It seems you want to achieve something similar to RDBMS

SELECT community_name, COUNT(*)
FROM table
GROUP BY community_name
HAVING COUNT(*) > 1

I don’t know if I understand it correctly. If this is the case, the corresponding approach should be to use the aggregation framework.

db.coll.aggregate([
    {$group: {_id: "$community_name", count: {$sum: 1}}}, //统计community_name重复出现的次数
    {$match: {count: {$gt: 1}}} //从中找出重复多于1次的记录
]);

This query can get results faster with the following indexes:

db.coll.createIndex({community_name: 1});

But even so, this query will traverse all records, and the speed will not be too fast.
In fact, it is wasteful to count all the records every time. It is best to cache the results after getting the results. How to cache depends on how you want to use the collected data.
A better way is to make a judgment before inserting. If the same community_name already exists, record it, such as

db.community_name_stat.update({
    community_name: 'xxx'
}, {
    '$set': {
        count: {'$inc': 1}
    },
    '$setOnInsert': {
        community_name: 'xxx',
        count: 1
    }
}, {
    upsert: true
});

In this way, you can directly get a community_name_stat collection to get how many times each community_name_stat集合得到每个community_name appears. Of course, the final approach depends on your needs. MongoDB is a very flexible thing, which is one of the important features that distinguishes it from relational databases. Understanding its various functions and customizing a most cost-effective solution for your needs is one of the biggest challenges in using MongoDB.

If you understand it correctly, you can use upsert directly: if the system already has a record with the same conditions, only update it, otherwise create a new record.

db.collection.update(query, update, {upsert: True, multi: <boolean>})

And you can also modify multiple records, if multi is set to true.