Finding closest key in defaultdict

Question

I am using defaultdicts to store lists of values where the keys are periods for which values could be observed. When looking up from a list of all periods of interest, I would like to find the closest period in my defaultdict (NB: not all periods are stored in the defaultdict).

As defaultdicts are not sorted however, the below approach does not return the correct value.

Is there a different way of returning the closest available key for defaultdicts?

from collections import defaultdict
import numpy as np

def_dict = defaultdict(list)
# entries that will be stored in the defaultdict
reg_dict = {0: ["a", "b"], 2: ["c", "d"], 5: ["k", "h"], -3: ["i", "l"]}

# store items from regular dict in defaultdict 
for k, v in reg_dict.items():
    def_dict[k] = v

# Lookup periods
periods = [-1, 0, 1, 2, 3, 4, 5, 6, 7, 8]

for period in periods:

    # this approach does not return the right keys as defaultdicts are not sorted
    closest_key = np.abs(np.array(list(def_dict.keys())) - period).argmin()

    print("period: ", period, " - looked up key: ", closest_key)

This returns the following:

period:  -1  - looked up key:  0
period:  0  - looked up key:  0
period:  1  - looked up key:  0
period:  2  - looked up key:  1
period:  3  - looked up key:  1
period:  4  - looked up key:  2
period:  5  - looked up key:  2
period:  6  - looked up key:  2
period:  7  - looked up key:  2
period:  8  - looked up key:  2

Answer 1

The way I understand, you want an output similar to this?

[0, 0, 0, 2, 2, 5, 5, 5, 5, 5]

For the above, the logic would be

closest_key = [min(def_dict.keys(), key = lambda x: abs(x - p)) for p in periods]

Specifying the optional key parameter to built in python functions is useful in situations like these.