Question

尝试使用C＃/ Linq甚至原始Mongodb查询本身如何将多个数组作为笛卡尔积加入。

比方说，我有一个集合，我过滤到以下两个文件：

[
{"movie":"starwars","showday":"monday"},
{"movie":"batman","showday":"thursday"},
{"movie":"sleepless","showday":"tuesday"}
]

[
{"actor":"angelina","location":"new york"},
{"actor":"jamie","location":"california"},
{"actor":"mcavoy","location":"arizona"}
]

如何连接每个数组中的每个项目以产生以下类型的结果？

[{"movie":"starwars","showday":"monday","actor":"angelina","location":"new york"},
{"movie":"batman","showday":"thursday","actor":"angelina","location":"new york"},
{"movie":"sleepless","showday":"tuesday","actor":"angelina","location":"new york"},
{"movie":"starwars","showday":"monday","actor":"jamie","location":"california"},
{"movie":"batman","showday":"thursday","actor":"jamie","location":"california"},
{"movie":"sleepless","showday":"tuesday","actor":"jamie","location":"california"},
{"movie":"starwars","showday":"monday","actor":"mcavoy","location":"arizona"},
{"movie":"batman","showday":"thursday","actor":"mcavoy","location":"arizona"},
{"movie":"sleepless","showday":"tuesday","actor":"mcavoy","location":"arizona"}]

我正在寻找可以处理任意数量文档的解决方案。因此，例如，如果在此示例中有第3个文档，它还有3个对象数组，它们将在数组中生成27个项目的结果集 - 或者原样为27行。

希望有一个如何使用C＃（Linq？）Mongodb驱动程序来查询和返回这样的数据的解决方案，但是对于mongodb特定查询也是开放的，因为我希望可以从那里反转逻辑。谢谢

Answer 1

您可以尝试以下聚合管道。

注意mergeObjects聚合运算符在3.5.6 +开发版本中可用，该版本将在即将发布的3.6版本中提供。

db.collection.find();
{
 "data" : [
  [
   {
    "movie" : "starwars",
    "showday" : "monday"
   },
   {
    "movie" : "batman",
    "showday" : "thursday"
   },
   {
    "movie" : "sleepless",
    "showday" : "tuesday"
   }
  ],
  [
   {
    "actor" : "angelina",
    "location" : "new york"
   },
   {
    "actor" : "jamie",
    "location" : "california"
   },
   {
    "actor" : "mcavoy",
    "location" : "arizona"
   }
  ]
 ]
}

使用条件表达式进行聚合。

aggregate({
 $project: {
  cp: {
   $reduce: {
    input: "$data",
    initialValue: {
     $arrayElemAt: ["$data", 0] // Set the initial value to the first element of the arrays.
    },
    in: {
     $let: {
      vars: {
       currentr: "$$this", // Current processing element
       currenta: "$$value" // Current accumulated value 
      },
      in: {
       $cond: [{ // Conditional expression to return the accumulated value as initial value for first element
        $eq: ["$$currentr", "$$currenta"]
       },
       "$$currenta",
       { // From second element onwards prepare the cartesian product
        $reduce: {
         input: {
          $map: {
           input: "$$currenta",
           as: a"a",
           in: {
            $map: {
             input: "$$currentr",
             as: r"r",
             in: {
              $mergeObjects: ["$$a", "$$r"] // Merge accumulated value with the current processing element
             }
            }
           }
          }
         },
         initialValue: [],
         in: {
         $concatArrays: ["$$value", "$$this"] // Reduce the merged values which will be used as accumulator for next element
         }
        }
       }]
      }
     }
    }
   }
  }
 }
});

聚合（使用$setUnion）。

仅添加此解决方案以抑制条件表达式以提供更易读的管道。

aggregate({
 $project: {
  cp: {
   $reduce: {
    input: "$data",
    initialValue: {
     $arrayElemAt: ["$data", 0] // Set the initial value to the first element of the arrays.
    },
    in: {
     $let: {
      vars: {
       currentr: "$$this", // Current processing element
       currenta: "$$value" // Current accumulated value 
      },
      in:{ 
       $reduce: {
        input: {
         $map: {
          input: "$$currenta",
          as: "a",
          in: {
           $map: {
            input: "$$currentr",
            as: "r",
            in: {
             $mergeObjects: ["$$a", "$$r"] // Merge accumulated value with the current processing element
            }
           }
          }
         }
        },
        initialValue: [],
        in: {
         $setUnion: ["$$value", "$$this"] // Reduce the merged values which will be used as accumulator for next element
        }
       }
      }
     }
    }
   }
  }
 }
});

<强>更新

由于第一个解决方案中的$cond不正确以及第二个解决方案中的$setUnion，Asya Kamsky在下面的评论中指出，上述解决方案都不适用于重复数组的值。

正确的解决方法是

以initialValue

的[ { } ]开头

或

更改input以排除第一个元素，例如input: {$slice:["$data", 1, {$subtract:[{$size:"$data"},1]}]},

完整的汇总管道

aggregate({
 $project: {
  cp: {
   $reduce: {
    input: {$slice:["$data", 1, {$subtract:[{$size:"$data"},1]}]},
    initialValue: {$arrayElemAt:["$data",0]},
    in: {
     $let: {
      vars: {
       currentr: "$$this", 
       currenta: "$$value" 
      },
      in:{ 
       $reduce: {
        input: {
         $map: {
          input: "$$currenta",
          as: "a",
          in: {
           $map: {
            input: "$$currentr",
            as: "r",
            in: {
             $mergeObjects: ["$$a", "$$r"] 
            }
           }
          }
         }
        },
        initialValue: [],
        in: {
         $concatArrays: ["$$value", "$$this"] 
        }
       }
      }
     }
    }
   }
  }
 }
});

参考： Cartesian product of multiple arrays in JavaScript

C＃Mongodb多个对象数组文档的笛卡尔积

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