mongodb $unwind 聚合框架中的对象
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$unwind an object in aggregation framework
提问by Brett Thomas
In the MongoDB aggregation framework, I was hoping to use the $unwind operator on an object (ie. a JSON collection). Doesn't look like this is possible, is there a workaround? Are there plans to implement this?
在 MongoDB 聚合框架中，我希望在对象（即 JSON 集合）上使用 $unwind 运算符。看起来这不可能，有解决方法吗？有计划实施吗？
For example, take the article collection from the aggregation documentation. Suppose there is an additional field "ratings" that is a map from user -> rating. Could you calculate the average rating for each user?
例如，从聚合文档中获取文章集合。假设有一个附加字段“ratings”，它是来自用户 -> rating 的地图。你能计算出每个用户的平均评分吗？
Other than this, I'm quite pleased with the aggregation framework.
除此之外，我对聚合框架非常满意。
Update: here's a simplified version of my JSON collection per request. I'm storing genomic data. I can't really make genotypes an array, because the most common lookup is to get the genotype for a random person.
更新：这是我每个请求的 JSON 集合的简化版本。我正在存储基因组数据。我不能真正将基因型作为数组，因为最常见的查找是获取随机人的基因型。
variants: [
{
name: 'variant1',
genotypes: {
person1: 2,
person2: 5,
person3: 7,
}
},
{
name: 'variant2',
genotypes: {
person1: 3,
person2: 3,
person3: 2,
}
}
]
回答by Asya Kamsky
It is not possible to do the type of computation you are describing with the aggregation framework - and it's notbecause there is no $unwindmethod for non-arrays. Even if the person:value objects were documents in an array, $unwindwould not help.
不可能使用聚合框架进行您所描述的计算类型 - 这不是因为没有$unwind非数组的方法。即使 person:value 对象是数组中的文档，$unwind也无济于事。
The "group by" functionality (whether in MongoDB or in any relational database) is done on the value of a field or column. We group by value of field and sum/average/etc based on the value of another field.
“分组依据”功能（无论是在 MongoDB 中还是在任何关系数据库中）是在字段或列的值上完成的。我们根据另一个字段的值按字段值和总和/平均值/等分组。
Simple example is a variant of what you suggest, ratings field added to the example article collection, but not as a map from user to rating but as an array like this:
简单示例是您建议的变体，将 ratings 字段添加到示例文章集合中，但不是作为从用户到 rating 的映射，而是作为这样的数组：
{ title : title of article", ...
ratings: [
{ voter: "user1", score: 5 },
{ voter: "user2", score: 8 },
{ voter: "user3", score: 7 }
]
}
Now you can aggregate this with:
现在您可以将其汇总为：
[ {$unwind: "$ratings"},
{$group : {_id : "$ratings.voter", averageScore: {$avg:"$ratings.score"} } }
]
But this example structured as you describe it would look like this:
但是这个例子如你所描述的那样结构化，它看起来像这样：
{ title : title of article", ...
ratings: {
user1: 5,
user2: 8,
user3: 7
}
}
or even this:
甚至这个：
{ title : title of article", ...
ratings: [
{ user1: 5 },
{ user2: 8 },
{ user3: 7 }
]
}
Even if you could $unwindthis, there is nothing to aggregate on here. Unless you know the complete list of all possible keys (users) you cannot do much with this. [*]
即使你能做到$unwind这一点，这里也没有什么可聚合的。除非您知道所有可能的密钥（用户）的完整列表，否则您对此无能为力。[*]
An analogous relational DB schema to what you have would be:
与您所拥有的类似的关系数据库架构是：
CREATE TABLE T (
user1: integer,
user2: integer,
user3: integer
...
);
That's not what would be done, instead we would do this:
这不是我们要做的，而是我们会这样做：
CREATE TABLE T (
username: varchar(32),
score: integer
);
and now we aggregate using SQL:
现在我们使用 SQL 进行聚合：
select username, avg(score) from T group by username;
select username, avg(score) from T group by username;
There is an enhancement request for MongoDB that may allow you to do this in the aggregation framework in the future - the ability to project values to keys to vice versa. Meanwhile, there is always map/reduce.
对 MongoDB 的增强请求可能允许您将来在聚合框架中执行此操作 - 将值投影到键的能力，反之亦然。同时，总是有map/reduce。
[*] There is a complicated way to do this if you know all unique keys (you can find all unique keys with a method similar to this) but if you know all the keys you may as well just run a sequence of queries of the form db.articles.find({"ratings.user1":{$exists:true}},{_id:0,"ratings.user1":1})for each userX which will return all their ratings and you can sum and average them simply enough rather than do a very complex projection the aggregation framework would require.
[*] 如果您知道所有唯一键（您可以使用与此类似的方法找到所有唯一键），则有一种复杂的方法可以做到这一点，但如果您知道所有键，您也可以运行一系列查询db.articles.find({"ratings.user1":{$exists:true}},{_id:0,"ratings.user1":1})每个 userX 的表单将返回他们的所有评分，您可以简单地对它们进行求和和平均，而不是进行聚合框架所需的非常复杂的投影。
回答by Adrian
Since 3.4.4, you can transform object to array using $objectToArray
从 3.4.4 开始，您可以使用 $objectToArray 将对象转换为数组
See: https://docs.mongodb.com/manual/reference/operator/aggregation/objectToArray/
请参阅：https: //docs.mongodb.com/manual/reference/operator/aggregation/objectToArray/
回答by saarp
This is an old question, but I've run across a tidbit of information through trial and error that people may find useful.
这是一个老问题，但我通过反复试验发现了一些人们可能会觉得有用的信息。
It's actually possible to unwind on a dummy value by fooling the parser this way:
实际上可以通过以这种方式欺骗解析器来解除虚拟值：
db.Opportunity.aggregate(
{ $project: {
Field1: 1, Field2: 1, Field3: 1,
DummyUnwindField: { $ifNull: [null, [1.0]] }
}
},
{ $unwind: "$DummyUnwindField" }
);
This will produce 1 row per document, regardless of whether or not the value exists. You may be able tinker with this to generate the results you want. I had hoped to combine this with multiple $unwinds to (sort of like emit() in map/reduce), but alas, the last $unwind wins or they combine as an intersection rather than union which makes it impossible to achieve the results I was looking for. I am sadly disappointed with the aggregate framework functionality as it doesn't fit the one use case I was hoping to use it for (and seems strangely like a lot of the questions on StackOverflow in this area are asking) - ordering results based on match rate. Improving the poor map reduce performance would have made this entire feature unnecessary.
无论该值是否存在，这都会为每个文档生成 1 行。您可以修改它以生成您想要的结果。我曾希望将它与多个 $unwinds 结合起来（有点像 map/reduce 中的 emit()），但是唉，最后一个 $unwind 获胜，或者它们组合为一个交集而不是联合，这使得我无法实现我的结果正在寻找。我对聚合框架功能感到遗憾，因为它不适合我希望使用它的一个用例（并且似乎很奇怪，这方面的 StackOverflow 上的很多问题都在问）-根据匹配对结果进行排序速度。改善糟糕的地图减少性能会使整个功能变得不必要。
回答by vijay
This is what I found & extended.
Lets create experimental database in mongo
让我们在 mongo 中创建实验数据库
db.copyDatabase('livedb' , 'experimentdb')
Now Use experimentdb & convert Array to object in your experimentcollection
现在使用 Experimentdb 并将数组转换为您的实验集合中的对象
db.getCollection('experimentcollection').find({}).forEach(function(e){
if(e.store){
e.ratings = [e.ratings]; //Objects name to be converted to array eg:ratings
db.experimentcollection.save(e);
}
})
Some nerdy js code to convert json to flat object
一些书呆子js代码将json转换为平面对象
var flatArray = [];
var data = db.experimentcollection.find().toArray();
for (var index = 0; index < data.length; index++) {
var flatObject = {};
for (var prop in data[index]) {
var value = data[index][prop];
if (Array.isArray(value) && prop === 'ratings') {
for (var i = 0; i < value.length; i++) {
for (var inProp in value[i]) {
flatObject[inProp] = value[i][inProp];
}
}
}else{
flatObject[prop] = value;
}
}
flatArray.push(flatObject);
}
printjson(flatArray);
