# 堆 heapd 的使用
import heapq
# 将 list 转换成 heapd # 此处还有疑问
heapq.heapify([1])
# 向 list_first 压入一个值
list_first = [2, 3, 5, 4]
heapq.heappush(list_first, 1)
print(list_first) # [1, 2, 5, 4, 3]
# 在 list_first 中弹出最小值 (list_first 为 None,则抛出异常)
min_value = heapq.heappop(list_first)
print(min_value) # 1
min_value = heapq.heappop(list_first)
print(min_value) # 2
# 在 list_first 中压入一个值 再去取最小值 (先推后弹)
min_values = heapq.heappushpop(list_first, 6)
print(min_values) # 3
# 在 list_first 取最小值, 再压入一个值 (先弹后推)
min_values = heapq.heapreplace(list_first, 1)
print(min_values)
# 将多个已经排序的的list 合并成一个已排序的 可迭代对象
list_second = [2, 3, 4, 5, 6]
list_third = [6, 7, 8, 8, 9]
list_sum = heapq.merge(list_second, list_third) # [2, 3, 4, 5, 6, 6, 7, 8, 8, 9]
while True:
try:
number = list_sum.__next__()
except StopIteration:
break
""" 取出 list_test 中 n 个最大值 """
list_test = [i for i in range(10, 200) if i % 2 != 0]
# 当 n = 1 时:
Max_value = max(list_test)
print(Max_value) # 199
# 当 n 较小时: # heapd.nlargest(n, list) 在 list 中取出 n 个最大值
Max_value_list = heapq.nlargest(10, list_test)
print(Max_value_list) # [199, 197, 195, 193, 191, 189, 187, 185, 183, 181]
# 当 n 较大时:
Max_value_list = sorted(list_test)[-10:]
print(Max_value_list) # [181, 183, 185, 187, 189, 191, 193, 195, 197, 199]
""" 取出 list_test 中 n 个最小值 """
list_test = [i for i in range(10, 200) if i % 2 != 0]
# 当 n = 1 时:
Min_value = min(list_test)
print(Min_value) # 11
# 当 n 较小时: nsmallest(n, list) 在 list 中取出 n 个最小值
Min_value_list = heapq.nsmallest(10, list_test)
print(Min_value_list) # [11, 13, 15, 17, 19, 21, 23, 25, 27, 29]
# 当 n 较大时:
Min_value_list = sorted(list_test)[0:10]
print(Min_value_list) # [11, 13, 15, 17, 19, 21, 23, 25, 27, 29]
原文:https://www.cnblogs.com/FutureHolmes/p/14898494.html