技術標籤：python

import heapq
nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
print(heapq.nlargest(3, nums)) # Prints [42, 37, 23]
print(heapq.nsmallest(3, nums)) # Prints [-4, 1, 2]


l = heapq.nsmallest(1,nums)
print(l)

portfolio = [
    {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'AAPL'
 
, 'shares': 50, 'price': 543.22},
    {'name': 'FB', 'shares': 200, 'price': 21.09},
    {'name': 'HPQ', 'shares': 35, 'price': 31.75},
    {'name': 'YHOO', 'shares': 45, 'price': 16.35},
    {'name': 'ACME', 'shares': 75, 'price': 115.65}
]
cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price']
 
)
expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])

print(cheap)

https://python3-cookbook.readthedocs.io/zh_CN/latest/c01/p04_find_largest_or_smallest_n_items.html