在MongoDB集合中按频率对文档中的数组进行排序

OP 指出，我们在问题中看到的文档是 （表面上是 ） 的结果，该结果可能将 、 和 值带到不断增长的数组（多达 50,000 个）上。如果目标是获得每种颜色的模式，则可以用作“多组”。假设每个单独的文档都有这样的形状（注意：这里只是为了简化而使用and;该方法扩展到文档中的任何其他字段）：$groupcolor$pushfruitvegetablemeat$facetfruitmeatvegetable

    {"color": "blue", "fruit": "A", "meat": "X"}

以下管道将生成我们寻求的模式：$facet

db.foo.aggregate([
    {$facet: {
        "most_fruit": [
            // Sum by color and fruit name:                                      
            {$group: {_id: {c:'$color', v:'$fruit'}, N: {$sum: 1}}}

            // Reorg by color only....                                           
            ,{$group: {_id: '$_id.c', X:{$push: {v:'$_id.v',N:'$N'}}}}

            // ...and now sort highest-to-lowest and take the highest one.       
            // Nice thing is if you really want, you are already set up to       
            // capture, for example, the highest *nd* the lowest.                
            ,{$project: {
                X: {$first: {$sortArray: {input: '$X', sortBy: {'N':-1} }} }
            }}
        ],

        // Same thing ... but for meat                                           
        "most_meat": [
            {$group: {_id: {c:'$color', v:'$meat'}, N: {$sum: 1}}}

            ,{$group: {_id: '$_id.c', X:{$push: {v:'$_id.v',N:'$N'}}}}
            ,{$project: {
                X: {$first: {$sortArray: {input: '$X', sortBy: {'N':-1} }} }
            }}
        ]
    }}
]);

生成具有此形状的东西：

{                                                                                
  most_fruit: [                                                                  
    {_id: 'blue', X: {v: 'A', N: 2} },                                           
    {_id: 'green',X: {v: 'F', N: 3} }                                            
  ],                                                                             
  most_meat: [                                                                   
    {_id: 'green',X: {v: 'Z', N: 4} },                                           
    {_id: 'blue', X: {x: 'X', N: 3} }                                            
  ]                                                                              
}    

返回包含所有信息的单个文档。虽然它不是按颜色组织的，但无法在数据库端（使用 MQL）进行进一步的处理，以使数据的分组或过滤更加高效;现在由客户端来设置信息。

下面是一个候选的客户端重组：

var oneDoc = c.next();

function processItem(obj, fname) {
    for(var n = 0; n < obj[fname].length; n++) {
        var cn = oneDoc[fname][n]['_id'];
        if(undefined == color_major[cn]) {
            // Set up ALL the possible modes.  -1 is our way of                  
            // signalling it has not yet been set.                               
            color_major[cn] = {'most_fruit':-1,'most_meat':-1};
        }
        color_major[cn][fname] = oneDoc[fname][n]['X'];
    }
}

var color_major = {};
processItem(oneDoc, 'most_fruit');
processItem(oneDoc, 'most_meat');

print(color_major);

收益 率：

{
  green: { most_fruit: { v: 'F', N: 3 }, most_meat: { v: 'Z', N: 4 } },
  blue: { most_fruit: { v: 'A', N: 2 }, most_meat: { v: 'X', N: 3 } }
}

如果你真的想打败DB..

(...但不是真的，因为上面$facet阶段的数据非常小......

下面是用于重新格式化数据的额外管道：

    // Turn $facet field names (lval) into values (rval):                                         
    ,{$project: {X: {$objectToArray: '$$ROOT'} }}

    // Double unwind is OK because there is only #color X 2 (fruit and meat) entries.             
    // Even if 1000 colors and fruit meat and veg and whatevs, still quite doable:                
    ,{$unwind: '$X'}
    ,{$unwind: '$X.v'}

    // Reorg on color:                                                                            
    ,{$group: {_id: '$X.v._id', W: {$push: {k: '$X.k', v: '$X.v.X'}} }}

    // ...and put it all back together:                                                           
    ,{$replaceRoot: { newRoot: {$mergeObjects: [ {color:'$_id'}, {$arrayToObject: '$W'} ]}
     }}

收益 率：

{
  color: 'green',
  most_fruit: {
    v: 'F',
    N: 3
  },
  most_meat: {
    v: 'Z',
    N: 4
  }
}
{
  color: 'blue',
  most_fruit: {
    v: 'A',
    N: 2
  },
  most_meat: {
    v: 'X',
    N: 3
  }
}