Question

给定一个常量，一个浮动的排序列表和一个2元素浮点列表，

CONST = 1.
lst1 = [1.2, 2.4, 3.1] #sorted
lst2 = [[2.0, 0.9], [3.1, 1.5], [1.0, 3.0], [2.5, 2.0]]

我为[CONST, a]

的所有元素构建了一个包含lst1对的新列表

我需要在list2列表中“投射”[CONST,a]：lst2中对的第二个值将更改为list1中最接近的值以及第一个值该对将是相同第二个值的所有第一个值的总和。

所以给出的例子的结果是：

[[6.1, 1.2], [3.5, 2.4], [2.0, 3.1]]

到目前为止，我有类似的内容：

from itertools import groupby
from bisect import bisect
from operator import itemgetter

for t in lst2:
    i = bisect(lst1, t[1])
    bounds = lst1[i-1:i+1] if i else [lst1[0]]
    t[1] = min(bounds, key=lambda x: abs(x-t[1]))

lst2 += [[CONST, a] for a in lst1]
lst2 = sorted(lst2, key=itemgetter(1))
res = [[sum([t[0] for t in group]), keys] for keys, group in groupby(lst2, itemgetter(1))]

但是列表（特别是lst1）可能很长（1e5 +），我觉得我可以在那里有更好的效率。有什么想法吗？

Answer 1

由于排序主导了运行时间，因此很难做得更快，但这是一个不需要求助于lst1的版本。但它会通过lst1迭代：

sorted_lst2 = sorted(lst2, key=itemgetter(1))
i = 0
res = []
k = 0
while i + 1 < len(lst1):
    before = lst1[i]
    const = CONST
    while i + 1 < len(lst1) and before == lst1[i + 1]:
        const += CONST
        i += 1
    after = lst1[i + 1]
    mid = before + (after - before) / 2
    sum_before = 0
    while k < len(sorted_lst2) and sorted_lst2[k][1] <= mid:
        sum_before += sorted_lst2[k][0]
        k += 1
    res.append([const + sum_before, before])
    i += 1
sum_before = 0
while k < len(sorted_lst2):
    sum_before += sorted_lst2[k][0]
    k += 1
res.append([CONST + sum_before, lst1[-1]])

Answer 2

我想我发现了一种有趣的方式：发电机发电机！ Python很棒！这大约是前一代码的两倍。

CONST = 1.
lst1 = [1.2, 2.4, 3.1] #sorted
lst2 = [[2.0, 0.9], [3.1, 1.5], [1.0, 3.0], [2.5, 2.0]]

def gen_generator(lst1):
    #Create iterators to form consecutive pairs
    it1_1, it1_2 = iter(lst1), islice(lst1+[float('inf')], 1, None)
    #Create iterator over items to insert and initialize p
    gen_generator.it2 = iter(sorted(lst2, key=itemgetter(1)))
    gen_generator.p=next(gen_generator.it2)
    for t1, t2 in zip(it1_1, it1_2):
        #Calculate the mid point of the pair t1, t2
        sup = (t1+t2)/2 if t2 != float('inf') else float('inf')
        #Create a generator for each item in lst1
        def generator():
            #All the previously inserted items have disappeared
            #So yield any p such that p[1] < sup
            while gen_generator.p is not None and gen_generator.p[1] < sup:
                yield gen_generator.p[0]
                #Move to next or None
                gen_generator.p = next(gen_generator.it2, None)
            #Add the constant
            yield CONST
        yield generator

res = [[sum(gen()), t] for gen, t in zip(gen_generator(lst1), lst1)]
return res

将python列表投影到现有值（使用生成器生成器）

2 个答案: