Python 寻找产生5个大和的指数
假设我有两个列表：Python 寻找产生5个大和的指数,python,Python,假设我有两个列表： a = [0.1,0.2,0.4,0.5] b = [0.5,0.6,0.7, 0.8] c = [all the sums of a and b] c = [] for a_ in a: for b_ in b: c.append(a_+b_) 现在，c将包含a和b之和的值_ 获取c中最大的5个元素 然后返回a和b指数产生的最大总和 我的想法是跟踪像c[index_of a][index_of b]=a_+b_ 所以，问题是a和b的什么指数返回k最大
a = [0.1,0.2,0.4,0.5]
b = [0.5,0.6,0.7, 0.8]
c = [all the sums of a and b]
c = []
for a_ in a:
for b_ in b:
c.append(a_+b_)
c[index_of a][index_of b]=a_+b_c = [0.6, 0.7, 0.7999999999999999, 0.9, 0.7, 0.8, 0.8999999999999999, 1.0, 0.9, 1.0, 1.1, 1.2000000000000002, 1.0, 1.1, 1.2, 1.3]
maximum=已排序（c，reverse=True）[:5]largest = [1.3, 1.2000000000000002, 1.2, 1.1, 1.1]
And then corresponding indices in a and b list are:
a_index = 3, b_index = 3 (1.3)
a_index = 2, b_index = 3 (1.2)
largest, a_index, b_index = [0] * 3
for i, n1 in enumerate(a):
for j, n2 in enumerate(b):
if n1 + n2 > largest:
largest = n1 + n2
a_index = i
b_index = j
a=[0.1,0.2,0.4,0.5]
b=[0.5,0.6,0.7,0.8]
#找出所有可能的和
#-与使用的项目索引的总和一起存储
#-enumerate将为我们提供索引和项目
c=[（x+y，x_i，y_i）表示枚举（a）中的x_i，x
对于y_i，y在枚举中（b）]
#将此列表排序
#-当我们对元组进行排序时，它将在第一个元组上进行排序
#首先是元组项（即总和），然后是其他项
#-那么就拿清单上的五个项目
索引=[（x_i，y_i）用于排序（c）[-5:]
枚举sorted([(x[0], y[0], x[1]+y[1]) for x in enumerate(a)
for y in enumerate(b)],
key=lambda x: x[2], reverse=True)
基于@donkopotamus的代码，但如果输入列表
abimport heapq
a = [0.1,0.2,0.4,0.5]
b = [0.5,0.6,0.7, 0.8]
# Change: generator expression for all of the possible sums (not realized yet)
# - store with the sum the item indices that were used
# - enumerate will give us the index and the item
c = ((x + y, x_i, y_i) for x_i, x in enumerate(a)
for y_i, y in enumerate(b))
# Use heapq.nlargest to get the indices that produce the largest sums
# without storing all sums in memory (it maintains a heap internally that
# contains only the five largest values seen to date at once)
# - as we are sorting tuples, it will sort on the first
# tuple item initially (ie the sum), and then the others
# - then just take the list five items
indices = [(x_i, y_i) for _, x_i, y_i in heapq.nlargest(5, c)]
问题中所示案例的预期输出是什么？如果你解释了你在做什么以及为什么它有用，我会支持这个答案。在幕后，
heapqheappushpop（）heapq.nlargestheappushpopO（log（n））n5log（5）*len（输入）log（len（输入））*len（输入）heapqsortedsortedheapqsortedsortednlargestsortedheapqabab