Question

给定l = [ [3,4,5], [30,40,50], [300,400,500] ]和数字45

foo(l, 45)应返回[30,40,50]，因为它的平均值最接近45。

写这样一个函数“foo”的好方法是什么？

编辑：

这是我的代码..（尚未测试）

def _avg(l):
    return reduce(lambda x,y: x+y, l)/len(l)

def foo(clusters, point):

    return min(clusters, key=lambda x: abs( _avg(x) - point) )

Answer 1

查找生成min值的生成器表达式的abs(sublist_average - target_average)。

l = [[3,4,5],[30,40,50],[300,400,500]]
avg = 45
result = min(((i,abs(sum(x)/len(x)-avg)) for i,x in enumerate(l)),key=lambda x:x[1])
#result = (1, 5.0)

print('the minimum delta, found at index {}, is {}'.format(*result))
#the minimum delta, found at index 1, is 5.0

Answer 2

你怎么在纸上做这个？你首先要计算第一个的平均值，取差值，然后保存到目前为止最小距离的索引。最简单的解决方案可能是最好，最易理解的解决方案。

Answer 3

写得非常冗长......

import sys
def closest_average_to_num (list_of_lists, num):
    closest_list = []
    lowest_diff = sys.maxint
    for list_of_interest in list_of_lists:
        avg = sum(list_of_interest) / float(len(list_of_interest))
        if (lowest_diff > abs(avg - num)):
             lowest_diff = abs(avg - num)
             closest_list = list_of_interest
    return closest_list

示例输出：

>>> closest_average_to_num ([ [3,4,5], [30,40,50], [300,400,500] ], 45)
[30, 40, 50]

从python中平均值最接近给定数字的列表列表中查找列表？

3 个答案: