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我尝试使用两次 MapReduce 聚合来获得每月唯一的用户数。

第一个 MR 函数计算出一个 mr_buyer_payment 集合，如下所示：

{ "_id" : { "u" : "01329f19-27b0-435b-9ca1-450984024a31", "tid" : ISODate("2013-09-01T00:00:00Z") }, "value" : { "payment" : 38, "count_pay" : 1 } }
{ "_id" : { "u" : "264dd104-b934-490b-988e-5822fd7970f6", "tid" : ISODate("2013-09-01T00:00:00Z") }, "value" : { "payment" : 4.99, "count_pay" : 1 } }
{ "_id" : { "u" : "27bb8f72-a13e-4676-862c-02f41fea1bc0", "tid" : ISODate("2013-09-01T00:00:00Z") }, "value" : { "payment" : 11.98, "count_pay" : 2 } }

第二个 MR 函数适用于小数据集，但是当查询增长超过 100 条记录时，它会得到错误的结果，一些值为 NaN。

调试日志显示了 Reduce 函数中的一些值，例如 v.payment、v.count_user 变得未定义。

date:Sun Jun 30 2013 17:00:00 GMT-0700 (PDT)  value:undefined / 162 / undefined

并且 MR 结果信息是有线的：

{
    "result" : "mr_buyer_all",
    "timeMillis" : 29,
    "counts" : {
        "input" : 167,
        "emit" : 167,
        "reduce" : 6,  // it should be 3, as same as "output" number
        "output" : 3
    },
    "ok" : 1,
}

这是第二个 MR 功能：

db.mr_buyer_payment.mapReduce(
    function(){
        var key = this._id.tid;
        var value = {
            payment:this.value.payment,
            count_pay:this.value.count_pay,
            count_user:1
        };
        if (value.count_pay>0)
        {
            print("date:"+key+"  u:"+this._id.u+"value:"+value.payment+" / "+value.count_pay+" / "+value.count_user);
            emit(key,value);
        }
    },
    function(key,values){
        var result = {revenue:0,count_pay:0,user:0};
        values.forEach(function(v){
            if (!v.count_user)
            {
                print("date:"+key+"  "+"value:"+v.payment+" / "+v.count_pay+" / "+v.count_user);
            } else
            {
                result.revenue += v.payment;
                result.count_pay += v.count_pay;
                result.user += v.count_user;
            }

        });
        return result;
    },
    {
        out:{replace:"mr_buyer_all"}
    }
)
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1 回答 1

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Reduce 函数中的子文档应使用与 Map 函数中相同的格式。所以解决方案是：

function(key,values){
    // the following key must be as same as the object in map
    var r = {payment:0,count_pay:0,count_user:0}
    values.forEach(function(v){
        r.payment += v.payment;
        r.count_pay += v.count_pay;
        r.count_user += v.count_user;
    });
    return r;
},
于 2013-10-04T20:51:17.297 回答