為什么為此for循環執行“ else”代碼塊？

Question

我想我明白第一句話。 這就是說，如果“單詞”中的項目不在“頻率”中，則將它們相加並為每個項目賦值1，對不對？ 我更困惑的是為什么執行“ else”塊及其工作方式。 我了解輸出，但不太清楚它是如何工作的。 它顯然認識到一個特定的單詞出現不止一次，但是它如何識別呢？ 再一次，如果第一個語句為真，為什么它將轉到“ else”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：{Apple = 1，Cherry = 1，Berry = 1，Banana = 2}

Answer 1

  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 2

我更困惑的是為什么執行“ else”塊及其工作方式。

if塊將映射條目顯式設置為1，因為它是與elements鍵關聯的第一個值。

else塊正在更新地圖條目。 它存在，具有一個值，並且需要增加。 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 ：您不能將else塊中的邏輯應用於if，因為您不能遞增null。

Answer 3

您的代碼表現正常。 else語句僅在需要時才執行（當給定項目已在列表中並且您只需要增加其計數時）。 在這里，我向您的代碼編寫了一些調試消息。 運行它，您將確切看到正在發生的事情。

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}