Question

我想要拆分两个heapqs（用作优先级队列），然后将它们一起添加，并将结果heapq与前两个heapqs相关联。

这在python中可行吗？

我目前的代码：

population = []
for i in range(0, 6):
    heappush(population, i)
new_population = []
for i in range(4, 9):
    heappush(new_population, i)

split_index = len(population) // 2
temp_population = population[:split_index]
population = new_population[:split_index] + temp_population
print(population)
print(heappop(population))

输出：

[4, 5, 6, 0, 1, 2]
4

通缉输出：

[0, 1, 2, 4, 5, 6]
0

Answer 1

使用nlargest代替切片，然后重新合并组合列表。

from heapq import nlargest, heapify
n = len(population) // 2
population = heapify(nlargest(population, n) +
                     nlargest(new_population, n))
print(heappop(population))

但是，您可能想要进行基准测试，如果对两个原始列表进行排序，然后合并结果，则会更快。对于几乎排序的列表，Python的sort例程速度很快，这可能会比heapq函数带来更少的开销。如果您实际上不需要优先级队列，则可能不需要最后一个heapify步骤（因为您无论如何都要对它们进行排序）。

from itertools import islice
from heapq import merge, heapify
n = len(population)  # == len(new_population), presumably
population = heapify(islice(merge(sorted(population), sorted(new_population)), n))

Python heapq：拆分并合并到有序的heapq中

1 个答案: