Reputation: 1117
Given collection:
{
"_id" : "1.1000038",
"recomendation" : [
"1.6739718"
]
}
/* 2 */
{
"_id" : "1.1000069",
"recomendation" : [
"1.9185509",
"1.9051998",
"1.9034279",
"1.8288046",
"1.8152670",
"1.858775",
"1.6224229",
"1.4591674",
"1.3862464",
"1.3427739",
"1.3080062",
"1.3003608",
"1.1694619",
"1.1634683",
"1.1590664",
"1.1524146",
"1.754599",
"1.700837",
"1.763617"
]
}
I need to query the MongoDB for a list of values and get the first element of the list of values
here is the query by mongo syntax
db.getCollection('similar_articles').find({"_id":{$in:["1.1000069","1.1000038"]}})
I don't want to filter it on the python side because it's can be too big.
I didn't find any documentation on it
desire output: Pandas DataFrame
_id recom
1.1000038 1.6739718
1.1000069 1.9185509
Upvotes: 0
Views: 196
Reputation: 15235
I don't know pymongo
so well, but you need this query:
First $match
by _id
s into the arreay (this is like the find you have).
And later use $project
to create the field recom
(you can use "recomendation" to overwrite the existing field) and set the value as the first into the array.
db.collection.aggregate([
{
"$match": { "_id": { "$in": [ "1.1000069", "1.1000038" ] } }
},
{
"$project": { "recom": { "$arrayElemAt": [ "$recomendation", 0 ] } }
}
])
Example here
Looking the doumentation it seems you only need to copy and paste this query.
Upvotes: 1