爲什麼這個for循環執行「else」代碼塊？

Question

我想我明白了第一個說法。這就是說，如果「單詞」中的項目不在「頻率」中，則添加它們併爲其賦值1，對不對？ 我更困惑的是爲什麼「else」塊被執行以及它是如何工作的。我理解輸出，但並不完全如此。它顯然認識到一個特定的詞出現不止一次，但它是如何認識到這一點的？再說一遍，爲什麼如果第一個陳述是真的，它會進入「其他」區塊？

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())); 
    } 
} 

輸出：{蘋果= 1，櫻桃= 1，貝里= 1，香蕉= 2}

Answer 1

您的代碼正常工作。 else語句只在需要時執行（當給定項目已經在列表中並且只需要增加其數量時）。在這裏，我寫了一些調試消息給你的代碼。運行它，你會看到究竟發生了什麼。

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())); 
    } 
} 

輸出：

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

什麼我比較困惑的是，爲什麼 「其他」 塊被執行以及它如何工作。

的if塊設置地圖項爲1明確，因爲它是第一個值與該elements鍵關聯。

else塊正在更新映射條目。它存在，它有一個價值，它需要增加。所以你說：

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：不能從其他區塊應用邏輯的，如果，因爲你不能增加空。