntryString, Integer e: l){
System.out.println(e.getKey()+" "+e.getValue());
}
代码仅供参考！希望对你有用
相关问答
Q1: 用JAVA语言设计一个类,统计一篇英文文章的词频,并按照词频由高到低输出。修改下面代码就行了。
这题目如果能增加一个类单词词频统计代码java的话会高效很多。。。如果非要在这个框框里面，代码麻烦 效率低下呢。
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;
public class Article {
//保存文章的内容
String content;
//保存分割后的单词集合
String[] rawWords;
//保存统计后的单词集合
String[] words;
//保存单词对应的词频
int[] wordFreqs;
//构造函数，输入文章内容
//提高部分单词词频统计代码java：从文件中读取
public Article() {
content = "kolya is one of the richest films i've seen in some time . zdenek sverak plays a confirmed old bachelor ( who's likely to remain so ) , who finds his life as a czech cellist increasingly impacted by the five-year old boy that he's taking care of . though it ends rather abruptly-- and i'm whining , 'cause i wanted to spend more time with these characters-- the acting , writing , and production values are as high as , if not higher than , comparable american dramas . this father-and-son delight-- sverak also wrote the script , while his son , jan , directed-- won a golden globe for best foreign language film and , a couple days after i saw it , walked away an oscar . in czech and russian , with english subtitles . ";
}
//对文章根据分隔符进行分词,将结果保存到rawWords数组中
public void splitWord(){
//分词的时候，因为标点符号不参与，所以所有的符号全部替换为空格
final char SPACE = ' ';
content = content.replace('\'', SPACE).replace(',', SPACE).replace('.', SPACE);
content = content.replace('(', SPACE).replace(')', SPACE).replace('-', SPACE);
rawWords = content.split("\\s+");//凡是空格隔开的都算单词，上面替换单词词频统计代码java了', 所以I've 被分成2个 //单词
}
//统计词，遍历数组
public void countWordFreq() {
//将所有出现的字符串放入唯一的set中，不用map,是因为map寻找效率太低了
SetString set = new TreeSetString();
for(String word: rawWords){
set.add(word);
}
Iterator ite = set.iterator();
ListString wordsList = new ArrayListString();
ListInteger freqList = new ArrayListInteger();
//多少个字符串未知，所以用list来保存先
while(ite.hasNext()){
String word = (String) ite.next();
int count = 0;//统计相同字符串的个数
for(String str: rawWords){
if(str.equals(word)){
count++;
}
}
wordsList.add(word);
freqList.add(count++);
}
//存入数组当中
words = wordsList.toArray(new String[0]);
wordFreqs = new int[freqList.size()];
for(int i = 0; i freqList.size(); i++){
wordFreqs[i] = freqList.get(i);
}
}
//根据词频，将词数组和词频数组进行降序排序
public void sort() {
class Word{
private String word;
private int freq;
public Word(String word, int freq){
this.word = word;
this.freq = freq;
}
}
//注意：此处排序，1）首先按照词频降序排列， 2）如果词频相同，按照字母降序排列，
//如 'abc' 'ab' 'aa'
class WordComparator implements Comparator{
public int compare(Object o1, Object o2) {
Word word1 = (Word) o1;
Word word2 = (Word) o2;
if(word1.freq word2.freq){
return 1;
}else if(word1.freq word2.freq){
return -1;
}else{
int len1 = word1.word.trim().length();
int len2 = word2.word.trim().length();
String min = len1 len2? word2.word: word1.word;
String max = len1 len2? word1.word: word2.word;
for(int i = 0; i min.length(); i++){
if(min.charAt(i) max.charAt(i)){
return 1;
}
}
return 1;
}
}
}
List wordList = new ArrayListWord();
for(int i = 0; i words.length; i++){
wordList.add(new Word(words[i], wordFreqs[i]));
}
Collections.sort(wordList, new WordComparator());
for(int i = 0; i wordList.size(); i++){
Word wor = (Word) wordList.get(i);
words[i] = wor.word;
wordFreqs[i] = wor.freq;
}
}
//将排序结果输出
public void printResult() {
System.out.println("Total " + words.length + " different words in the content!");
for(int i = 0; i words.length; i++){
System.out.println(wordFreqs[i] + " " + words[i]);
}
}
//测试类的功能
public static void main(String[] args) {
Article a = new Article();
a.splitWord();
a.countWordFreq();
a.sort();
a.printResult();
}
}
-----------------------
Total 99 different words in the content!
5 and
4 the
4 i
4 a
3 as
2 with
2 who
2 to
2 time
2 sverak
2 son
2 s
2 old
2 of
2 it
2 in
2 his
2 czech
1 zdenek
1 year
1 wrote
1 writing
1 won
1 whining
1 while
1 wanted
1 walked
1 ve
1 values
1 though
1 this
1 these
1 that
1 than
1 taking
1 subtitles
1 spend
1 some
1 so
1 seen
1 script
1 saw
1 russian
1 richest
1 remain
1 rather
1 production
1 plays
1 oscar
1 one
1 not
1 more
1 m
1 likely
1 life
1 language
1 kolya
1 jan
1 is
1 increasingly
1 impacted
1 if
1 higher
1 high
1 he
1 golden
1 globe
1 foreign
1 for
1 five
1 finds
1 films
1 film
1 father
1 english
1 ends
1 dramas
1 directed
1 delight
1 days
1 couple
1 confirmed
1 comparable
1 characters
1 cellist
1 cause
1 care
1 by
1 boy
1 best
1 bachelor
1 away
1 are
1 an
1 american
1 also
1 after
1 acting
1 abruptly
Q2: java程序统计单词词频
1.其实你这个功能用lucene加IK Analyer分词器很好解决。
把文档读入，然后用lucene创建索引，在索引里你的每个单词都会被分出来，词频就是命中次数totalHits，然后读取索引，放进map就行。这只是一个大体思路。
2.如果不用lucene，就用正则表达式匹配英文单词之间的空格，把空格去掉放进map的key，同时设置value=1。这样你的每个英文单词（有连词符号的如I'm根据你的需求另考虑）就都放进了map的key，且value都等于1。当然这些key值是有重复的，所以就要开始遍历比较key值，用几个嵌套循环把后面元素的key值与前面元素的key值比较，如果有相同的则把该元素remove，前面元素的value+1。考虑到map没有index可能遍历起来不容易实现比较，你也可以在匹配的时候把单词存进一个list，然后再新建一个同样长度的list用来存放词频（初值设为1），然后再按上面的思路来操作。也是大体思路，希望对你有所帮助
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