在循环有序列表中查找最接近值的优雅方法

Question

Given a sorted list such as [1.1, 2.2, 3.3] and a bounding value such as math.pi*2 , return the closest value for any given value from [0 - math.pi*2)

The function should return the index of the value, so that f(1.2) returns 0 while f(2.1) returns 1 , and f(6.0) should wrap around at math.pi*2 and return 0 , being closer to 1.1 than to 3.3 given the bounding value. Just to be entirely explicit, this function should also wrap around on on the lower end, so that f(1.0, [5.0, 6.0], bound = math.pi*2) returns 1 .

The use case is to map an arbitrary angle in radians to the nearest existing valid angle in the list. I've written this kind of function a couple of times in python with bisect , but the code always ends up offending my aesthetic senses. The high complexity and number of edge cases seems out of proportion with the intuitive simplicity of the function. So I am asking if anyone can come up with a pleasing implementation, both in terms of efficiency and elegance.

Answer 1

Here's a more elegant approach. Eliminate the edge cases by wrapping the number line around:

from bisect import bisect_right

def circular_search(points, bound, value):
    ##
    ## normalize / sort input points to [0, bound)
    points = sorted(list(set([i % bound for i in points])))
    ##
    ## normalize search value to [0, bound)
    value %= bound
    ##
    ## wrap the circle left and right
    ext_points = [i-bound for i in points] + points + [i+bound for i in points]
    ##
    ## identify the "nearest not less than" point; no
    ## edge cases since points always exist above & below
    index = bisect_right(ext_points, value)
    ##
    ## choose the nearest point; will always be either the
    ## index found by bisection, or the next-lower index
    if abs(ext_points[index]-value) >= abs(ext_points[index-1]-value):
        index -= 1
    ##
    ## map index to [0, npoints)
    index %= len(points)
    ##
    ## done
    return points[index]

As written, works unless inputs are wonky like no points, or bound==0.

Answer 2

Use the bisect module as a base:

from bisect import bisect_left
import math

def f(value, sorted_list, bound=math.pi * 2):
    value %= bound
    index = bisect_left(sorted_list, value)
    if index == 0 or index == len(sorted_list):
        return min((abs(bound + sorted_list[0] - value), 0), (abs(sorted_list[-1] - value), len(sorted_list) - 1))[1]
    return min((index - 1, index), 
        key=lambda i: abs(sorted_list[i] - value) if i >= 0 else float('inf'))

Demo:

>>> sorted_list = [1.1, 2.2, 3.3]
>>> f(1.2, sorted_list)
0
>>> f(2.1, sorted_list)
1
>>> f(6.0, sorted_list)
0
>>> f(5.0, sorted_list)
2

Answer 3

The easiest way would be just to use min:

def angular_distance(theta_1, theta_2, mod=2*math.pi):
    difference = abs(theta_1 % mod - theta_2 % mod)
    return min(difference, mod - difference)

def nearest_angle(L, theta):
    return min(L, key=lambda theta_1: angular_distance(theta, theta_2))

In [11]: min(L, key=lambda theta: angular_distance(theta, 1))
Out[11]: 1.1

Making use of the sorted-ness of the list you can use the bisect module:

from bisect import bisect_left

def nearest_angle_b(theta, sorted_list, mod=2*math.pi):
    i1 = bisect_left(sorted_list, theta % mod)
    if i1 == 0:
        i1, i2 = len(sorted_list) - 1, 0
    elif i1 == len(sorted_list):
        i1, i2 = i1 - 1, 0
    else:
        i2 = (i1 + 1) % len(sorted_list)
    return min((angular_distance(theta, L[i], mod), i, L[i])
                 for i in [i1, i2])

Returns the distance, the index and the angle in the list to which it's closest to.