Get closest dictionary of data with Python

Question

Considering you have a list of dictionaries with different elements eg:

{
    "acousticness": 0.0681,
    "energy": 0.724,
    "loudness": -5.941,
    "tempo": 132.056,
    "valence": 0.676
},
{
    "acousticness": 0.2754,
    "energy": 0.866,
    "loudness": -7.874,
    "tempo": 180.056,
    "valence": 0.540
},    
{
    "acousticness": 0.0681,
    "energy": 0.724,
    "loudness": -5.941,
    "tempo": 132.056,
    "valence": 0.676
}

And you give the user the ability to enter a dictionary themselves eg

{
    "acousticness": 0.1382,
    "energy": 0.7274,
    "loudness": -5.8246,
    "tempo": 122.6412,
    "valence": 0.6153
}

How would you iterate through the dictionary in Python3 to get the closest dictionary back?

I found this explanation on how to find the nearest element in a normal array but can't get around how to do the same with array comparison.

Thanks for any help in advance!

Answer 1

Assuming you want the minimum absolute difference among the values of the dictionaries, you could do something like this:

data = [{
    "acousticness": 0.0681,
    "energy": 0.724,
    "loudness": -5.941,
    "tempo": 132.056,
    "valence": 0.676
},
    {
        "acousticness": 0.2754,
        "energy": 0.866,
        "loudness": -7.874,
        "tempo": 180.056,
        "valence": 0.540
    },
    {
        "acousticness": 0.0681,
        "energy": 0.724,
        "loudness": -5.941,
        "tempo": 132.056,
        "valence": 0.676
    }]

target = {
    "acousticness": 0.1382,
    "energy": 0.7274,
    "loudness": -5.8246,
    "tempo": 122.6412,
    "valence": 0.6153
}


def key(d, t=target):
    return sum(abs(t[k] - v) for k, v in d.items())


result = min(data, key=key)
print(result)

Output

{'tempo': 132.056, 'loudness': -5.941, 'acousticness': 0.0681, 'valence': 0.676, 'energy': 0.724}

The key of the answer, is to use the key parameter of min . Note that this answer can be tweaked to accommodate multiple closeness measures. For example you could change key to compute the euclidean distance between the dictionary values:

import math

def key(d, t=target):
    return math.sqrt(sum((t[k] - v)**2 for k, v in d.items())