Will indexing make find_one() any faster?

Question

If I have a pymongo query, in a collection with around 4000 documents, like the following:

mong  =  pymongo.Connection()['ASD_2']['APS2'] 
py_mong = mong.find_one({'plate':'123456'})

Considering there is no .explain()["cursor"] or .explain()["nscanned"] allowed on find_one() , and therefore no method at hand to find out the nature of the scan, can anyone tell me if it's worth indexing the collection when I am only going to be using find_one() ?

Answer 1

是的，建立索引将避免线性搜索所需要的板并在对数时间内工作（快得多）。

Answer 2

Yes, absolutely.

How do I know?

First, it would be utterly unacceptable if it didn't. People would have complained about the surprising and unnecessary slowness of find_one as compared to find , and the dev team would have fixed it (or risk being viewed as dumb).

Second, I checked the code. The implementation is exactly what you'd expect, meaning find_one is a mere wrapper around find :

def find_one(self, spec_or_id=None, *args, **kwargs):
    if spec_or_id is not None and not isinstance(spec_or_id, dict):
        spec_or_id = {"_id": spec_or_id}
    for result in self.find(spec_or_id, *args, **kwargs).limit(-1):
        return result
    return None

(pymongo version 2.4.2)