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
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));
其他回答
和往常一样，apache-commons有合适的方法MapUtils。putAll(地图、对象[]):
例如，要创建一个彩色地图:
Map<String, String> colorMap = MapUtils.putAll(new HashMap<String, String>(), new String[][] {
{"RED", "#FF0000"},
{"GREEN", "#00FF00"},
{"BLUE", "#0000FF"}
});
Java 8与流:
private static final Map<String, TemplateOpts> templates = new HashMap<>();
static {
Arrays.stream(new String[][]{
{CUSTOMER_CSV, "Plantilla cliente", "csv"}
}).forEach(f -> templates.put(f[0], new TemplateOpts(f[1], f[2])));
}
它也可以是Object[][]，用于在forEach循环中放入任何东西并将其映射
我喜欢使用静态初始化“技术”，当我有一个抽象类的具体实现，它定义了一个初始化构造函数，但没有默认构造函数，但我希望我的子类有一个默认构造函数。
例如:
public abstract class Shape {
public static final String COLOR_KEY = "color_key";
public static final String OPAQUE_KEY = "opaque_key";
private final String color;
private final Boolean opaque;
/**
* Initializing constructor - note no default constructor.
*
* @param properties a collection of Shape properties
*/
public Shape(Map<String, Object> properties) {
color = ((String) properties.getOrDefault(COLOR_KEY, "black"));
opaque = (Boolean) properties.getOrDefault(OPAQUE_KEY, false);
}
/**
* Color property accessor method.
*
* @return the color of this Shape
*/
public String getColor() {
return color;
}
/**
* Opaque property accessor method.
*
* @return true if this Shape is opaque, false otherwise
*/
public Boolean isOpaque() {
return opaque;
}
}
以及这个类的具体实现——但它想要/需要一个默认构造函数:
public class SquareShapeImpl extends Shape {
private static final Map<String, Object> DEFAULT_PROPS = new HashMap<>();
static {
DEFAULT_PROPS.put(Shape.COLOR_KEY, "yellow");
DEFAULT_PROPS.put(Shape.OPAQUE_KEY, false);
}
/**
* Default constructor -- intializes this square to be a translucent yellow
*/
public SquareShapeImpl() {
// the static initializer was useful here because the call to
// this(...) must be the first statement in this constructor
// i.e., we can't be mucking around and creating a map here
this(DEFAULT_PROPS);
}
/**
* Initializing constructor -- create a Square with the given
* collection of properties.
*
* @param props a collection of properties for this SquareShapeImpl
*/
public SquareShapeImpl(Map<String, Object> props) {
super(props);
}
}
然后要使用这个默认构造函数，只需执行以下操作:
public class StaticInitDemo {
public static void main(String[] args) {
// create a translucent, yellow square...
Shape defaultSquare = new SquareShapeImpl();
// etc...
}
}
注意:这个答案实际上属于问题如何直接初始化一个HashMap(在字面上)?但由于在写这篇文章时，它被标记为这篇文章的副本……
在Java 9的Map.of()之前(它也被限制为10个映射)，你可以扩展你选择的Map实现，例如:
public class InitHashMap<K, V> extends HashMap<K, V>
重新实现HashMap的构造函数:
public InitHashMap() {
super();
}
public InitHashMap( int initialCapacity, float loadFactor ) {
super( initialCapacity, loadFactor );
}
public InitHashMap( int initialCapacity ) {
super( initialCapacity );
}
public InitHashMap( Map<? extends K, ? extends V> map ) {
super( map );
}
并添加一个额外的构造函数，它受到Aerthel的答案的启发，但通过使用Object…和<K, V>类型:
public InitHashMap( final Object... keyValuePairs ) {
if ( keyValuePairs.length % 2 != 0 )
throw new IllegalArgumentException( "Uneven number of arguments." );
K key = null;
int i = -1;
for ( final Object keyOrValue : keyValuePairs )
switch ( ++i % 2 ) {
case 0: // key
if ( keyOrValue == null )
throw new IllegalArgumentException( "Key[" + (i >>> 1) + "] is <null>." );
key = (K) keyOrValue;
continue;
case 1: // value
put( key, (V) keyOrValue );
}
}
Run
public static void main( final String[] args ) {
final Map<Integer, String> map = new InitHashMap<>( 1, "First", 2, "Second", 3, "Third" );
System.out.println( map );
}
输出
{1=First, 2=Second, 3=Third}
你也可以扩展Map接口:
public interface InitMap<K, V> extends Map<K, V> {
static <K, V> Map<K, V> of( final Object... keyValuePairs ) {
if ( keyValuePairs.length % 2 != 0 )
throw new IllegalArgumentException( "Uneven number of arguments." );
final Map<K, V> map = new HashMap<>( keyValuePairs.length >>> 1, .75f );
K key = null;
int i = -1;
for ( final Object keyOrValue : keyValuePairs )
switch ( ++i % 2 ) {
case 0: // key
if ( keyOrValue == null )
throw new IllegalArgumentException( "Key[" + (i >>> 1) + "] is <null>." );
key = (K) keyOrValue;
continue;
case 1: // value
map.put( key, (V) keyOrValue );
}
return map;
}
}
Run
public static void main( final String[] args ) {
System.out.println( InitMap.of( 1, "First", 2, "Second", 3, "Third" ) );
}
输出
{1=First, 2=Second, 3=Third}
这是我最喜欢的
不想(或不能)使用Guava的ImmutableMap.of() 或者我需要一个可变Map 或者我需要从JDK9+的Map.of()中超过10个条目限制
public static <A> Map<String, A> asMap(Object... keysAndValues) {
return new LinkedHashMap<String, A>() {{
for (int i = 0; i < keysAndValues.length - 1; i++) {
put(keysAndValues[i].toString(), (A) keysAndValues[++i]);
}
}};
}
它非常紧凑，并且忽略了杂散值(即没有值的最终键)。
用法:
Map<String, String> one = asMap("1stKey", "1stVal", "2ndKey", "2ndVal");
Map<String, Object> two = asMap("1stKey", Boolean.TRUE, "2ndKey", new Integer(2));
