本篇内容介绍了“使用stream的Collectors.toMap()方法常见问题如何解决”的有关知识，在实际案例的操作过程中，不少人都会遇到这样的困境，接下来就让小编带领大家学习一下如何处理这些情况吧！希望大家仔细阅读，能够学有所成！
使用stream的Collectors.toMap()方法常见问题
java8开始的流式编程很大程度上简化了我们的代码，提高了开发效率。
我们经常会使用到stream的Collectors.toMap(）来将List转换Map
在使用过程中有两个小坑需要注意
1、java.lang.IllegalStateException: Duplicate key
2、java.lang.NullPointerException
第一个是由于在List转Map过程中Map集合的key重复导致的；
第二个是由于在List转Map过程中Map集合的value有null导致的（当存在value值为空时，使用Collectors.toMap()会报NPE，因为底层调用了Map的merge方法，而map方法规定了此处的vlue不能为null，从而抛出空指针异常）；
解决方案
1、Collectors.toMap(dto ->key值 , dto -> dto,(v1,v2) -> v1)
在后面添加(v1,v2)->v1 指定选取第一个值 当key值重复的时候，根据情况而定选取第一个还是第二个）
2、自定义一个Map来接收，不使用Collectors.toMap()
第一种情况示例：
import com.google.common.collect.Lists;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import lombok.Data;
public class Test {
private static List<User> userList = Lists.newArrayList();
@Data
public static class User {
private String userCode;
private String userName;
}
/**
* 初始化数据
* (这里的userCode=10002重复)
*/
public static void initData() {
User user1 = new User();
user1.setUserCode("10001");
user1.setUserName("张三");
User user2 = new User();
user2.setUserCode("10002");
user2.setUserName("李四");
User user3 = new User();
user3.setUserCode("10002");
user3.setUserName("王五");
userList.add(user1);
userList.add(user2);
userList.add(user3);
}
public static void main(String[] args) {
initData();
//反例
// Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName));
//正例，在后面添加(u1,u2)->u1 指定选取第一个值 当key值重复的时候，根据情况而定选取第一个还是第二个
Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName, (u1, u2) -> u1));
System.out.println(userMap);
}
}第二种情况示例：
import com.google.common.collect.Lists;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import lombok.Data;
public class Test {
private static List<User> userList = Lists.newArrayList();
@Data
public static class User {
private String userCode;
private String userName;
}
/**
* 初始化数据
* (这里的userCode=10003的userName为空)
*/
public static void initData() {
User user1 = new User();
user1.setUserCode("10001");
user1.setUserName("张三");
User user2 = new User();
user2.setUserCode("10002");
user2.setUserName("李四");
User user3 = new User();
user3.setUserCode("10003");
user3.setUserName(null);
userList.add(user1);
userList.add(user2);
userList.add(user3);
}
public static void main(String[] args) {
initData();
//反例
// Map<String, String> userMap = userList.stream().collect(Collectors.toMap(User::getUserCode, User::getUserName));
//正例 (如果对转换后的顺序有要求，这里还可以使用LinkedHashMap)
Map<String, String> userMap = userList.stream().collect(HashMap::new, (map, user) -> map.put(user.getUserCode(), user.getUserName()), HashMap::putAll);
System.out.println(userMap);
}
}Stream ToMap(Collectors.toMap) 实践
Requirements
List TO Map
List Stream 转换 Map时向collect()方法中传递Collector对象，对象由Collectors.toMap()方法返回。
如下实现List转换为Map
List<GroupBrandCateBO> list = new ArrayList<>(
Arrays.asList(
new GroupBrandCateBO("v1", "g1", "b1"),
new GroupBrandCateBO("v1", "g1", "b1"),
new GroupBrandCateBO("v3", "g3", "b3")
)
);
Map<String, String> map = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal, LinkedHashMap::new));
System.out.println(map.getClass());
Map<String, String> map0 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal));
System.out.println(map0.getClass());
System.out.println(map0.toString());
Map<String, String> map1 = list.stream().collect(Collectors.toMap(GroupBrandCateBO::getVersion, GroupBrandCateBO::getGroupCode));
System.out.println(map1.toString());Console
class java.util.LinkedHashMap
class java.util.HashMap
{v1=g1, v3=g3}
Exception in thread “main” java.lang.IllegalStateException: Duplicate key g1
at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)
…
问题分析
toMap()函数重载：
未指定合并函数mergeFunction情况下，传入throwingMerger()返回BinaryOperator对象，当出现key重复时，调用合并函数！
未指定Supplier实例情况下，默认生成HashMap实例。
public static <T, K, U>
Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
Function<? super T, ? extends U> valueMapper) {
return toMap(keyMapper, valueMapper, throwingMerger(), HashMap::new);
}
public static <T, K, U>
Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
Function<? super T, ? extends U> valueMapper,
BinaryOperator<U> mergeFunction) {
return toMap(keyMapper, valueMapper, mergeFunction, HashMap::new);
}
public static <T, K, U, M extends Map<K, U>>
Collector<T, ?, M> toMap(Function<? super T, ? extends K> keyMapper,
Function<? super T, ? extends U> valueMapper,
BinaryOperator<U> mergeFunction,
Supplier<M> mapSupplier) {
BiConsumer<M, T> accumulator
= (map, element) -> map.merge(keyMapper.apply(element),
valueMapper.apply(element), mergeFunction);
return new CollectorImpl<>(mapSupplier, accumulator, mapMerger(mergeFunction), CH_ID);
}
private static <T> BinaryOperator<T> throwingMerger() {
return (u,v) -> { throw new IllegalStateException(String.format("Duplicate key %s", u)); };
}补充
关于合并函数
List<GroupBrandCateBO> list = new ArrayList<>(
Arrays.asList(
new GroupBrandCateBO("v1", "g1", "b1"),
new GroupBrandCateBO("v1", "g2", "b2"),
new GroupBrandCateBO("v1", "g2", "b2"),
new GroupBrandCateBO("v3", "g3", "b3")
)
);
Map<String, String> map00 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> currVal));
Map<String, String> map01 = list.stream().collect(Collectors.toMap(item -> item.getVersion(), item -> item.getGroupCode(), (oldVal, currVal) -> oldVal + currVal));
System.out.println(map00.toString());
System.out.println(map01.toString());Console
{v1=g2, v3=g3}
{v1=g1g2g2, v3=g3}
传入Lambda表达式将转化为BinaryOperator<U> mergeFunction对象，合并处理value，非Key！！！
比如：
(oldVal, currVal) -> currVal) // key相同时当前值替换原始值
(oldVal, currVal) -> oldVal + currVal //key相同时保留原始值和当前值