Count multiple fields in MongoDB using python

Question

I have this simple mongodb query:

db.checks_reports_data_df8.find({"report_id": str(report["_id"])})

The collection checks_reports_data_df8 is having some fields like "success" (can be 0 or 1) and is_listed (can be 0 or 1).

What I need is to be able to get the counts of all fields where success = 1 as count_success and all fields where is_listed = 1 as count_listed. The final result will be something like all the results from the query above + count_success and count_listed

I suppose that this can be done using the aggregation framework.

EDIT: To clarify, here is the return from the query above:

{'is_listed': 0, 'success': 1, '_id': ObjectId('54dca9920e13a771a44433d4'), 'report_id': u'54dca97758a5d3a37c8b4567'}
{'is_listed': 0, 'success': 1, '_id': ObjectId('54dca9920e13a771a44433a3'), 'report_id': u'54dca97758a5d3a37c8b4567'}
{'is_listed': 1, 'success': 1, '_id': ObjectId('54dca9920e13a771a44433c2'), 'report_id': u'54dca97758a5d3a37c8b4567'}

What I need, is to have count_success = 3 (the sum of the success=1 fields) and count_listed = 1 (the sum of the is_listed=1 fields)

For clarity, could you post a sample document and your expected output? — wdberkeley
– wdberkeley, Commented Feb 12, 2015 at 20:12

thegreenogre · Accepted Answer · 2015-04-08 13:23:52Z

1

Something like this should work.

db.checks_reports_data_df8.aggregate([
        {
                '$match' : 
                    {
                        "report_id": str(report["_id"])
                    }
        },
        {
            '$group' : 
                {
                    '_id': '$report_id',
                    count_success :{'$sum': '$success'},
                    count_is_listed:{'$sum':'$is_listed'}
                }
        }
])

answered Apr 8, 2015 at 13:23

thegreenogre

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