java 如何在java中计算字符串数组中的单词？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14936107/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to count words in array of strings in java?
提问by user1835208
I am learning about arrays and I wanted to make a program count words. Given: String myWords = {"soon; hi; also; soon; job; also"};, I have to create a method like countWrods(myWords);The printed result should be the words printed alphabetical order, the number of unique words and total words.
here is my code:
我正在学习数组，我想让一个程序计算单词。鉴于：String myWords = {"soon; hi; also; soon; job; also"};，我必须创建一个方法，如countWrods(myWords);打印结果应该是按字母顺序打印的单词，唯一单词的数量和总单词数。这是我的代码：
public class Words {
public static void main(String[] args){
String[] myWords = {"soon; hi; also; soon; job; mother; job; also; soon; later"};
Words myW= new Words();
myW.countWords();
System.out.println("\tWords \tFreq");
}
public static String[] countWords(myWords){
for (int i=0; i<myWords.length; i++){
String temp = myWords[i];
//System.out.println(temp + " ");
for(int j=i+1; j<myWords.length; j++){
String temp2= myWords[j];
System.out.println("No. of unique words: " );
}
}
}
}
What should I do next?
我接下来该怎么做？
回答by Shubham
import java.io.*;
import java.util.*;
public class Count_Words_Scan
{
void main()throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("ENTER A STRING ");
String str = br.readLine();
str= str.toLowerCase();
int c=0;
Scanner sc = new Scanner(str);
while(sc.hasNext())
{
sc.next();
c++;
}
System.out.println("NO.OF WORDS = "+c);
}
}
Input: the word counter
Output: NO.OF WORDS = 3
输入：字计数器
输出：NO.OF WORDS = 3
回答by Ivaylo Strandjev
回答by amicngh
I am assuming you want to count the words in a string .
我假设您想计算 string 中的单词数。
String: "soon hi also soon job mother job also soon later"
字符串："soon hi also soon job mother job also soon later"
public class Words {
Map<String , Integer> dictionary=new HashMap<String,Integer>();
public static void main(String[] args) {
String myWords = "soon hi also soon job mother job also soon later";
Words myW = new Words();
String[] array=myWords.split("\s+");
myW.countWords(array);
System.out.println(myW.dictionary);
}
private void countWords(String[] myWords) {
for(String s:myWords){
if(dictionary.containsKey(s))
dictionary.put(s, dictionary.get(s)+1);
else
dictionary.put(s, 1);
}
}
}
O/P : {mother=1, later=1, job=2, hi=1, also=2, soon=3}
O/P : {mother=1, later=1, job=2, hi=1, also=2, soon=3}
回答by Boris the Spider
First you need to split your String, presumably on ";" - then you can whack that into a TreeSetto sort it and make then words unqiue. Add a counter to count the total words. You could also use a TreeMapto keep a count of each word, override the put method on the map to aggregate as you go...
首先，您需要拆分您的String，大概是在“;”上 - 然后你可以TreeSet把它打成一个来排序，然后让单词unqiue。添加一个计数器来计算总字数。您还可以使用 aTreeMap来记录每个单词的数量，覆盖地图上的 put 方法以在您进行时进行聚合...
final String myString = {"soon; hi; also; soon; job; mother; job; also; soon; later"};
final String[] myStrings = myString.split(";");
final Map<String, Integer> myStringMap = new TreeMap<>(){
@override
public String put(final String key, final Integer value) {
if(contains(key)) {
return put(key, get(key) + 1);
} else {
return put(key, 1);
}
}
};
for(final String string : myStrings) {
myStringMap.put(string.trim(), 1);
}
Now myStringMap.size()is the number of unique words, myStringMap.keys()is a alphabetically sorted Setof all unquie words and if you want the total you just need to add up the values:
现在myStringMap.size()是唯一词的数量，myStringMap.keys()是按字母顺序排列Set的所有 unquie 词，如果您想要总数，您只需要将值相加：
int totalWords = 0;
for(final Integer count : myStringMap.values()) {
totalWorks += count;
}
