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
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
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
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
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
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
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
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
edited Nov 23 '18 at 10:37
Anthony Winzlet
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
edited Nov 23 '18 at 10:37
Anthony Winzlet
13.9k41239
edited Nov 23 '18 at 10:37
Anthony Winzlet
13.9k41239
edited Nov 23 '18 at 10:37
Anthony Winzlet
13.9k41239
13.9k41239
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
asked Nov 21 '18 at 11:48
Lakshman Pilaka
6472722
6472722
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 Winzlet
13.9k41239
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53411390%2fcount-distinct-in-mongodb-not-giving-distinct-results%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 Winzlet
13.9k41239
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": "$deviceCount" }
}}
])
Output
[
{
"city": "city1",
"deviceCount": 1,
"deviceType": 0
},
{
"city": "city1",
"deviceCount": 2,
"deviceType": 1
}
]
answered Nov 21 '18 at 19:24
Anthony Winzlet
13.9k41239
add a comment |
1
1
1
You can use below aggregation
db.collection.aggregate([
{ "$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 Winzlet
13.9k41239
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 Winzlet
13.9k41239
answered Nov 21 '18 at 19:24
Anthony Winzlet
13.9k41239
answered Nov 21 '18 at 19:24
Anthony Winzlet
13.9k41239
answered Nov 21 '18 at 19:24
Anthony Winzlet
13.9k41239
13.9k41239
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53411390%2fcount-distinct-in-mongodb-not-giving-distinct-results%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
stackoverflow.com/questions/24761266/… read last answer.
– Sameer
Nov 21 '18 at 12:30