Question

首先，我的数据库中有两个可能的文档的例子

{

  "_id" : "0"

  "data": { 
             "lettersCollection" : [
               {"lettersList" : ["A","B","C"] },
               {"lettersList" : ["D","T","E"] },
               {"lettersList" : ["X","Y","Z"] },
             ]
          }
}



{

  "_id" : "1"

  "data": { 
            "lettersCollection" : [
              {"lettersList" : ["A","B","D"] },
              {"lettersList" : ["X","Y","Z"] },
              {"lettersList" : ["E","C","M"] },
            ]
          }
}

我想要完成什么？

我想列出一个字母及其出现次数的下降列表。

例如：如果数据库仅包含上述2个文档，则结果为：

A - 2
B - 2
C - 2
D - 2
E - 2
X - 2
Y - 2
Z - 2
T - 1
M - 1

注意：一封信只能在文档中出现一次。例如，字母“B”不可能在特定文档的任何lettersList中出现多次

到目前为止我尝试了什么？

我是MongoDB的新手，我越接近这个查询，它只适用于单个值（例如：_id，它始终是唯一的，因此计数总是1），并且不会迭代嵌套数组，不按降序排列：

db.test.group({
    "key": {
        "_id": true
    },
    "initial": {
        "countstar": 0
    },
    "reduce": function(obj, prev) {
        if (true != null) if (true instanceof Array) prev.countstar += true.length;
        else prev.countstar++;
    }
});

Answer 1

您可以使用aggregation pipeline（通常应该是您的首选聚合）来执行此操作：

db.test.aggregate([
    // Duplicate the docs, once per lettersCollection element
    {$unwind: '$data.lettersCollection'},
    // Duplicate the docs again, this time once per lettersList
    {$unwind: '$data.lettersCollection.lettersList'},
    // Group them back together by letter and count them
    {$group: {_id: '$data.lettersCollection.lettersList', count: {$sum: 1}}},
    // Sort by count descending
    {$sort: {count: -1}}
])

输出：

{
    "result" : [ 
        {
            "_id" : "Z",
            "count" : 2
        }, 
        {
            "_id" : "X",
            "count" : 2
        }, 
        {
            "_id" : "Y",
            "count" : 2
        }, 
        {
            "_id" : "E",
            "count" : 2
        }, 
        {
            "_id" : "D",
            "count" : 2
        }, 
        {
            "_id" : "C",
            "count" : 2
        }, 
        {
            "_id" : "B",
            "count" : 2
        }, 
        {
            "_id" : "A",
            "count" : 2
        }, 
        {
            "_id" : "M",
            "count" : 1
        }, 
        {
            "_id" : "T",
            "count" : 1
        }
    ],
    "ok" : 1
}

在迭代嵌套列表时使用mongodb中的count

1 个答案: