为什么这个for循环执行“else”代码块？

Question

我想我明白了第一个说法。这就是说，如果“单词”中的项目不在“频率”中，则添加它们并为其赋值1，对不对？ 我更困惑的是为什么“else”块被执行以及它是如何工作的。我理解输出，但并不完全如此。它显然认识到一个特定的词出现不止一次，但它是如何认识到这一点的？再说一遍，为什么如果第一个陈述是真的，它会进入“其他”区块？

public class Testing { 

    static List<String> list() { 

    List<String> words = new ArrayList<String>(); 

    words.add("Cherry"); 
    words.add("Banana"); 
    words.add("Apple"); 
    words.add("Banana"); 
    words.add("Berry"); 

    return words; 
    } 

    static Map<String, Integer> ArrayFrequencies(List<String> words) { 

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

    for (String elements : words) { 

     if (!frequencies.containsKey(elements)) { 
     frequencies.put(elements, 1); 
     } else { 
     frequencies.put(elements, frequencies.get(elements) + 1); 
     } 
    } 

    return frequencies; 
    } 

    public static void main(String[] args) { 

    System.out.println(ArrayFrequencies(list())); 
    } 
} 

输出：{苹果= 1，樱桃= 1，贝里= 1，香蕉= 2}

Answer 1

您的代码正常工作。 else语句只在需要时执行（当给定项目已经在列表中并且只需要增加其数量时）。在这里，我写了一些调试消息给你的代码。运行它，你会看到究竟发生了什么。

package strings; 
import java.util.ArrayList; 
import java.util.HashMap; 
import java.util.List; 
import java.util.Map; 

public class Testing { 

    static List<String> list() { 

    List<String> words = new ArrayList<String>(); 

    words.add("Cherry"); 
    words.add("Banana"); 
    words.add("Apple"); 
    words.add("Banana"); 
    words.add("Berry"); 
    words.add("Banana"); 
    words.add("Berry"); 
    words.add("Banana"); 

    return words; 
    } 

    static Map<String, Integer> ArrayFrequencies(List<String> words) { 

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

    for (String element : words) { 

     if (!frequencies.containsKey(element)) { 
      System.out.println("Seems like => " + element + " is not in the list yet. Adding it"); 
     frequencies.put(element, 1); 
     } else { 

      System.out.println("Seems like => " + element + " is in the list already. Incrementing its count from : " + 
           frequencies.get(element) + " => to : " + (frequencies.get(element) + 1)); 
     frequencies.put(element, frequencies.get(element) + 1); 
     } 
    } 

    return frequencies; 
    } 

    public static void main(String[] args) { 

    System.out.println(ArrayFrequencies(list())); 
    } 
} 

输出：

Seems like => Cherry is not in the list yet. Adding it 
Seems like => Banana is not in the list yet. Adding it 
Seems like => Apple is not in the list yet. Adding it 
Seems like => Banana is in the list already. Incrementing its count from : 1 => to : 2 
Seems like => Berry is not in the list yet. Adding it 
Seems like => Banana is in the list already. Incrementing its count from : 2 => to : 3 
Seems like => Berry is in the list already. Incrementing its count from : 1 => to : 2 
Seems like => Banana is in the list already. Incrementing its count from : 3 => to : 4 
{Apple=1, Cherry=1, Berry=2, Banana=4} 

Answer 2

if(!frequencies.containsKey(elements)) 
    { 
     //put the data in map for the first time 
     //Suppose Element "Apple" entering for the first time 
     frequencies.put(elements, 1); 
    } else { 
     //put the data in map if map already contains that element 
     //Element "Apple entering for the second time". Here it will get previous count and increase by one 
     frequencies.put(elements, frequencies.get(elements) + 1); 
    } 

Answer 3

什么我比较困惑的是，为什么 “其他” 块被执行以及它如何工作。

的if块设置地图项为1明确，因为它是第一个值与该elements键关联。

else块正在更新映射条目。它存在，它有一个价值，它需要增加。所以你说：

frequencies.get(elements) + 1 //get the prior value and add one to it 
frequencies.put(^^^^^^^)  //update the map entry with the new value from above. 

注意：不能从其他区块应用逻辑的，如果，因为你不能增加空。