CODEWITHSUNDEEP
pythonmongodbpymongo
sanyassh
sanyassh

Reputation: 8530

Pymongo aggregate() $project

I want to exclude the '_id' field and replace it with 'action_id' field when using aggregate. The problem that i must specify all other fields in $project that i dont want to exclude. This is my query:

cursor = db.aggregate(
        {
            '$group': {
                '_id': '$somefield',
                'count': {'$sum': 1},
                'average_latency': {'$avg': '$latency'}
            }
        },
        {
            '$project': {
                'action_id': '$_id',
                '_id': False,
                'count': True,
                'average_latency': True
                }
        }

What shall i do to avoid writing every 'field': True in $project?

Upvotes: 1

Views: 965

Answers (1)

Ashh
Ashh

Reputation: 46471

You can use $addFields to add an extra field and then $project to remove _id

cursor = db.aggregate(
  { '$group': {
    '_id': '$somefield',
    'count': { '$sum': 1 },
    'average_latency': { '$avg': '$latency' }
  }},
  { '$addFields': { 'action_id': '$_id' }},
  { '$projects': { '_id': False }}
)

Upvotes: 1

Related Questions