Swipe through data once, with a comb that will pick up the points you're looking for.
Think single pass search.
Think k nearest neighbors, but brute force, but with multiple targets.
Extremely light weight: Pure python -- no other dependencies.
The main function of swipe is highest_score_swipe.
The other objects are there to support you in making your own kind of
swiping function.
What highest_score_swipe does is
get the k items from the iterable it who score the highest with score_of.
(think of score_of an inverse of key argument of the sorted python
function).
As far as the output is concerned, you can acheive something about the same with
highest_score_swipe(it, score_of, k, output)
as with
output(sorted(it, key=score_of, reverse=True)[:k])
(With slightly different output functions)
The difference is that in the last one,
- you have to fit all of the data in memory
- you have to sort all of the data
But to get the top k elements you don't have to. You just have to scan though
the data once while maintaining a list of the top items.
So when there's a lot of data, highest_score_swipe will save you both memory and
computation.
>>> from swipe import highest_score_swipe
>>>
>>> data = [('Christian', 12), ('Seb', 88), ('Thor', 27), ('Sylvain', 42)]Let's see what you get out of the box (i.e. only specifying what's required,
using defaults for all the rest).
We'll iter(data) just to this once to show that data only has to be iterable.
>>> highest_score_swipe(iter(data))
[(('Thor', 27), ('Thor', 27))]Now, out of the box, you don't get much, and looks a bit strange.
Reason is if you don't specify k you just get the
top item, and if you don't specify what score should be used to measure the "top",
it'll just use python's default comparison operator which here brings
('Thor', 27) on the top because it's lexicographically the last.
And why is ('Thor', 27) repeated twice?
Because it acts both as a score (the first) and a data item (the second).
Where it becomes interesting (and useful) is when you specify what score function it should use. So let's.
>>> length_of_name = lambda x: len(x[0])
>>> by_age = lambda x: x[1]
>>> highest_score_swipe(data, by_age)
[(88, ('Seb', 88))]
>>> highest_score_swipe(data, length_of_name)
[(9, ('Christian', 12))]
>>> highest_score_swipe(data, length_of_name, k=2)
[(7, ('Sylvain', 42)), (9, ('Christian', 12))]Now let's see about that output argument.
It's used to specify how you want the result to be processed before returning.
>>> highest_score_swipe(data, length_of_name, k=2, output='top_tuples')
[(9, ('Christian', 12)), (7, ('Sylvain', 42))]
>>> highest_score_swipe(data, length_of_name, k=2, output='items')
[('Sylvain', 42), ('Christian', 12)]
>>> highest_score_swipe(data, length_of_name, k=2, output='scores')
[7, 9]
>>> highest_score_swipe(data, length_of_name, k=2, output='top_score_items')
[('Christian', 12), ('Sylvain', 42)]You can also specify a custom function:
>>> highest_score_swipe(
... data, length_of_name, k=2,
... output=lambda km: [f"{name} (whose name has {score} letters), is {age}" for score, (name, age) in km]
... )
['Sylvain (whose name has 7 letters), is 42', 'Christian (whose name has 9 letters), is 12']What if you wanted the indices (that is, the integer indexing the data) of the top 2 as your output? Here's a recipe for that:
>>> highest_score_swipe(
... enumerate(data), # enumerate the data to get a (i, item) iterator
... lambda x: length_of_name(x[1]), # apply your scoring function to the item
... k=2,
... output=lambda km: [x[1][0] for x in km] # extract the indices
... )