英文:
Is there a way to split a list of objects into n number of lists depending on the number of unique values of an object field?
问题
I have a method that returns a list of objects.
In that object, there is a String field orderType.
Is there a way for me to separate that list into n number of lists depending on the unique values of orderType?
For example, if the method returns a list containing 10 objects, then 4 of them have an orderType value of x, 3 have an orderType value of y, and another 4 have a value of z.
Then the result will be 3 lists containing 4, 3, 4 objects respectively.
orderType does not only have x, y, and z as possible values.
I am having a hard time putting this into actual code, but my idea is like this:
List<Orders> orderList = getOrder();for (int i = 0; i < orderList.size(); i++) {if (orderList.get(i).orderType is unique) {create a new list} else {add to an existing list having the same orderType}}
英文:
I have a method that returns a list of objects.
In that object, there is a String field orderType.
Is there a way for me to separate that list into n number of lists depending on the unique values of orderType?
For example the method returns a list containing 10 objects, then 4 of them has orderType value of x, 3 has orderType value of y, then another 4 has a value of z.
then the result will be 3 lists containing 4,3,4 objects respectively
orderType does not only have x, y and z as possible values.
i am having a hard time putting this into an actual code but my idea is like this
List<Orders> orderList = getOrder();for(int i=0; i<orderList.size();i++){if(orderList.get(i).orderType is unique){create a new list}else{add to an existing list having the same orderType}}
答案1
得分: 1
以下是翻译后的代码部分：
我建议使用这个相对通用的函数：```javapublic static <T, U> Map<U, List<T>> getLists(List<T> values, Function<T, U> orderBy) {Map<U, List<T>> lists = new HashMap<>();for (T value : values) {U type = orderBy.apply(value);if (!lists.containsKey(type)) {lists.put(type, new ArrayList<>());}lists.get(type).add(value);}return lists;}
订单类：
class Order {private String orderType;private String name;public Order(String orderType, String name) {this.orderType = orderType;this.name = name;}public String getOrderType() {return orderType;}public String getName() {return name;}}
如何使用：
public static void main(String args[]){List<Order> orders = new ArrayList<>();orders.add(new Order("TYPE1", "Order1"));orders.add(new Order("TYPE2", "Order2"));orders.add(new Order("TYPE1", "Order3"));orders.add(new Order("TYPE1", "Order4"));orders.add(new Order("TYPE2", "Order5"));orders.add(new Order("TYPE1", "Order6"));orders.add(new Order("TYPE1", "Order7"));orders.add(new Order("TYPE1", "Order8"));Map<String, List<Order>> map = getLists(orders, Order::getOrderType);for (String key : map.keySet()) {System.out.println(key);for (Order order : map.get(key)) {System.out.println(order.getOrderType() + " " + order.getName());}System.out.println();}}
请注意，代码中的HTML实体（如<和"）已被移除，以便更好地显示代码。
英文:
I'd suggest this rather general function:
public static <T, U> Map<U, List<T>> getLists(List<T> values, Function<T, U> orderBy) {Map<U, List<T>> lists = new HashMap<>();for (T value : values) {U type = orderBy.apply(value);if (!lists.containsKey(type)) {lists.put(type, new ArrayList<>());}lists.get(type).add(value);}return lists;}
Order Class:
class Order {private String orderType;private String name;public Order(String orderType, String name) {this.orderType = orderType;this.name = name;}public String getOrderType() {return orderType;}public String getName() {return name;}}
How to use:
public static void main(String args[]){List<Order> orders = new ArrayList<>();orders.add(new Order("TYPE1", "Order1"));orders.add(new Order("TYPE2", "Order2"));orders.add(new Order("TYPE1", "Order3"));orders.add(new Order("TYPE1", "Order4"));orders.add(new Order("TYPE2", "Order5"));orders.add(new Order("TYPE1", "Order6"));orders.add(new Order("TYPE1", "Order7"));orders.add(new Order("TYPE1", "Order8"));Map<String, List<Order>> map = getLists(orders, Order::getOrderType);for (String key : map.keySet()) {System.out.println(key);for (Order order : map.get(key)) {System.out.println(order.getOrderType() + " " + order.getName());}System.out.println();}}
答案2
得分: 0
我认为更容易将元素提取到一个Map中，然后您可以利用Collectors.toMap的第三个参数，将相同键的值合并起来。
public void tryIt() {// 给定类class Foo {String x;Foo(String x) {this.x = x;}@Overridepublic String toString() {return this.x;}}List<Foo> foos = new ArrayList<>();IntStream.range(0, 4).forEach(i -> {foos.add(new Foo("x"));foos.add(new Foo("z"));});IntStream.range(0, 3).forEach(i -> foos.add(new Foo("y")));// 创建一个元素到原始对象列表的映射，根据相同的元素键进行分组Map<String, List<Foo>> collect = foos.stream().collect(Collectors.toMap(f -> f.x,Collections::singletonList,(l1, l2) -> Stream.concat(l1.stream(), l2.stream()).collect(Collectors.toList())));System.out.println(collect);}
输出
{x=[x, x, x, x], y=[y, y, y], z=[z, z, z, z]}
英文:
I think it's easier to pull out the element to a Map, then you can take advantage of the Collectors.toMap third parameter that combines the values of the same key.
public void tryIt() {// givenclass Foo {String x;Foo(String x) {this.x = x;}@Overridepublic String toString() {return this.x;}}List<Foo> foos = new ArrayList<>();IntStream.range(0, 4).forEach(i -> {foos.add(new Foo("x"));foos.add(new Foo("z"));});IntStream.range(0, 3).forEach(i -> foos.add(new Foo("y")));// create a mapping of element, to list of original object, grouping by same element keyMap<String, List<Foo>> collect = foos.stream().collect(Collectors.toMap(f -> f.x,Collections::singletonList,(l1, l2) -> Stream.concat(l1.stream(), l2.stream()).collect(Collectors.toList())));System.out.println(collect);}
Output
{x=[x, x, x, x], y=[y, y, y], z=[z, z, z, z]}
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