如何在Java中初始化一个静态Map ?
方法一:静态初始化器 方法二:实例初始化器(匿名子类) 或 还有别的方法吗?
它们各自的优点和缺点是什么?
下面是一个例子来说明这两种方法:
import java.util.HashMap;
import java.util.Map;
public class Test {
private static final Map<Integer, String> myMap = new HashMap<>();
static {
myMap.put(1, "one");
myMap.put(2, "two");
}
private static final Map<Integer, String> myMap2 = new HashMap<>(){
{
put(1, "one");
put(2, "two");
}
};
}
当前回答
这一个使用Apache common -lang，它很可能已经在你的类路径上了:
Map<String, String> collect = Stream.of(
Pair.of("hello", "world"),
Pair.of("abc", "123"),
Pair.of("java", "eight")
).collect(Collectors.toMap(Pair::getKey, Pair::getValue));
其他回答
使用Eclipse Collections，以下所有功能都可以工作:
import java.util.Map;
import org.eclipse.collections.api.map.ImmutableMap;
import org.eclipse.collections.api.map.MutableMap;
import org.eclipse.collections.impl.factory.Maps;
public class StaticMapsTest
{
private static final Map<Integer, String> MAP =
Maps.mutable.with(1, "one", 2, "two");
private static final MutableMap<Integer, String> MUTABLE_MAP =
Maps.mutable.with(1, "one", 2, "two");
private static final MutableMap<Integer, String> UNMODIFIABLE_MAP =
Maps.mutable.with(1, "one", 2, "two").asUnmodifiable();
private static final MutableMap<Integer, String> SYNCHRONIZED_MAP =
Maps.mutable.with(1, "one", 2, "two").asSynchronized();
private static final ImmutableMap<Integer, String> IMMUTABLE_MAP =
Maps.mutable.with(1, "one", 2, "two").toImmutable();
private static final ImmutableMap<Integer, String> IMMUTABLE_MAP2 =
Maps.immutable.with(1, "one", 2, "two");
}
您还可以使用Eclipse Collections静态地初始化原始映射。
import org.eclipse.collections.api.map.primitive.ImmutableIntObjectMap;
import org.eclipse.collections.api.map.primitive.MutableIntObjectMap;
import org.eclipse.collections.impl.factory.primitive.IntObjectMaps;
public class StaticPrimitiveMapsTest
{
private static final MutableIntObjectMap<String> MUTABLE_INT_OBJ_MAP =
IntObjectMaps.mutable.<String>empty()
.withKeyValue(1, "one")
.withKeyValue(2, "two");
private static final MutableIntObjectMap<String> UNMODIFIABLE_INT_OBJ_MAP =
IntObjectMaps.mutable.<String>empty()
.withKeyValue(1, "one")
.withKeyValue(2, "two")
.asUnmodifiable();
private static final MutableIntObjectMap<String> SYNCHRONIZED_INT_OBJ_MAP =
IntObjectMaps.mutable.<String>empty()
.withKeyValue(1, "one")
.withKeyValue(2, "two")
.asSynchronized();
private static final ImmutableIntObjectMap<String> IMMUTABLE_INT_OBJ_MAP =
IntObjectMaps.mutable.<String>empty()
.withKeyValue(1, "one")
.withKeyValue(2, "two")
.toImmutable();
private static final ImmutableIntObjectMap<String> IMMUTABLE_INT_OBJ_MAP2 =
IntObjectMaps.immutable.<String>empty()
.newWithKeyValue(1, "one")
.newWithKeyValue(2, "two");
}
注意:我是Eclipse Collections的提交者
我喜欢匿名类语法;只是代码更少。然而，我发现的一个主要缺点是，您将无法通过远程序列化该对象。您将得到一个关于无法在远程端找到匿名类的异常。
Java 9
我们可以用地图。ofEntries，调用Map。条目(k, v)来创建每个条目。
import static java.util.Map.entry;
private static final Map<Integer,String> map = Map.ofEntries(
entry(1, "one"),
entry(2, "two"),
entry(3, "three"),
entry(4, "four"),
entry(5, "five"),
entry(6, "six"),
entry(7, "seven"),
entry(8, "eight"),
entry(9, "nine"),
entry(10, "ten"));
我们也可以使用Map。如Tagir在他的回答中所建议的，但我们不能使用Map.of有超过10个条目。
Java 8
我们可以创建一个映射条目流。在java.util.AbstractMap中我们已经有两个Entry的实现，它们是SimpleEntry和SimpleImmutableEntry。在这个例子中，我们可以使用former as:
import java.util.AbstractMap.*;
private static final Map<Integer, String> myMap = Stream.of(
new SimpleEntry<>(1, "one"),
new SimpleEntry<>(2, "two"),
new SimpleEntry<>(3, "three"),
new SimpleEntry<>(4, "four"),
new SimpleEntry<>(5, "five"),
new SimpleEntry<>(6, "six"),
new SimpleEntry<>(7, "seven"),
new SimpleEntry<>(8, "eight"),
new SimpleEntry<>(9, "nine"),
new SimpleEntry<>(10, "ten"))
.collect(Collectors.toMap(SimpleEntry::getKey, SimpleEntry::getValue));
在这种情况下，我绝不会创建匿名子类。静态初始化器同样有效，如果你想让映射不可修改，例如:
private static final Map<Integer, String> MY_MAP;
static
{
Map<Integer, String>tempMap = new HashMap<Integer, String>();
tempMap.put(1, "one");
tempMap.put(2, "two");
MY_MAP = Collections.unmodifiableMap(tempMap);
}
如果你可以使用字符串表示你的数据，这也是一个选项在Java 8:
static Map<Integer, String> MAP = Stream.of(
"1=one",
"2=two"
).collect(Collectors.toMap(k -> Integer.parseInt(k.split("=")[0]), v -> v.split("=")[1]));
