BaseQuery.paginate() return inconsistent results - python-2.7

Question

When I manually execute this query in MySQL, the number of results I get is 28:
SELECT *
FROM position_reporting_structures prs
INNER JOIN positions AS pos ON pos.id = prs.reports_to
INNER JOIN jobs ON jobs.id = pos.job_id
INNER JOIN position_fulfillments AS pf ON pf.position_id = pos.id
INNER JOIN parties ON parties.id = pf.party_id
WHERE jobs.title LIKE '%QA/QC%' OR parties.name LIKE '%QA/QC%';
This is the interpreted SQLAlchemy value of the query above:
query = (
PositionReportingStructure.query
.join(Position,
Position.id == PositionReportingStructure.reports_to)
.join(Job,
Job.id == Position.job_id)
.join(PositionFulfillment,
PositionFulfillment.position_id == Position.id)
.join(Party,
Party.id == PositionFulfillment.party_id)
.filter(db.or_(
Job.title.like('%QA/QC%'),
Party.name.like('%QA/QC%')))
)
I have a pagination decorator that wraps a function whose returning value is a BaseQuery. Inside the decorator, these are the code snippets that I've used:
query = f(*args, **kwargs)
page = 1 // just a dummy value
per_page = 5 // just a dummy value
paginate = query.paginate(page=page, per_page=per_page)
I'm expecting that it will return more or less 6 pages and 28 items in total, but it's not. The outcome is 1 page, 4 items per page, 4 total items and no previous/next pages which is incorrect. For further investigation, I have tried changing the value of page variable to compare each one:
page = 1 : 1 page, 4 items, 4 total items
page = 2 : 6 pages, 3 items, 28 total items
page = 3 : 6 pages, 2 items, 28 total items
page = 4 : 6 pages, 3 items, 28 total items
page = 5 : 6 pages, 5 items, 28 total items
page = 6 : 6 pages, 3 items, 28 total items
As you may notice, per page items is too inconsistent. Anybody who can explain what causes this contingency? I am currently using Flask-SQLAlchemy-2.0.

Answer 1

I already knew the root cause of this problem. My query returns some duplicate rows which I think is not acceptable to SQLAlchemy so in return automatically gives the distinct ones (even without using GROUP BY or DISTINCT).
As the context of my problem stated, I passed a count of 5 items in the per_page parameter of the paginate function, at the first page, it would only return 4 items since it groups the duplicate ones, thus, satisfies the paginate function's conditional statement (the one that has an emphasis):
**if page == 1 and len(items) < per_page:
total = len(items)**
else:
total = self.order_by(None).count()