在已排序的python列表中，找到最接近目标值的值及其在列表中的索引 [英] In a python list which is sorted, find the closest value to target value and its index in the list

问题描述

我正在尝试在python的排序列表中获取最接近的值及其索引.

I am trying to get both the closest value and its index in a sorted list in python.

在MATLAB中，可以通过以下方式实现:

In MATLAB this is possible with:

[closest_val,index] = min(abs(array - target))

我想知道是否可以通过python实现类似的方法.

I was wondering if there is a similar way this can be implemented in python.

我见过一个可以做一个或另一个的帖子，但是我还没有看到两者都在一起做.

I've seen posts which do one or the other, but I haven't seen both done together.

Link to finding closest value in list post.

推荐答案

bisect，因为该列表未排序.在这里，我们没有相同的问题，我们可以使用bisect作为其提供的速度:

bisect wasn't used in the linked question because the list was not sorted. Here, we don't have the same problem, and we can use bisect for the speed it provides:

import bisect

def find_closest_index(a, x):
    i = bisect.bisect_left(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x > x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index([1, 2, 3, 7, 10, 11], 0)   # => 0, 1
find_closest_index([1, 2, 3, 7, 10, 11], 7)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 8)   # => 3, 7
find_closest_index([1, 2, 3, 7, 10, 11], 9)   # => 4, 10
find_closest_index([1, 2, 3, 7, 10, 11], 12)  # => 5, 11

编辑:如果数组降序:

def bisect_left_rev(a, x, lo=0, hi=None):
    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] > x: lo = mid+1
        else: hi = mid
    return lo

def find_closest_index_rev(a, x):
    i = bisect_left_rev(a, x)
    if i >= len(a):
        i = len(a) - 1
    elif i and a[i] - x < x - a[i - 1]:
        i = i - 1
    return (i, a[i])

find_closest_index_rev([11, 10, 7, 3, 2, 1], 0)   # => 5, 1
find_closest_index_rev([11, 10, 7, 3, 2, 1], 7)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 8)   # => 2, 7
find_closest_index_rev([11, 10, 7, 3, 2, 1], 9)   # => 1, 10
find_closest_index_rev([11, 10, 7, 3, 2, 1], 12)  # => 0, 11

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