正如您所猜测的，$project允许您这样做:

db.col.aggregate([
{
    $project : 
    {
        _id: "$command._id",
        name: "$command.name", 
        strike: "$command.strike", 
        duration: "$command.duration"
    }
}
]).pretty()

我插入了您以前的结果，上面的查询返回了以下结果:

{
    "_id" : ObjectId("57ec6b6f6c61e919b578fe7c"),
    "name" : "Run",
    "strike" : 15,
    "duration" : 123
}
{
    "_id" : ObjectId("573d688d080cc2cbe8aecbbc"),
    "name" : "Run",
    "strike" : 12,
    "duration" : 597
}

所以piping用这个$product进行查询应该会得到你想要的结果.

Update after comments

如果您主要关心的不是确切的 struct ，而是排除几个字段(不必列出要包含的所有字段)，那么您可以使用find()而不是aggregate().

aggregate's product only lets you exclude _id. This means you need to manually list all fields to include.
Note: Since version 3.4 of MongoDB it is possible to exclude fields in $project phase (https://docs.mongodb.com/manual/reference/operator/aggregation/project/#exclude-fields)

但是，find允许您列出要隐藏的字段.

可供替代的

(1)您可以使用$out将聚合结果重定向到另一个集合:

{ $out : "commands" }

(2)即使 struct 不完全符合您的要求，您也可以执行find次查询并隐藏字段:

db.commands.find({}, {_id:0, "command.score":0, eventname:0}).pretty()

它会返回这个，这与你想要的非常接近:

{
    "command" : {
        "_id" : ObjectId("57ec6b6f6c61e919b578fe7c"),
        "name" : "Run",
        "strike" : 15,
        "duration" : 123
    }
}