计算元组列表中唯一元组的发生率

Question

For a classifieds Django website project, I have a list of tuples made up of (user_id, ad_id) pairs. This denotes the clicker's user_id , alongwith the relevant ad_id .

For example:

gross_clicks = [(1, 13),(1, 12), (1, 13), (2, 45), (2, 13), (1, 15), ...(n, m)]

The elements in this list are by no means unique - each click gets pushed into this list regardless of whether it's by the same user and/or it's on the same ad.

Now I can get all unique clicks by doing:

unique_clicks = []
import operator
gross_click_ids = map(operator.itemgetter(0), gross_clicks)
return len(set(gross_click_ids))

But how do I get unique clicks per ad ? Ie if same user clicked on two different ads, that would be counted as 2 separate clicks.

Performance matters too - it's a large data set - so would prefer the most efficient solution, alongwith an illustrative example.

Answer 1

Use the distinct method on the queryset instead. Let's say your model is User and you want to get unique user_id , ad_id pairs.

User.objects.all().values_list('id', 'ad_id').distinct('id', 'ad_id')

This performs the work on the database level, so I expect it would be faster than doing it in Python as Willem mentioned.

I may have misunderstood your question. Please let me know if that's the case so I can try to provide an alternate solution.

Answer 2

Just take unique tuples:

unique_clicks = set(gross_clicks)

This gives you the set of unique user impressions per ad.

In your sample input, (1, 13) appears twice, but in a set it would appear just once:

>>> gross_clicks = [(1, 13), (1, 12), (1, 13), (2, 45), (2, 13), (1, 15)]
>>> set(gross_clicks)
{(1, 12), (1, 13), (1, 15), (2, 45), (2, 13)}

Using sets to track unique elements is as efficient as it can get, given a large list of tuples as input (testing if any given tuple is already in the set is a O(1) constant time operation).

However, if this data came from your database, it is more efficient to ask it to give you unique pairs instead.