给定以下数据集：
{ "_id" : ObjectId("510458b188ce1d16e616129b"), "codes" : [ "oxtbyr", "xstute" ], "name" : "Ciao Mambo", "permalink" : "ciaomambo", "visits" : 1 }
{ "_id" : ObjectId("510458b188ce1d16e6161296"), "codes" : [ "zpngwh", "odszfy", "vbvlgr" ], "name" : "Anthony's at Spokane Falls", "permalink" : "anthonysatspokanefalls", "visits" : 0 }
如何将这个python/pymongo排序转换为可以与mongodb聚合框架一起使用的东西？我正在根据
codes数组中的代码数对结果进行排序。z = [(x['name'], len(x['codes'])) for x in restaurants]
sorted_by_second = sorted(z, key=lambda tup: tup[1], reverse=True)
for x in sorted_by_second:
print x[0], x[1]
这在python中有效，我只想知道如何在mongodb查询端实现相同的目标。
最佳答案
> db.z.aggregate({ $unwind:'$codes'},
{ $group : {_id:'$_id', count:{$sum:1}}},
{ $sort :{ count: 1}})