Question

我有一只红树收集,一只希望获得特定地点名称、时装和变式补贴的总数。由于我们拥有大量数据,因此它将更好地利用地图/uce。难道有人会向我提出任何建议吗?

这里我收集的json格式

{ "_id" : ObjectId( "4e3c280ecacbd1333b00f5ff" ),
  "timestamp" : "20110805",
  "variants" : { "94" : { "number_of_ad_clicks" : 41,
      "number_of_search_keywords" : 9,
      "total_duration" : 0,
      "os" : { "os_2" : 2,
        "os_1" : 1,
        "os_0" : 0 },
      "countries" : { "ge" : 6,
        "ca" : 1,
        "fr" : 8,
        "uk" : 4,
        "us" : 6 },
      "screen_resolutions" : { "(320, 240)" : 1,
        "(640, 480)" : 5,
        "(1024, 960)" : 5,
        "(1280, 768)" : 5 },
      "widgets" : { "widget_1" : 1,
        "widget_0" : 0 },
      "languages" : { "ua_uk" : 8,
        "ca_en" : 2,
        "ca_fr" : 2,
        "us_en" : 5 },
      "search_keywords" : { "search_keyword_8" : 8,
        "search_keyword_5" : 5,
        "search_keyword_4" : 4,
        "search_keyword_7" : 7,
        "search_keyword_6" : 6,
        "search_keyword_1" : 1,
        "search_keyword_3" : 3,
        "search_keyword_2" : 2 },
      "number_of_pageviews" : 18,
      "browsers" : { "browser_4" : 4,
        "browser_0" : 0,
        "browser_1" : 1,
        "browser_2" : 2,
        "browser_3" : 3 },
      "keywords" : { "keyword_5" : 5,
        "keyword_4" : 4,
        "keyword_1" : 1,
        "keyword_0" : 0,
        "keyword_3" : 3,
        "keyword_2" : 2 },
      "number_of_keyword_clicks" : 83,
      "number_of_visits" : 96 } },
  "site_name" : "fonter.com",
  "number_of_variants" : 1 }

Here is my try. but failed. He is my try.

m = function() {
    emit(this.query, {variants: this.variants});
}

r = function(key , vals) {
    var clicks = 0 ;
    for(var i = 0; i < vals.length(); i++){
        clicks = vals[i][ number_of_ad_clicks ];
    }
    return clicks;
}
res = db.variant_daily_collection.mapReduce(m, r, {out : "myoutput", "query":{"site_name":  fonter.com ,  timestamp :  20110805 }})
db.myoutput.find()

could somebody any suggestion?

Thank you very much, i try you solution but nothing return. I invoke the mapreduce in the following, is there any thing wrong?

res = db.variant_daily_collection.mapReduce(map, reduce, {out : "myoutput", "query":{"site_name":  facee.com ,  timestamp :  20110809 ,  variant_id :  305 }})
db.myoutput.find()

Answer 1

The emit function emits both a key and a value.

如果您将<代码>key作为GROUP BY和的<>作为<编码>SUM()、AVG(()等。

In your case you want to "group by": site_name, timestamp and variant id. It looks like you may have more than one variant, so you will need to loop through the variants, like this:

map = function() {
  for(var i in variants){
    var key = {};
    key.timestamp = this.timestamp;
    key.site_name = this.site_name;
    key.variant_id = i; // that s the "94" string.

    var value = {};
    value.clicks = this.variants[i].number_of_ad_clicks;

    emit(key, value);
  }
}

降低功能将获得各种价值,如<代码>{点击:41 }。该功能必须归还一个看上去的物体。

So if you get values = [ {clicks:21}, {clicks:10}, {clicks:5} ] you must output {clicks:36}.

So you do something like this:

reduce = function(key , vals) {
    var returnValue = { clicks: 0 }; // initializing to zero
    for(var i = 0; i < vals.length(); i++){
        returnValue.clicks += vals[i].clicks;
    }
    return returnValue;
}

Note that the value from map has the same shape as the return from reduce.

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