Conditional $sum in MongoDB

MongodbMongodb QueryAggregation Framework

Mongodb Problem Overview


My collection in mongodb is similar to the following table in SQL:

Sentiments(Company,Sentiment)

Now, I need to execute a query like this:

SELECT
  Company, 
  SUM(CASE WHEN Sentiment >0 THEN Sentiment ELSE 0 END) AS SumPosSenti, 
  SUM(CASE WHEN Sentiment <0 THEN Sentiment ELSE 0 END) AS SumNegSenti
FROM Sentiments
GROUP BY Company

What should I do to write this query in Mongo? I am stuck at the following query:

db.Sentiments.aggregate(
{ $project: {_id:0, Company:1, Sentiment: 1} },
{ $group: {_id: "$Company", SumPosSenti: {$sum: ? }, SumNegSenti: {$sum: ? } } }
);

Mongodb Solutions


Solution 1 - Mongodb

As Sammaye suggested, you need to use the $cond aggregation projection operator to do this:

db.Sentiments.aggregate(
    { $project: {
        _id: 0,
        Company: 1,
        PosSentiment: {$cond: [{$gt: ['$Sentiment', 0]}, '$Sentiment', 0]},
        NegSentiment: {$cond: [{$lt: ['$Sentiment', 0]}, '$Sentiment', 0]}
    }},
    { $group: {
        _id: "$Company",
        SumPosSentiment: {$sum: '$PosSentiment'},
        SumNegSentiment: {$sum: '$NegSentiment'}
    }});

Solution 2 - Mongodb

Starting from version 3.4, we can use the $switch operator which allows logical condition processing in the $group stage. Of course we still need to use the $sum accumulator to return the sum.

db.Sentiments.aggregate(
    [
        { "$group": { 
            "_id": "$Company",  
            "SumPosSenti": { 
                "$sum": { 
                    "$switch": { 
                        "branches": [ 
                            { 
                                "case": { "$gt": [ "$Sentiment", 0 ] }, 
                                "then": "$Sentiment"
                            }
                        ], 
                        "default": 0 
                    }
                }
            }, 
            "SumNegSenti": {
                "$sum": { 
                    "$switch": { 
                        "branches": [ 
                            { 
                                "case": { "$lt": [ "$Sentiment", 0 ] }, 
                                "then": "$Sentiment"
                            }
                        ], 
                        "default": 0 
                    } 
                }
            }
        }}
    ]
)

If you have not yet migrated your mongod to 3.4 or newer, then note that the $project stage in this answer is redundant because the $cond operator returns a numeric value which means that you can $group your documents and apply $sum to the $cond expression.

This will improve the performance in your application especially for large collection.

db.Sentiments.aggregate(
    [
        { '$group': {
            '_id': '$Company',
            'PosSentiment': { 
                '$sum': {
                    '$cond': [
                        { '$gt': ['$Sentiment', 0]}, 
                        '$Sentiment', 
                        0
                    ]
                }
            },
            'NegSentiment': { 
                '$sum': {
                    '$cond': [
                        { '$lt': ['$Sentiment', 0]}, 
                        '$Sentiment', 
                        0
                    ]
                }
            }
        }}
    ]
)

Consider a collection Sentiments with the following documents:

{ "Company": "a", "Sentiment" : 2 }
{ "Company": "a", "Sentiment" : 3 }
{ "Company": "a", "Sentiment" : -1 }
{ "Company": "a", "Sentiment" : -5 }

The aggregation query produces:

{ "_id" : "a", "SumPosSenti" : 5, "SumNegSenti" : -6 }

Solution 3 - Mongodb

Explaining the snippets above, that uses the array syntax:

PosSentiment: {$cond: [{$gt: ['$Sentiment', 0]}, '$Sentiment', 0]}

is equal to:

PosSentiment: {$cond: { if: {$gt: ['$Sentiment', 0]}, then: '$Sentiment', else: 0} }

The array syntax summarizes the long syntax to just { $cond: [if, then, else] }

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAafreen SheikhView Question on Stackoverflow
Solution 1 - MongodbJohnnyHKView Answer on Stackoverflow
Solution 2 - MongodbstyvaneView Answer on Stackoverflow
Solution 3 - MongodbdpolicastroView Answer on Stackoverflow