MongoDb Aggregation: How can I group an array-1 based on another array-2 when given array-1 and array-2?

Question

EDIT: My original question was

MongoDb Aggregation: Can you $unwind an input document variable in the pipline of a $lookup stage?

Consider the code below:

{$lookup: {
    from:"mydoc", 
    let: {"c":"$myArray"}, 
    pipeline: [ 
        {$unwind: "$$c"},
    ]
    as:"myNewDoc"
 }}

How would I unwind c if I wanted to?

/////END OF ORIGINAL QUESTION

-----EDIT-----

From Tom Slabbaert's comment we now know that it is possible to $unwind an input document variable in the pipline of a $lookup stage. But it is not recommended.

What am I trying to achieve?

Consider these collections, poll and castedvote from this answer from a question I had asked.

I am trying to get an output like below:

numberOfVotes: 6,
hasThisUserVoted: true,
numberOfComments: 12,
castedVotesPerChoice:{
    "choiceA": [
        {"_id": ObjectId("..."), "voter": "Juzi", "choice": 0, "pollId": 100 },
        {"_id": ObjectId("..."), "voter": "Juma", "choice": 0, "pollId": 100 },
        {"_id": ObjectId("..."), "voter": "Jane", "choice": 0, "pollId": 100 },
    ],
    "choiceB": [
        {"_id": ObjectId("..."), "voter": "Jamo", "choice": 1, "pollId": 100 },
        {"_id": ObjectId("..."), "voter": "Juju", "choice": 1, "pollId": 100 },
        {"_id": ObjectId("..."), "voter": "Jana", "choice": 1, "pollId": 100 }
    ],
    "choiceC": [ ]
}

my current implementation:

db.poll.aggregate([
    {"$match": {"_id": 100}}, 
    // ...lookup to get comments
    {"$lookup": {
        "from":"castedvotes", 
        "let": {"pollId":"$_id"}, 
        "pipeline":[
            {"$match":
                {"$expr":
                    {"$eq": ["$pollId", "$$pollId"]},
            }},
        ], 
        "as":"votes" // will use this to get number of votes and find out if the authenticated user has voted.
    }},
    {"$unwind":"$choices"},
    {"$lookup": {
        "from":"castedvotes", 
        "let": {"c":"$choices"}, 
        "pipeline":[
            {"$match":
                {"$expr":
                    {"$eq": ["$choice", "$$c.id"]},
            }},
        ], 
        "as":"votesPerChoice"
    }},
])

The issue I have with my current implementation is that it is doing a lookup on the same collection twice I feel like this is unnecessary and it makes the code not dry. With $unwind I know I can un-$unwind as described here.

So my question is how can I get my desired output with one $lookup to the casted vote collection? Since both lookups return the same data.

Or to ask the question differently how can I group an array-1 based on another array-2 in mongodb aggregation when given array-1 and array-2?

This question answers how to group arrays based on another array in mongodb aggregation by structuring the $lookup stage a certain way. It does not answer my question.

Answer 1

If I've understood the "puzzle" post correctly (Post title and EDIT are different use cases), we can get the desired result with a single $lookup:

db.poll.aggregate([
  {
    "$match": {
      "_id": 100
    }
  },
  {
    "$lookup": {
      "from": "castedvotes",
      "localField": "pollId",
      "foreignField": "choices.id",
      "as": "voters"
    }
  },
  {
    $project: {
      numberOfVotes: {
        $size: "$voters"
      },
      hasThisUserVoted: {
        $in: [
          "$_id",
          "$voters.pollId"
        ]
      },
      /**How to calculate it?*/
      numberOfComments: {
        $multiply: [
          {
            $size: "$voters"
          },
          2
        ]
      },
      castedVotesPerChoice: {
        $arrayToObject: {
          $map: {
            input: "$choices",
            as: "choice",
            in: {
              k: "$$choice.name",
              v: {
                $filter: {
                  input: "$voters",
                  as: "voter",
                  cond: {
                    $eq: [
                      "$$voter.choice",
                      "$$choice.id"
                    ]
                  }
                }
              }
            }
          }
        }
      }
    }
  }
])

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