Question

myArr = array([4,1,88,44,3])
myNumber = 25
FindClosest(myArr, myNumber)
...
4, 44

有没有办法在列表中找到最接近的2个数字给某个数字，使其中一个更高而另一个更低？

我可以通过以下方式找到最接近的号码：

min(myArr.tolist(), key=lambda x:abs(x-myNumber))

Answer 1

排序不是必需的，并且当它应该只是O（n）时，使这个时间复杂度为O（n logn）。

我相信这是你正在寻找的，利用numpy数组索引：

{{1}}

Answer 2

first = min([for i in myArr if a>i])
second = max([for i in myArr if a<i])

此外，正确答案为44, 4而非44, 88

Answer 3

upper = min([ i for i in myArr.tolist() if i >= myNumber], key=lambda x:abs(x-myNumber))
lower = min([ i for i in myArr.tolist() if i < myNumber], key=lambda x:abs(x-myNumber))

Answer 4

我有一个答案，找到输入值最接近的2个数字，请参阅下面的程序： -

from collections import OrderedDict


def find_closer():
    a = [9, 19, 87, 43, 10]
    b = int(input("Enter number :"))
    # diff_list = []
    diff_dict = OrderedDict()
    if b in a:
        b = input("Number already exists, please enter another number ")
    else:
        for x in a:
            diff = x - b
            if diff < 0:
                # diff_list.append(diff*(-1))
                diff_dict[x] = diff*(-1)
            else:
                # diff_list.append(diff)
                diff_dict[x] = diff
    print("diff_dict", diff_dict)
    # print(diff_dict[9])
    sort_dict_keys = sorted(diff_dict.keys())
    print(sort_dict_keys)
    closer_less = 0
    closer_more = 0
    for closer in sort_dict_keys:
        if closer < b:
            closer_less = closer
        else:
            closer_more = closer
            break
    print("closer less =", closer_less, "and closer more =", closer_more)


find_closer()

这是一个纯粹的python程序。希望它有所帮助!!

找到给定数字上方和下方最接近的元素

4 个答案: