取列表或字典最大/最小的前几个

import heapq

a_list = [3, 4, 2, 5, 1, 6]
c_dict = {'A': 3, 'B': 4, 'C': 5}
topNum = 2

print(heapq.nlargest(topNum, a_list))
print(heapq.nlargest(topNum, c_dict))
print(heapq.nlargest(topNum, c_dict.keys()))
print(heapq.nlargest(topNum, c_dict.values()))
print('=' * 15)
print(heapq.nsmallest(topNum, a_list))
print(heapq.nsmallest(topNum, c_dict))
print(heapq.nsmallest(topNum, c_dict.keys()))
print(heapq.nsmallest(topNum, c_dict.values()))

[6, 5]
['C', 'B']
['C', 'B']
[5, 4]
===============
[1, 2]
['A', 'B']
['A', 'B']
[3, 4]