收集基于相似索引的列表集合

Question

我有一个像这样的集合： List<List<Object>> firstList

我想将一个类似的模式列表组合在一起： List<List<Object>> secondList 但按索引分组。

说

firstList [1]:
   0 = {Object A}"
   1 = {Object B}"
   2 = {Object C}"

firstList [2]:
   0 = {Object A}"
   1 = {Object B}"
   2 = {Object C}"

我想将此集合分组为

secondList [1]:
   0 = {Object A}"
   1 = {Object A}"

secondList [2]:
   0 = {Object B}"
   1 = {Object B}"

secondList [3]:
   0 = {Object C}"
   1 = {Object C}"

到目前为止我尝试过的是

for (int i = 0; i <firstList.size() ; i++) {
    List<Object> list = firstList.get(i);
    List<Object> rlPr = new ArrayList<>();

    for (int j = 0; j <list.size()-1; j++) {
        rlPr.add(list.get(i));
    }
    secondList.add(rlPr);
}

但我没有得到预期的结果。 我正在使用 java 8。

编辑：所有列表的大小都相同

Answer 1

您可以使用Map 按索引对值进行分组：

Map<Integer, List<Object>> map = new TreeMap<>();
for (List<Object> objects : firstList) {
    for (int i = 0, l = objects.size(); i < l; i++) {
        map.computeIfAbsent(i, k -> new ArrayList<>()).add(objects.get(i));
    }
}

然后返回List：

List<List<Object>> secondList = new ArrayList<>(map.values());

Answer 2

首先，获取子列表的最大长度。在您的情况下，两者都是 3。遍历所有这些索引，从该索引的每个子列表中获取值并将这些值收集到一个新的子列表中。

final int maxLengthOfSublist = list1.stream().mapToInt(List::size).max().orElse(0);
final List<List<Integer>> list2 = IntStream.range(0, maxLengthOfSublist)
    .mapToObj(idx -> list1.stream()
        .filter(sublist -> idx < sublist.size())
        .map(sublist -> sublist.get(idx))
        .collect(Collectors.toList())
    )
    .collect(Collectors.toList());

即使子列表的长度不同，这也会起作用。

Answer 3

假设主列表中的列表大小相同，您可以执行以下操作...

    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.Collections;
    import java.util.List;

    public class OhCrikey {

        public static void main(String[] args) {
            List<List<String>> lists = Arrays.asList(
                    Arrays.asList("1a", "2a", "3a"),
                    Arrays.asList("1b", "2b", "3b"),
                    Arrays.asList("1c", "2c", "3c")
            );

            List<List<String>> result = transform(lists);

            result.forEach(System.out::println);
        }

        private static <T> List<List<T>> transform(List<List<T>> lists) {
            if(lists == null || lists.isEmpty()) {
                return Collections.emptyList();
            }
            // Check each sub-list contains the same number of elements
            checkAllSameSize(lists);

            // Initialise the results list
            List<List<T>> result = new ArrayList<>();

            // Get the size of each sub-list
            int totalLists = lists.get(0).size();

            // Fill up the results list with 'totalLists' empty lists
            for(int k = 0; k < totalLists; k++) {
                result.add(new ArrayList<>());
            }

            // For each input list...
            lists.forEach(list -> {
                // Iterate over it and add the kth element to the kth result list.
                for(int k = 0; k < list.size(); k++) {
                    result.get(k).add(list.get(k));
                }
            });

            return result;
        }

        private static <T> void checkAllSameSize(List<List<T>> lists) {
            int size = lists.get(0).size();

            // Make sure each list has the same size as the first list
            lists.forEach(list -> {
                if(list.size() != size) {
                    throw new AssertionError("All child lists must have same number of elements.");
                }
            });
        }
    }

打印...

    [1a, 1b, 1c]
    [2a, 2b, 2c]
    [3a, 3b, 3c]

Answer 4

一种流式处理方式是（将Object 替换为CustomObject）：

List<List<CustomObject>> secondList = new ArrayList<>( // Collection to ArrayList
       firstList.stream() // Stream<List<CustomObject>>
                .flatMap(Collection::stream) // Stream<CustomObject>
                .collect(Collectors.groupingBy(Function.identity())) // Map<CustomObject, List<CustomObject>>
                .values()); // Collection<List<CustomObject>>