Python Rank values from closest to target-CodePudding

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?

CodePudding user response：

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.

CodePudding user response：

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}}

CodePudding user response：

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)