字符串的计数频率 [英] Counting frequency of a string

问题描述

我基本上要搜索的字符串的频率。例如，如果我通过在我，然后在下面的句子中词的频率：我去了海滩和我只见三人应该是2，我已经构造中，我采取（任意长度的）文本这样的方法，将它分成数组的空白，并遍历数组，如果每个指标匹配单词搜索。然后，我递增频率计数器，并返回数为字符串。这里的方法：

I essentially want to search the frequency of a string. For example, if I pass in the word "I", then the frequency of the word in the following sentence: "I went to the beach and I saw three people" should be 2. I've constructed such method in which I take a text (of any length), split it into an array by the white space, and loop through the array, searching if each index matches the word. Then, I increment the frequency counter and return the number as a string. Here's the method:

private int freq() {
String text = "I went to the beach and I saw three people";
String search = "I";
String[] splitter = text.split("\\s+");
int counter = 0;
   for (int i=0; i<splitter.length; i++)
   {
       if (splitter[i]==search) 
       {
           counter++;
       }
       else
       {

       }
   }
   return counter;
       }

}  

这是法外：

String final = Integer.toString(freq());
System.out.println(final);

但正如我运行它，我不断收到0作为结果。我不知道我做错了。

But as I run this, I keep getting 0 as the result. I don't know what I'm doing wrong.

编辑：你是正确的！什么是一个问题:(。

You're all correct! What a waste of a question :(.

推荐答案

使用等于而不是 ==

if (text[i].equals(search) )
   {
       counter++;
   }

更好的解决方案

使用一个Map的话地图＆LT地图，字符串，整数方​​式＆gt; 频率

Use a Map to map the words Map<String,Integer> with frequency.

String [] words = line.split(" ");

Map<String,Integer> frequency = new HashMap<String,Integer>();

for (String word:words){

    Integer f = frequency.get(word);
    //checking null
    if(f==null) f=0;
    frequency.put(word,f+1);
}

然后你就可以找出一个特定的词：

Then you can find out for a particular word with:

frequency.get(word)

这篇关于字符串的计数频率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！