Count distinct in mongodb not giving distinct results
1
a proper sql guy struggling with no-sql:
i am looking for an alternative in mongodb for a query like
select city, count(distinct device) from collection group by city, devicetype
My collection is like
{
"city":"city1",
"device": "device1"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
},
{
"city":"city1",
"device": "device3"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
}
After grouping, i need distinct devices used by their devicetype of each city
means, the output would be like
{
"city":"city1",
"deviceType":0,
"deviceCount":1 //device2
},
{
"city":"city1",
"deviceType":1,
"deviceCount":2 //device1, device3
}
I tried
"$group": {
_id: { "city": "$city", "devicetype": "$devicetype" },
"devicetype": { "$first": "$devicetype" },
"count": { "$sum": 1 }
}
I am getting the total count of devices, not the distinct count.
please suggest!!
mongodb mongodb-query aggregation-framework
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
add a comment |
1
a proper sql guy struggling with no-sql:
i am looking for an alternative in mongodb for a query like
select city, count(distinct device) from collection group by city, devicetype
My collection is like
{
"city":"city1",
"device": "device1"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
},
{
"city":"city1",
"device": "device3"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
}
After grouping, i need distinct devices used by their devicetype of each city
means, the output would be like
{
"city":"city1",
"deviceType":0,
"deviceCount":1 //device2
},
{
"city":"city1",
"deviceType":1,
"deviceCount":2 //device1, device3
}
I tried
"$group": {
_id: { "city": "$city", "devicetype": "$devicetype" },
"devicetype": { "$first": "$devicetype" },
"count": { "$sum": 1 }
}
I am getting the total count of devices, not the distinct count.
please suggest!!
mongodb mongodb-query aggregation-framework
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30
add a comment |
1
1
1
a proper sql guy struggling with no-sql:
i am looking for an alternative in mongodb for a query like
select city, count(distinct device) from collection group by city, devicetype
My collection is like
{
"city":"city1",
"device": "device1"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
},
{
"city":"city1",
"device": "device3"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
}
After grouping, i need distinct devices used by their devicetype of each city
means, the output would be like
{
"city":"city1",
"deviceType":0,
"deviceCount":1 //device2
},
{
"city":"city1",
"deviceType":1,
"deviceCount":2 //device1, device3
}
I tried
"$group": {
_id: { "city": "$city", "devicetype": "$devicetype" },
"devicetype": { "$first": "$devicetype" },
"count": { "$sum": 1 }
}
I am getting the total count of devices, not the distinct count.
please suggest!!
mongodb mongodb-query aggregation-framework
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
a proper sql guy struggling with no-sql:
i am looking for an alternative in mongodb for a query like
select city, count(distinct device) from collection group by city, devicetype
My collection is like
{
"city":"city1",
"device": "device1"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
},
{
"city":"city1",
"device": "device3"
"deviceType":1
},
{
"city":"city1",
"device": "device2"
"deviceType":0
}
After grouping, i need distinct devices used by their devicetype of each city
means, the output would be like
{
"city":"city1",
"deviceType":0,
"deviceCount":1 //device2
},
{
"city":"city1",
"deviceType":1,
"deviceCount":2 //device1, device3
}
I tried
"$group": {
_id: { "city": "$city", "devicetype": "$devicetype" },
"devicetype": { "$first": "$devicetype" },
"count": { "$sum": 1 }
}
I am getting the total count of devices, not the distinct count.
please suggest!!
mongodb mongodb-query aggregation-framework
mongodb mongodb-query aggregation-framework
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
edited Nov 23 '18 at 10:37
Anthony Winzlet
15.6k41339
15.6k41339
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
asked Nov 21 '18 at 11:48
Lakshman PilakaLakshman Pilaka
6522722
6522722
2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30
add a comment |
2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30
2
2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30
add a comment |
1 Answer
1
active
oldest
votes
1
You can use below aggregation
db.collection.aggregate([
{ "$group": {
_id: {
"city": "$city",
"deviceType": "$deviceType"
},
"deviceCount": { "$addToSet": "$device" },
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": 0,
"city": "$_id.city",
"deviceType": "$_id.deviceType",
"deviceCount": { "$size": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You can use below aggregation
db.collection.aggregate([
{ "$group": {
_id: {
"city": "$city",
"deviceType": "$deviceType"
},
"deviceCount": { "$addToSet": "$device" },
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": 0,
"city": "$_id.city",
"deviceType": "$_id.deviceType",
"deviceCount": { "$size": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
add a comment |
1
You can use below aggregation
db.collection.aggregate([
{ "$group": {
_id: {
"city": "$city",
"deviceType": "$deviceType"
},
"deviceCount": { "$addToSet": "$device" },
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": 0,
"city": "$_id.city",
"deviceType": "$_id.deviceType",
"deviceCount": { "$size": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
add a comment |
1
1
1
You can use below aggregation
db.collection.aggregate([
{ "$group": {
_id: {
"city": "$city",
"deviceType": "$deviceType"
},
"deviceCount": { "$addToSet": "$device" },
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": 0,
"city": "$_id.city",
"deviceType": "$_id.deviceType",
"deviceCount": { "$size": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
You can use below aggregation
db.collection.aggregate([
{ "$group": {
_id: {
"city": "$city",
"deviceType": "$deviceType"
},
"deviceCount": { "$addToSet": "$device" },
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": 0,
"city": "$_id.city",
"deviceType": "$_id.deviceType",
"deviceCount": { "$size": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
answered Nov 21 '18 at 19:24
Anthony WinzletAnthony Winzlet
15.6k41339
15.6k41339
add a comment |
add a comment |
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2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30