从 Python 中的大型字典中获取前 3 个最高值的快速有效方法是什么？

Question

I've been trying for several hours to go through a dictionary and get three highest-value keys to be presented in the output.

So here's an example dictionary:

dict1 = {
    "a": 5,
    "b": 1,
    "c": 20,
    "d": 15,
    "e": 100,
    "f": 75
}

And I want to parse out the three highest values and present them something like this:

'e' has a value of 100
'f' has a value of 75
'c' has a value of 20

So it should be key-value pairs, not just the values.

I tried something like this but it didn't seem convenient at all.

v = list(dict1.values())
k = list(dict1.keys())

print(f"{k[v.index(max(v))]} has a value of {max(v)}.")

# and so forth. not to mention this is only the one highest value, I need three

I'm sure there's a really simple way right under my nose. Thank you.

Answer 1

The standard library Counter class can make short work of what you are trying to do:

from collections import Counter

dict1 = {
    "a": 5,
    "b": 1,
    "c": 20,
    "d": 15,
    "e": 100,
    "f": 75
}

for k, v in Counter(dict1).most_common(3):
    print(f"{k} has a value of {v}")

Output:

e has a value of 100
f has a value of 75
c has a value of 20

Answer 2

Try to sort the dictionary items and print first 3 items:

dict1 = {"a": 5, "b": 1, "c": 20, "d": 15, "e": 100, "f": 75}

for k, v in sorted(dict1.items(), key=lambda k: k[1], reverse=True)[:3]:
    print(f"'{k}' has a value of {v}")

Prints:

'e' has a value of 100
'f' has a value of 75
'c' has a value of 20