Как через функцию max вывести 3 наибольших значения

my_dict = {'one': 10000, 'two': 5000, 'three': 2000, 'four': 7000}
max_val = max(my_dict.values())
final_dict = {k:v for k, v in my_dict.items() if v == max_val}
print(final_dict)

>>> my_dict = {'one': 10000, 'two': 5000, 'three': 2000, 'four': 7000}
>>> sorted(my_dict.values(), reverse=True)[:3]
[10000, 7000, 5000]

my_dict = {'one': 10000, 'two': 5000, 'three': 2000, 'four': 7000}
for i in range(3):
    print(my_dict.pop(max(my_dict, key=my_dict.get)))

m = {'one': 10000, 'two': 5000, 'three': 2000, 'four': 7000}
for _ in range(3):
    k = max(m, key=m.get)
    print(k, m.pop(k))

In [1]: def test1(dict_):
   ...:     return sorted(dict_.values(), reverse=True)[:3]
   ...:
 
In [2]: def test2(dict_):
   ...:     d = dict(dict_)
   ...:     return tuple(d.pop(max(d, key=d.get)) for i in range(3))
   ...:
 
In [3]: def test3(dict_):
   ...:     from heapq import nlargest
   ...:     return nlargest(3, dict_.values())
   ...:
 
In [4]: from random import randrange
 
In [5]: dict_ = {str(i) : randrange(-100_000, 100_000) for i in range(100_000)}
 
In [6]: timeit test1(dict_)
67 ms ± 3.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
 
In [7]: timeit test2(dict_)
49.7 ms ± 4.32 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
 
In [8]: timeit test3(dict_)
10.7 ms ± 2.09 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)