count multiple colmns and true results only mongodb query - mongodb

Question

I'm new to mongodb and I've been trying to figure this out but I haven't so far. I think I'm close. I have records similar to this.
{
"_id" : ObjectId("5bfdf4385a37e507cff0ff62"),
"Search_word" : "job01",
"hadoop" : 0,
"hive" : 0,
"javascript" : 0,
"mongodb" : 0,
"sql" : 1,
"java" : 0,
"sas" : 0,
"powerbi" : 0,
"python" : 1,
"pig" : 0,
"scala" : 0
}
I'm trying to create groups by search word and count the 1s in all the other values. Ive never used mongodb before.
The end result would look like this
job01, sql : 100, hive: 205, etc...
job02, sql : 121, hive 10, etc...
In python its literally like this
skill_data = df.groupby(by='Search_word').sum()
I tried doing something like this just to get sql count where it is 1
db.data_final.aggregate(
{"$group" :
{_id : {Search_word:"$Search_word", sql: {"$eq":["$sql",1]} }, count : { $sum : 1} } } )
but it gives me 2 counts one for false where sql not equal 1 and a count that I want for sql that is 1. how can I get rid of the false ? This is what I get now:
{ "_id" : { "Search_word" : "job01", "sql" : true }, "count" : 124 } // this is what i want only and do it for all other values in one query
{ "_id" : { "Search_word" : "job01", "sql" : false}, "count" : 279 }
Any help is appreciated

Answer 1

You could try this since the value of each field is either 0 or 1.
db.collection.aggregate({$group: {_id: "$Search_word", sql: {$sum: "$sql"}}});