How to check if a pymongo cursor has query results

Question

I need to check if a find statement returns a non-empty query.

What I was doing was the following:

query = collection.find({"string": field})
if not query: #do something

Then I realized that my if statement was never executed because find returns a cursor, either the query is empty or not.

Therefore I checked the documentation and I find two methods that can help me:

1. count(with_limit_and_skip=False) which (from the description):

   Returns the number of documents in the results set for this query.

   It seems a good way to check, but this means that I need to count all the results in cursor to know if it is zero or not, right? A little bit expensive?

2. retrieved which (from the description):

   The number of documents retrieved so far.

   I tested it on an empty query set and it returns zero, but it's not clear what it does and I don't know if it's right for me.

So, which is the best way (best practice) to check if a find() query returns an empty set or not? Is one of the methods described above right for this purpose? And what about performance? Are there other ways to do it?

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Just to be clear: I need to know if the query is empty and I'd like to find the best way with the cursor with respect to performance and being pythonic.

Answer 1

EDIT: While this was true in 2014, modern versions of pymongo and MongoDB have changed this behaviour. Buyer beware:

.count() is the correct way to find the number of results that are returned in the query. The count() method does not exhaust the iterator for your cursor, so you can safely do a .count() check before iterating over the items in the result set.

Performance of the count method was greatly improved in MongoDB 2.4. The only thing that could slow down your count is if the query has an index set on it, or not. To find out if you have an index on the query, you can do something like

query = collection.find({"string": field})
print query.explain()

If you see BasicCursor in the result, you need an index on your string field for this query.

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EDIT: as @alvapan pointed out, pymongo deprecated this method in pymongo 3.7+ and now prefers you to use count_documents in a separate query.

item_count = collection.count_documents({"string": field})

The right way to count the number of items you've returned on a query is to check the .retreived counter on the query after you iterate over it, or to enumeratethe query in the first place:

# Using .retrieved
query = collection.find({"string": field})
for item in query:
    print(item)

print('Located {0:,} item(s)'.format(query.retrieved))

Or, another way:

# Using the built-in enumerate
query = collection.find({"string": field})
for index, item in enumerate(query):
    print(item)

print('Located {0:,} item(s)'.format(index+1))

Answer 2

How about just using find_one instead of find ? Then you can just check whether you got a result or None. And if "string" is indexed, you can pass fields = {"string":1, "_id" :0}, and thus make it an index-only query, which is even faster.

Answer 3

Another solution is converting cursor to list, if the cursor doesn't have any data then empty list else list contains all data.

 doc_list = collection.find({}); #find all data
 have_list = True if len(list(doc_list)) else False;

Answer 4

From my tests, the quickest way is

if query.first():
    # do something

In [51]: %timeit query = MyMongoDoc.objects(); query.first()
100 loops, best of 3: 2.12 ms per loop

In [52]: %timeit query = MyMongoDoc.objects(); query.count()
100 loops, best of 3: 4.28 ms per loop

(Using MongoDB 2.6.7, 2015-03-26)

Answer 5

I end up using simple counter, since i don't want calling server twice for no reason:

cursor = someCollection.find( query )

ct = 0

for doc in cursor:

    ct += 1

    # some code

if ct == 0:
    
    # cursor was empty