Count number of times string matches a field across multiple documents - mongodb

Question

Say I have a number of documents that look like this:
{
'domain': 'www.stackoverflow.com',
'time': 1380860676457
}
{
'domain': 'www.google.com',
'time': 1380860678001
}
{
'domain': 'www.stackoverflow.com',
'time': 1380860657233
}
What's the best way to end up with the following output?
{
'domain': 'www.stackoverflow.com',
'count': 2
}
Are there any performance considerations (a different way to store the logs?) if the initial collection contains, say, a million or more documents?

Answer 1

You could use aggregation. Something like:
db.sites.aggregate([{
$group: {
_id: '$domain',
count: {$sum: 1}
}
}]);
This groups on the domain field and adds 1 to count for each document it finds. To make it look like the output you want you could also add a projection operation to your aggregation:
$project: {
domain: '$_id',
count: 1,
_id: 0
}

Answer 2

What you need is just group by domain and sum chunks. You could do it by collection method aggregate like this:
db.cls.aggregate(
{$group:{_id:"$domain", count: {$sum : 1}}},
{$project:{_id:0, domain:"$_id", count:"$count"}}
)
First, $group gives you:
{
"result" : [
{
"_id" : "www.google.com",
"count" : 1
},
{
"_id" : "www.stackoverflow.com",
"count" : 2
}
],
"ok" : 1
}
And the second, $project gives you:
{
"result" : [
{
"count" : 1,
"domain" : "www.google.com"
},
{
"count" : 2,
"domain" : "www.stackoverflow.com"
}
],
"ok" : 1
}
Or you could do it just by collection method group:
db.cls.group({
key: {domain:1},
reduce: function(curr,result){ result.count += 1 },
initial:{count:0}
})
And to speed up process you should have an index on domain field, as was mentioned by #AnujAneja.