我正在尝试在元组列表上实现最小堆。 例如:
A=[('a',2),('b',1)]
如何基于这些元组的第二个元素对A进行堆放,以便将A堆放到[('b',1),('a',2)]
上? (我必须保持一个最小的堆。)
答案 0 :(得分:2)
根据@JimMischel的评论,将您的元组放在优先级为第一个元素的元组中。然后使用heapq
:
import heapq
list = [('a', 2), ('b', 1), ('c', 0), ('d', 1)]
heap_elts = [(item[1], item) for item in list]
heapq.heapify(heap_elts) # you specifically asked about heapify, here it is!
while len(heap_elts) > 0:
print(heapq.heappop(heap_elts)[1]) # element 1 is the original tuple
产生:
('c', 0)
('b', 1)
('d', 1)
('a', 2)
答案 1 :(得分:0)
import heapq
A=[('a',2),('b',1), ('d', 0), ('c', 2), ('a', 2)]
h = []
for el in A:
heapq.heappush(h, (el[1], el[0]))
print(h)
结果:
[(0, 'd'), (2, 'a'), (1, 'b'), (2, 'c'), (2, 'a')]