Map reduce to count the unique count - mongodb

Question

I want a map reduce function to draw the below output from the below input collection which satisfy the below condition.
Input collection:
[{
a:1,
b:'test',
indices:[1,2,4,5]
}, {
a:2,
b:'test',
indices:[2, 3, 5]
}, {
a:2,
b:'test',
indices:[1, 2, 4]
}, {
a:3,
b:'apple',
indices:[1, 2]
}, {
a:4,
b:'apple',
indices:[1, 3, 5]
}, {
a:5,
b:'orange',
indices:[232]
}, {
a:5,
b:'dummy',
indices:[2]
}, {
a:6,
b:'dummy',
indices:[11, 2, 4]
}, {
a:6,
b:'dummy',
indices:[11, 3, 2]
}, {
a:6,
b:'dummy',
indices:[1, 2, 3, 4, 5]
}]
Conditions are:
select only which has indices array has 2. This can be send as
query. i.e, query:{indices:{$in:2}}
Group by b
If there are duplicates a, then it should be considered as 1 eg: Document having a=2 are present in two times satisfying the condition indices
has 2.
My input collection always satisfies the condition of if a
prsents in "test", it will not present in dummy/apple/etc. but a
can be duplicate.
Here is what I tried:
db.x.mapReduce(function(){
emit(this.b, 1);
}, function(key, reducable){
return Array.sum(reducable);
}, {
out: {inline: 1},
query:{
'indices':{$in:2}
}
});
Output:
[
{
"_id" : test",
"value" : {
"count" : 3 -> It should be 2
}
},{
"_id" : apple",
"value" : {
"count" : 2
}
},{
"_id" : dummy",
"value" : {
"count" : 4 -> It should be 2
}
}]
Expected Output:
[{
"_id" : test",
"value" : {
"count" : 2
}
},{
"_id" : apple",
"value" : {
"count" : 2
}
},{
"_id" : dummy",
"value" : {
"count" : 2
}
}]

Answer 1

No need for map/reduce. Use aggregation:
> db.crawler_status.aggregate([
{ "$match" : { "indices" : 2 } },
{ "$group" : { "_id" : { "b" : "$b", "a" : "$a" } } },
{ "$group" : { "_id" : "$_id.b", "count" : { "$sum" : 1 } } }
])
{ "_id" : "test", "count" : 2 }
{ "_id" : "apple", "count" : 1 } // your sample output was mistaken
{ "_id" : "dummy", "count" : 2 }