Custom dictionary lookup in Python

Question

if I have a dictionary like this

>>> d = {10: 3, 100: 2, 1000: 1}

I can type something like:

>>> d.get(10), d.get(100), d.get(1000)
(3, 2, 1)

Though I want that if the given key is not found, the value corresponding to the nearest key respect the given key is returned:

>>> d.get(20), d.get(60), d.get(200)
(3, 2, 2)

Instead the result in Python is

(None, None, None)

What's a Pythonic way to implement the behavior I described?

Thanks

Answer 1

You can derive from dict to change the behaviour of the get() method:

class ClosestDict(dict):
    def get(self, key):
        key = min(self.iterkeys(), key=lambda x: abs(x - key))
        return dict.get(self, key)

d = ClosestDict({10: 3, 100: 2, 1000: 1})
print (d.get(20), d.get(60), d.get(200))

prints

(3, 2, 2)

Note that the complexity of get() no longer is O(1), but O(n).

Answer 2

bisect module allows fast lookup of insertion position in a sorted list.

from bisect import bisect_right

def closest_matches(data, query):
    keys = sorted(data)
    return [data[i] for i in (min(map(abs, (keys[p-1], keys[p]))) for p in (bisect_right(keys, k) for k in query))]

>>> d = {10: 3, 100: 2, 1000: 1}
>>> closest_matches(d, [20, 60, 200])
[3, 3, 2]

Answer 3

Checkout this recipe Fuzzy matching dictionary .