In MongoDB's pymongo, how do I do a count()?

If you're using pymongo version 3.7.0 or higher, see this answer instead.

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If you want results_count to ignore your limit():

results = db.datasets.find({"test_set":"abc"}).sort("abc",pymongo.DESCENDING).skip((page-1)*num).limit(num)
results_count = results.count()

for post in results:

If you want the results_count to be capped at your limit(), set applySkipLimit to True:

results = db.datasets.find({"test_set":"abc"}).sort("abc",pymongo.DESCENDING).skip((page-1)*num).limit(num)
results_count = results.count(True)

for post in results:

Since pymongo version 3.7.0 and above count() is deprecated. Instead use Collection.count_documents. Running cursor.count or collection.count will result in following warning message:

DeprecationWarning: count is deprecated. Use Collection.count_documents instead.

To use count_documents the code can be adjusted as follows

import pymongo

db = pymongo.MongoClient()
col = db[DATABASE][COLLECTION]

filter = {"test_set":"abc"}
sort = [("abc",pymongo.DESCENDING)]
skip = 10
limit = 10

doc_count = col.count_documents(filter, skip=skip)
results = col.find(filter).sort(sort).skip(skip).limit(limit)

for doc in result:
   //Process Document

Note: count_documents method performs relatively slow as compared to count method. In order to optimize you can use collection.estimated_document_count. This method will return estimated number of docs(as the name suggested) based on collection metadata.

Not sure why you want the count if you are already passing limit 'num'. Anyway if you want to assert, here is what you should do.

results = db.datasets.find({"test_set":"abc"}).sort("abc",pymongo.DESCENDING).skip((page-1)*num).limit(num)

results_count = results.count(True)

That will match results_count with num