通过map reduce的MongoDB唯一值聚合

Question

我在MongoDB中看到很多关于聚合的问题,但是,我还没有找到完整的解决方案.

这是我的数据示例:

{
    "fruits" : {
        "apple" : "red",
        "orange" : "orange",
        "plum" : "purple"
    }
}
{
    "fruits" : {
        "apple" : "green",
        "plum" : "purple"
    }
}
{
    "fruits" : {
        "apple" : "red",
        "orange" : "yellow",
        "plum" : "purple"
    }
}

现在,我的目标是确定每种水果的每种颜色的受欢迎程度,所以这样的东西就是输出集合:

{
    "_id" : "apple"
    "values" : {
        "red" : 2,
        "green" : 1
    }
}
{
    "_id" : "orange"
    "values" : {
        "orange" : 1,
        "yellow" : 1
    }
}
{
    "_id" : "plum"
    "values" : {
        "purple" : 3
    }
}

我尝试了各种M/R功能,最后它们要么不起作用,要么它们需要指数长.在示例(水果)的背景下,我有大约1,000种不同的水果和100,000种颜色,超过10,000,000份总文件.我目前的工作M/R是这样的:

map = function() {
    if (!this.fruits) return;
    for (var fruit in this.fruits) {
        emit(fruit, {
            val_array: [
                {value: this.fruits[fruit], count: 1}
            ]
        });
    }
};

reduce = function(key, values) {
    var collection = {
        val_array: []
    };
    var found = false;
    values.forEach(function(map_obj) {
        map_obj.val_array.forEach(function(value_obj) {
            found = false;
            // if exists in collection, inc, else add
            collection.val_array.forEach(function(coll_obj) {
                if (coll_obj.value == value_obj.value) {
                    // the collection already has this object, increment it
                    coll_obj.count += value_obj.count;
                    found = true;
                    return;
                }
            });
            if (!found) {
                // the collection doesn't have this obj yet, push it
                collection.val_array.push(value_obj);
            }
        });
    });
    return collection;
};

现在,这确实有效,对于100条记录,它只需要一秒钟左右,但时间非线性增加,因此100M记录需要很长时间.问题是我在使用collection数组的reduce函数中做了一个穷人子聚合,因此需要我迭代两个collection和我的map函数中的值.现在我只需要弄清楚如何有效地做到这一点(即使它需要多次减少).欢迎任何建议!

---

编辑由于缺少更好的发布地点,这是我的解决方案.

---

首先,我创建了一个名为的文件mr.js:

map = function() {
    if (!this.fruits) return;
    var skip_fruits = {
        'Watermelon':1,
        'Grapefruit':1,
        'Tomato':1 // yes, a tomato is a fruit
    }
    for (var fruit in this.fruits) {
        if (skip_fruits[fruit]) continue;
        var obj = {};
        obj[this.fruits[fruit]] = 1;
        emit(fruit, obj);
    }
};

reduce = function(key, values) {
    var out_values = {};
    values.forEach(function(v) {
        for(var k in v) { // iterate values
            if (!out_values[k]) {
                out_values[k] = v[k]; // init missing counter
            } else {
                out_values[k] += v[k];
            }
        }
    });
    return out_values;
};

var in_coll = "fruit_repo";
var out_coll = "fruit_agg_so";
var total_docs = db[in_coll].count();
var page_size = 100000;
var pages = Math.floor(total_docs / page_size);
print('Starting incremental MR job with '+pages+' pages');
db[out_coll].drop();
for (var i=0; i<pages; i++) {
    var skip = page_size * i;
    print("Calculating page limits for "+skip+" - "+(skip+page_size-1)+"...");
    var start_date = db[in_coll].find({},{date:1}).sort({date:1}).skip(skip).limit(1)[0].date;
    var end_date = db[in_coll].find({},{date:1}).sort({date:1}).skip(skip+page_size-1).limit(1)[0].date;
    var mr_command = {
        mapreduce: in_coll,
        map: map,
        reduce: reduce,
        out: {reduce: out_coll},
        sort: {date: 1},
        query: {
            date: {
                $gte: start_date,
                $lt: end_date
            }
        },
        limit: (page_size - 1)
    };
    print("Running mapreduce for "+skip+" - "+(skip+page_size-1));
    db[in_coll].runCommand(mr_command);
}

该文件迭代我的整个集合,逐步映射/减少100k文档(按date哪个必须有一个索引!),并将它们减少为单个输出集合.它的使用方式如下:mongo db_name mr.js.

然后,几个小时后,我收集了所有信息.为了弄清楚哪些水果颜色最多,我使用mongo shell来打印前25个:

// Show number of number of possible values per key
var keys = [];
for (var c = db.fruit_agg_so.find(); c.hasNext();) {
    var obj = c.next();
    if (!obj.value) break;
    var len=0;for(var l in obj.value){len++;}
    keys.push({key: obj['_id'], value: len});
}
keys.sort(function(a, b){
    if (a.value == b.value) return 0;
    return (a.value > b.value)? -1: 1;
});
for (var i=0; i<20; i++) {
    print(keys[i].key+':'+keys[i].value);
}

这种方法真的很酷,因为它是增量的,我可以在mapreduce运行时处理输出数据.

Answer 1

看来你真的不需要val_array.为什么不使用简单的哈希？试试这个:

map = function() {
    if (!this.fruits) return;
    for (var fruit in this.fruits) {
        emit(fruit, 
             {this.fruits[fruit]: 1});
    }
};

reduce = function(key, values) {
  var colors = {};

  values.forEach(function(v) {
    for(var k in v) { // iterate colors
      if(!colors[k]) // init missing counter
        colors[k] = 0

      color[k] += v[k];
    }
  });

  return colors;
}