java 如何将集合转换为映射

how to convert a set to map

提问by oshai

I have a set and I want to convert it to map in order to use it later in guava's Maps.difference(). I only care about the keys in the difference.
Came up with this version:

我有一套，我想将它转换为地图，以便稍后在番石榴的Maps.difference(). 我只关心差异中的键。
想出了这个版本：

private <T> Map<T, T> toMap(Set<T> set) {
  return set.stream().collect(Collectors.toMap(Function.identity(), Function.identity()));
}

However, I know that usually, a set has a backing field of map. This is the method I use to create the map:

但是，我知道通常，一个集合有一个 map 的支持字段。这是我用来创建地图的方法：

public static <E> Set<E> newConcurrentHashSet() {
  return Collections.newSetFromMap(new ConcurrentHashMap<E, Boolean>());
}

Since I only need the keys I thought maybe I can get a view of this field somehow. any idea?

因为我只需要钥匙，我想也许我可以以某种方式了解这个领域。任何的想法？

回答by Shane

I ended up with a fairly straight-forward one line solution with Java 8 as follows:

我最终得到了一个相当简单的 Java 8 单行解决方案，如下所示：

Map<String, Foo> map = fooSet.stream().collect(Collectors.toMap(Foo::getKey, e -> e));

• fooSet is a set of objects of type Foo, i.e. Set<Foo> fooSet
• Foo has a getter called getKeywhich returns a String

• fooSet 是一组 Foo 类型的对象，即 Set<Foo> fooSet
• Foo 有一个 getter 调用getKey它返回一个字符串

回答by developer

You can convert a Setto a Map(same key and value taken from elements of Set) as shown below:

您可以将 a 转换Set为 a Map（取自 的元素的相同键和值Set），如下所示：

private <T> Map<T, T> toMap(Set<T> set) {
    Map<T, T> map = new ConcurrentHashMap<>();
    set.forEach(t -> map.put(t, t));//contains same key and value pair
    return map;
}

回答by Andreas

From comment:

来自评论：

I would like to know which items only on left, which only on right, which in common (similar to map difference)

我想知道哪些项目只在左边，哪些只在右边，哪些是共同点（类似于地图差异）

Use removeAll()and [retainAll()][3].

使用removeAll()和 [ retainAll()][3].

Example:

例子：

Set<Integer> set1 = new HashSet<>(Arrays.asList(1,3,5,7,9));
Set<Integer> set2 = new HashSet<>(Arrays.asList(3,4,5,6,7));

Set<Integer> onlyIn1 = new HashSet<>(set1);
onlyIn1.removeAll(set2);

Set<Integer> onlyIn2 = new HashSet<>(set2);
onlyIn2.removeAll(set1);

Set<Integer> inBoth = new HashSet<>(set1);
inBoth.retainAll(set2);

System.out.println("set1: " + set1);
System.out.println("set2: " + set2);
System.out.println("onlyIn1: " + onlyIn1);
System.out.println("onlyIn2: " + onlyIn2);
System.out.println("inBoth : " + inBoth);

Output

输出

set1: [1, 3, 5, 7, 9]
set2: [3, 4, 5, 6, 7]
onlyIn1: [1, 9]
onlyIn2: [4, 6]
inBoth : [3, 5, 7]

---

---

Now, if you want to know all values and where they were found, you can do this (Java 8):

现在，如果您想知道所有值以及它们的位置，您可以这样做（Java 8）：

Set<Integer> setA = new HashSet<>(Arrays.asList(1,3,5,7,9));
Set<Integer> setB = new HashSet<>(Arrays.asList(3,4,5,6,7));

Map<Integer, String> map = new HashMap<>();
for (Integer i : setA)
    map.put(i, "In A");
for (Integer i : setB)
    map.compute(i, (k, v) -> (v == null ? "In B" : "In Both"));

System.out.println("setA: " + setA);
System.out.println("setB: " + setB);
map.entrySet().stream().forEach(System.out::println);

Output

输出

setA: [1, 3, 5, 7, 9]
setB: [3, 4, 5, 6, 7]
1=In A
3=In Both
4=In B
5=In Both
6=In B
7=In Both
9=In A

回答by Mantas

See similar answer here.

在这里看到类似的答案。

Assuming that your original set is a set of values (no keys in original data!), you will need to specify keys for the newly created map. Guava's Maps.uniqueIndexmight be helpful (see here)

假设您的原始集合是一组值（原始数据中没有键！），您将需要为新创建的映射指定键。番石榴Maps.uniqueIndex可能会有所帮助（请参阅此处）

Otherwise, if your original set is a set of keys (no values in original data!) that you want to retain, you will need to specify default or specific values for the newly created map. Guava's Maps.toMapmight be helpful here. (See more here)

否则，如果您的原始集合是要保留的一组键（原始数据中没有值！），您将需要为新创建的映射指定默认值或特定值。番石榴Maps.toMap可能会在这里有所帮助。（在这里查看更多）

回答by Bj?rn Kahlert

Improvement of developer's answer:

改进开发人员的回答：

Map<String, Foo> map = fooSet.stream().collect(Collectors.toMap(Foo::getKey, Function.identity()));

or if you statically import Collectors.toMapand Function.identity:

或者，如果您静态导入Collectors.toMap和Function.identity：

Map<String, Foo> map = fooSet.stream().collect(toMap(Foo::getKey, identity()));