Question

我有一个List并需要将其转换为嵌套（2级）地图，其中list由2个不同的对象属性分组，值应使用第三个属性。例如：

[A("1","2","3"),A("1","2","4"),A("1","3","4"),A("2","3","5")]

应转换为

{1={2=[3, 4], 3=[4]}, 2={3=[5]}}

现在得到这个结果我使用以下结构：

    Map<String,Map<String,List<A>>> grouppedRecords = list.stream().collect(groupingBy(A::getA, groupingBy(A::getB)));

    Map<String,Map<String,List<String>>> results =  grouppedRecords.entrySet().stream().collect(Collectors.toMap(
            Map.Entry::getKey,
            e12 -> e12.getValue().entrySet().stream().collect(Collectors.toMap(
                    Map.Entry::getKey,
                    e22 -> e22.getValue().stream().map(A::getC).collect(Collectors.toList())
            )))
    );

但它看起来过于复杂。你能帮助我更好地解决同样的问题吗？

Answer 1

您可以使用const Post = require('../models/post'); const Comment = require('../models/comment'); getpost: async (req, res, next) => { const { post_id } = req.params; const post = await Post.findById(post_id); const comments_id = post._comment; var comments = []; comments_id.forEach(async id => { comments.push(await Comment.findById(id)); console.log(comments) // The output are elements which exist in Comment model }); console.log(comments); // The output is [] comments.forEach(comment => { console.log(comment); });将Collectors.mapping个实例映射到A：

getC()

如何使用Java 8流

1 个答案: