如何获得给定字符串上的单词对示例
快速,快速的棕色,棕色的狐狸,狐狸的跳跃跳跃等,等等。
然后算出它出现了多少次?
下面的代码只能计算单个单词。
import java.util.*;
import java.util.Map;
import java.util.HashMap;
public class Tokenizer
{
public static void main(String[] args)
{
int index = 0; int tokenCount; int i =0;
Map<String,Integer> wordCount = new HashMap<String,Integer>();
Map<Integer,Integer> letterCount = new HashMap<Integer,Integer>();
String message="The Quick brown fox jumps over the lazy brown dog the quick";
StringTokenizer string = new StringTokenizer(message);
tokenCount = string.countTokens();
System.out.println("Number of tokens = " + tokenCount);
while (string.hasMoreTokens()) {
String word = string.nextToken().toLowerCase();
Integer count = wordCount.get(word);
Integer lettercount = letterCount.get(word);
if(count == null) {
wordCount.put(word, 1);
}
else {
wordCount.put(word, count + 1);
}
}
for (String words : wordCount.keySet())
{System.out.println("Word : " + words + " has count :" +wordCount.get(words));
}
int first ,second;
first = second = Integer.MIN_VALUE;
String firstword ="";
String secondword="";
for(Map.Entry<String, Integer> entry : wordCount.entrySet())
{
int count = entry.getValue();
String word = entry.getKey();
if(count>first){
second = first;
secondword = firstword;
first = count;
firstword = word;
}
else if(count>second && count ==first){
second = count;
secondword = word;
}
}
System.out.println(firstword + "" + first);
System.out.println(secondword + " " + second);
for(i = 0; i < message.length(); i++){
char c = message.charAt(i);
if (c != ' ') {
int value = letterCount.getOrDefault((int) c, 0);
letterCount.put((int) c, value + 1);
}
}
for(int key : letterCount.keySet()) {
System.out.println((char) key + ": " + letterCount.get(key));
}
}
}
好的,因此从这个问题中我了解到,您需要检查是否必须在整个字符串中计算字符串中的一对单词。我看到了您的代码,并认为它比所需的要复杂得多。请参见下面的代码段。
如果找到,请从映射中获取单词对的值,然后递增并将其重新设置。
String message = "The Quick brown fox jumps over the lazy brown dog the quick";
String[] split = message.split(" ");
Map<String, Integer> map = new HashMap<>();
int count = 0;
for (int i = 0; i < split.length - 1; i++) {
String temp = split[i] + " " + split[i + 1];
temp = temp.toLowerCase();
if (message.toLowerCase().contains(temp)) {
if (map.containsKey(temp))
map.put(temp, map.get(temp) + 1);
else
map.put(temp, 1);
}
}
System.out.println(map);
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