Question

我想编写一个代码，它将两个最接近的数字定位到给定的参数，而不使用任何模块。以下是我目前的情况。

 list1 = (1,3,5,8,12)
 x = 9
 for value in list1:         
     point = list1[x - 1], list1[x + 1]

预期输出为[8,12]

Answer 1

您可以使用已排序的平方dist-list来实现它。通过计算差异并平方，你可以摆脱负距离。接近目标值的东西，即

 9 - x =  0 ==> squared 0
10 - x =  1 ==> squared 1
 8 - x = -1 ==> squared 1
12 - x =  3 ==> squared 9  etc.

保持距离，远处的东西更加遥远。

# your "list" was a tuple - make it a list
data = [1,3,5,8,12]
x = 9

# calculate the difference between each value and your target value x
diffs = [t-x for t in data]  
print(diffs)

# sort all diffs by the squared difference 
diffsSorted = sorted([t-x for t in data], key = lambda x:x**2) 
print(diffsSorted)

# take the lowes 2 of them
diffVals = diffsSorted[0:2]
print(diffVals)

# add x on top again
values = [t + x for t in diffVals]

print(values)


# Harder to understand, but you could reduce it into a oneliner: 
allInOne=[k+x for k in sorted([t-x for t in data], key = lambda x:x**2)][:2]
print(allInOne)

输出：

[-8, -6, -4, -1, 3]  # difference between x and each number in data 
                     # [64, 36, 16, 1, 9] are the squared distances   
[-1, 3, -4, -6, -8]  # sorted differences
[-1, 3]              # first 2 - just add x back
[8, 12]              # result by stepwise
[8, 12]              # result allInOne

中间步骤（未打印）：

[64, 36, 16, 1, 9]   # squared differences - we use that as key to 
                     # sort  [-8, -6, -4, -1, 3] into [-1, 3, -4, -6, -8]

比较heapq和list-approach的一些测量（改为abs（）而不是平方）：

import timeit

setup = """ 
import random

random.seed(42)
rnd = random.choices(range(1,100),k={})

import heapq

x = 42
n = 5 """

h = "fit = heapq.nsmallest(n, rnd, key = lambda L: abs(x - L))"
l = "allInOne=[k+x for k in sorted([t-x for t in rnd], key = lambda x:abs(x))][:n]"


rt = {}
for k in range(1,6):
    s = setup.format(10**k)

    rt[10**k] = (timeit.timeit(l,setup=s,number=1000),timeit.timeit(h,setup=s,number=1000))

print(rt)

输出：

#   rnd-size      list approch             heapq
{
        10: (  0.06346651086960813, 0.11704596144812314), 
       100: (  0.5278085906813885,  0.8281634763797711), 
      1000: (  5.032436315978541,   7.462741343986483), 
     10000: ( 54.45165343575938,   79.96112521267483), 
    100000: (577.708372381287,    835.539905495399)
}

list总是比heapq更快，heapq（特别是对于更大的列表）更好的空间复杂性。

Answer 2

试试这个：

409 Conflict

返回所需值t= [abs(i-x) for i in list1] sorted_list = sorted(enumerate(t), key=lambda i:i[1]) (list1[sorted_list[0][0]], (list1[sorted_list[1][0]]))

的元组

如何找到序列中参数的两个最接近的值？

2 个答案: