比较数组并返回差值 [英] Compare arrays and Return the Difference

问题描述

我在运行时在内存中创建了一个数组A，另一个数组B保存在mongo数据库中.如何有效地从A中获得所有不在B中的元素?

I have an array A in memory created at runtime and another array B saved in a mongo database. How can I efficiently get all the elements from A that are not in B?

您可以假设存储在mongodb中的数组比运行时创建的数组大几个数量级，因此，我认为从mongo获取完整数组并计算结果效率不高，但是我没有在mongo中发现了任何查询操作，可以让我计算出想要的结果.

You can assume that the array stored in mongodb is several orders of magnitude bigger than the array created at runtime, for that reason I think that obtaining the full array from mongo and computing the result would not be efficient, but I have not found any query operation in mongo that allows me to compute the result I want.

请注意， $ nin 运算符的作用与之相反.我想要的，即它从B中检索不在A中的元素.

Note that the $nin operator does the opposite of what I want, i.e., it retrieves the elements from B that are not in A.

示例:

在运行时在我的应用中创建的数组A为[2, 3, 4].

Array A, created in my appliction at runtime, is [2, 3, 4].

存储在mongodb中的数组B是[1, 3, 5, 6, 7, 10].

Array B, stored in mongodb, is [1, 3, 5, 6, 7, 10].

我期望的结果是[2, 4].

推荐答案

在响应中修改"文档的唯一内容是 .mapReduce() ，其中前者是更好的选择.

The only things that "modify" the document in response are .aggregate() and .mapReduce(), where the former is the better option.

在这种情况下，您需要的 $setDifference 比较集合"并返回两者之间的差异".

In that case you are asking for $setDifference which compares the "sets" and returns the "difference" between the two.

因此用数组表示一个文档:

So representing a document with your array:

db.collection.insert({ "b": [1, 3, 5, 6, 7, 10] })

运行聚合:

db.collection.aggregate([{ "$project": { "c": { "$setDifference": [ [2,3,4], "$b" ] } } }])

哪个返回:

{ "_id" : ObjectId("596005eace45be96e2cb221b"), "c" : [ 2, 4 ] }

如果您不希望集合"，而是想提供像[2,3,4,4]这样的数组，则可以与 $in ，如果您至少具有MongoDB 3.4:

If you do not want "sets" and instead want to supply an array like [2,3,4,4] then you can compare with $filter and $in instead, if you have MongoDB 3.4 at least:

db.collection.aggregate([
  { "$project": {
    "c": {
      "$filter": {
        "input": [2,3,4,4],
        "as": "a",
        "cond": {
          "$not": { "$in": [ "$$a", "$b" ]  }
        }
      }
    }   
  }}
])

或使用 $filter 和

Or with $filter and $anyElementTrue in earlier versions:

db.collection.aggregate([
  { "$project": {
    "c": {
      "$filter": {
        "input": [2,3,4,4],
        "as": "a",
        "cond": {
          "$not": {
            "$anyElementTrue": {
              "$map": {
                "input": "$b",
                "as": "b",
                "in": {
                  "$eq": [ "$$a", "$$b" ]    
                }
              }    
            }
          }
        }    
      }
    }    
  }}
])

两者都会返回:

{ "_id" : ObjectId("596005eace45be96e2cb221b"), "c" : [ 2, 4, 4 ] }

由于4作为输入两次"提供，因此当然不是一组"，因此也返回两次".

Which is of course "not a set" since the 4 was provided as input "twice" and is therefore returned "twice" as well.

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