将元组列表中给出的坐标转换为复数

Question

I have written the following code for calculating the closest pair in a plane. The code works fine but I have two questions. First, is there a shorter/better way to convert the representation of points from tuples to complex in the first function? and second, is the way I have passed the first function to the second function correct?

import sys

x = [1, 4]
y = [1, 5]


def to_complex(hor, ver):
    list_of_points = list(zip(hor, ver))
    list_of_points.sort(key=lambda el: (el[0], el[1]))
    complex_points = [complex(item[0], item[1]) for item in list_of_points]
    return complex_points


# Brute force algorithm.


def brute_force(points=to_complex(x, y)):
    n = len(points)
    if n < 2:
        return sys.maxsize
    else:
        min_distance = sys.maxsize
        for i in range(n):
            for j in range(i + 1, n):
                if abs(points[i] - points[j]) < min_distance:
                    min_distance = abs(points[i] - points[j])
                    closest_pair = (points[i], points[j])

return min_distance, closest_pair


print(brute_force())

Answer 1

A shorter version of complex(item[0], item[1]) is complex(*item) , which splats a sequence into separate arguments. And I think this is a little better because it will explicitly fail on tuples longer than 2 items.

But it might be even clearer to just write item[0] + item[1] * 1j . Besides looking like the way you write a complex number, this also avoids the implicit conversion to float that allows complex(…) to silently work on things like Decimal objects or sympy constant-valued expressions—you'll get an error in the first case, and a sympy complex constant-valued expression in the second. (Of course if you want to coerce things to complex values, that's a negative rather than a positive.)

Answer 2

You can write your algorithm as follows

# Brute force algorithm.
def brute_force(hor, ver):
    # Convert points to complex
    points = [complex(x, y) for x, y in zip(hor, ver)]
    n = len(points)

    min_distance = sys.maxsize
    closest_pair = None
    if n < 2: return min_distance
    else:
        for i in range(n):
            for j in range(i+1, n):
                temp = abs(points[i] - points[j])
                if temp < min_distance:
                    min_distance = temp
                    closest_pair = (points[i], points[j])
    return min_distance, closest_pair

x = [1, 4, 3, 1]
y = [1, 5, 1, 5]

brute_force(x,y)