java - 用于显示 Java ArrayList 中唯一值出现次数的字符串

Question

我有一个字符串，正在将其转换为 ArrayList，目的是计算列表中特定值的数量。

我的字符串可能如下所示:

String myString = "Living Room, Bedroom, Bedroom, Bedroom, Bathroom, Bathroom, Family Room."

我目前正在通过这种方式查找唯一值出现的次数:

Map<String, Long> count = Stream.of(myString.split(",")).collect(Collectors.groupingBy(w -> w, Collectors.counting()));

最好的方法是什么，以便我可以打印类似“我的家包括 1 间客厅、3 间卧室、2 间浴室和 1 间家庭 Activity 室”的内容。

<小时/>

目前，我的解决方案如下。我还没有接受答案，因为我认为我遵循的方式有点笨拙:

String newDescription = "My Home includes ";

public String getDescription(){
    String myDescription = home.getDescription();

    Map<String, Long> count = Stream.of(myDescription.split(",")).collect(Collectors.groupingBy(w -> w, Collectors.counting()));

    for (Map.Entry<String, Long> entry : count.entrySet()){
        newDescription = newDescription + entry.getValue() + " " + entry.getKey() + "(s), ";
    }
    return newDescription;
}

<小时/>

我之前是这样处理的，但另一个 SO 用户建议进行更改:

public ArrayList convertDescription(){
    //getDescription is inherited from another class to provide a string like the myString example above
    String myDescription = home.getDescription(); 

    ArrayList<String>Map<String, listDescriptionLong> =count new= ArrayList<String>(ArraysStream.asListof(myDescription.split(",")));

    int livingRoomOccurrences = Collections.frequencycollect(listDescription, "Living Room");
    int bedroomOccurrences = CollectionsCollectors.frequencygroupingBy(listDescription, "Bedroom");
    int bathroomOccurrencesw =-> Collections.frequency(listDescriptionw, "Bathroom");
    int familyRoomOccurrences = CollectionsCollectors.frequencycounting(listDescription, "Family Room");
    int kitchenOccurrences = Collections.frequency(listDescription, "Kitchen");
    int walkInClosetOccurrences = Collections.frequency(listDescription, "Walk-in Closet");
    //not sure what I should return to make it easy to override getDescription with the proper string formatting
    return listDescription;
}

public String getDescription(){
    return home.getDescription();
}

Answer 1

我建议你获取所有元素的频率并提取你需要的内容，而不是一遍又一遍地扫描集合。

Map<String, Long> count = Stream.of(myString.split(","))
                            .collect(Collectors.groupingBy(w -> w, Collectors.counting());