Python Rank values from closest to target

Question

I have a problem. In my code I have the following dictionary:

dict = {} dict['A'] = {'slope': -51, 'score': 0} dict['B'] = {'slope': 12, 'score': 0} dict['C'] = {'slope': -4, 'score': 0} dict['D'] = {'slope': -31, 'score': 0} target = -21

The dictionary isn't sorted! Now what I am trying to do is, give each item in the dict a score (1 to 4). The dict closest to the target gets 4 points, the next one 3 points, etc.

First I thought I iterate over the dict values, but then I realised that I don't know which if there are values closer than the current one, so I can't keep the score about that.

In the end, it needs to look like this:

{'A': {'slope': -51, 'score': 2}, 'B': {'slope': 12, 'score': 1}, 'C': {'slope': -4, 'score': 3}, 'D': {'slope': -31, 'score': 4}}

What can I do to achieve this?

Why is -51 -> 2 and -4 -> 3? Also, don't name your variable dict as you're shadowing builtin name. — CristiFati
– CristiFati, Commented Apr 25, 2022 at 18:26
@CristiFati because -31 is closest to -21, -4 is second closest to -21, -51 is third closest to -21 and 12 is farthest from -21. More Points are assigned if you are closer. — Abhyuday Vaish
– Abhyuday Vaish, Commented Apr 25, 2022 at 18:30
@AbhyudayVaish: Thx, I didn't carefully read the question, it's in reversed order! — CristiFati
– CristiFati, Commented Apr 25, 2022 at 18:32

BrokenBenchmark · Accepted Answer · 2022-04-25 18:34:18Z

Sort the items of the dictionary by distance from the target in descending order. Then, the index of the item plus one is the number of points it should get:

sorted_items = sorted(state.items(), key=lambda x: abs(x[1]['slope'] - target), reverse=True) for idx, (key, _) in enumerate(sorted_items, start=1): state[key]['score'] = idx print(state)

This outputs:

{ 'A': {'slope': -51, 'score': 2}, 'B': {'slope': 12, 'score': 1}, 'C': {'slope': -4, 'score': 3}, 'D': {'slope': -31, 'score': 4} }

I've renamed dict to state here, because dict is the name of a Python builtin.

I like your approach of sorting in reverse and using enumerate() for scores better!

pho · Accepted Answer · 2022-04-25 18:24:15Z

First, don't call your dictionaries dict, because dict is the name of the builtin dictionary class.

d = {} d['A'] = {'slope': -51, 'score': 0} d['B'] = {'slope': 12, 'score': 0} d['C'] = {'slope': -4, 'score': 0} d['D'] = {'slope': -31, 'score': 0} target = -21

Now, you can sort the keys of the dict based on how far the slope value is from target:

>>> sorted_keys = sorted(d.keys(), key=lambda k: abs(d[k]['slope'] - target)) ['D', 'C', 'A', 'B']

Then, you can assign scores:

num_keys = len(sorted_keys) for key, score in zip(sorted_keys, range(num_keys, 0, -1)): d[key]['score'] = score

Which gives you this d:

{'A': {'slope': -51, 'score': 2}, 'B': {'slope': 12, 'score': 1}, 'C': {'slope': -4, 'score': 3}, 'D': {'slope': -31, 'score': 4}}

trincot · Accepted Answer · 2022-04-25 18:38:02Z

You can sort the values of the dict.

Note: don't call your dict like that, since it is the name of a native function.

d = { 'A': {'slope': -51, 'score': 0}, 'B': {'slope': 12, 'score': 0}, 'C': {'slope': -4, 'score': 0}, 'D': {'slope': -31, 'score': 0} } target = -21 for i, item in enumerate(sorted(d.values(), key=lambda x: abs(target - x["slope"]), reverse=True), 1): item["score"] = i print(d)

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