MapReduce di PyMongo

Anda dapat menggunakan kerangka kerja agregasi

import pymongo
conn = pymongo.MongoClient()
db = conn.test
col =  db.collection

for doc in col.aggregate([{'$unwind': '$impressions'}, 
    {'$match': {'impressions.service': 'furniture'}}, 
    {'$group': {'_id': '$impressions.id', 'impressions_count': {'$sum': 1}}}, 
    ]):
    print(doc)

Atau lebih efisien menggunakan $map dan $setDifference operator.

col.aggregate([
    { "$project": { "impressions": {"$setDifference": [{ "$map": { "input": "$impressions", "as": "imp", "in": { "$cond": { "if": { "$eq": [ "$$imp.service", "furniture" ] }, "then": "$$imp.id", "else": 0 }}}}, [0]]}}}, 
    { "$unwind": "$impressions" }, 
    { "$group": { "_id": "$impressions", "impressions_count": { "$sum": 1 }}}
])

Yang menghasilkan:

{'_id': 122.0, 'impressions_count': 1}
{'_id': 124.0, 'impressions_count': 1}
{'_id': 127.0, 'impressions_count': 1}
{'_id': 123.0, 'impressions_count': 2}