将一对元组与元组对列表进行比较

Question

所以我正在做这个编程作业，目前我一直在比较一个元组中的值对与列表中的元组对。

这些对基本上是 x 和 y 坐标，我需要从列表中找到最接近元组对的一个。例如，给定点 (-4, 3) 和列表 [(10, 6), (1, 7), (6, 3), (1, 9)]，最接近的是 (1, 7).

这些数字总是随着编程的随机部分而变化，但上面的定义是一个函数。这是全部内容：

def nearest(point, more_points):
    '''
    Finds the nearest point to the base point
    '''
    (x, y) = point
    for i, j in more_points:
        a = math.fabs(x - i)
        tempI = i
        b = math.fabs(y - j)
        tempJ = j
        tempA, tempB = a , b
        if min(tempA) <  a:

point = () 
my_points = []
c = 0
lpoint = list(point)
while c < 2:
     lpoint.append(random.randrange(-5,5,1)) # generate the "point"
     c += 1
tpoint = tuple(lpoint)
c = 0
colx = [] # x points
coly = [] # y points
# generate the points
while c < 4:
    colx.append(random.randint(0,10))
    coly.append(random.randint(0,10))
    c += 1

my_point = list(zip(colx,coly))
print(my_point)
the_nearest = nearest(tpoint,my_point)
print(the_nearest)

我想做的是取点中的 x,y，然后取 "other" 点并求出差值，然后用它来求出 "nearest"，但我输了我被困住了。重点是用户定义的函数。

Answer 1

假设以下函数计算2点以内的距离：

def distance(point_a, point_b):
    """Returns the distance between two points."""
    x0, y0 = point_a
    x1, y1 = point_b
    return math.fabs(x0 - x1) + math.fabs(y0 - y1)

您可以遍历所有点并找到最小距离：

def nearest(point, all_points):
    closest_point, best_distance = None, float("inf")
    for other_point in all_points:
        d = distance(point, other_point)
        if d < best_distance:
             closest_point, best_distance = other_point, d
    return closest_point

虽然我可以想出一个更 pythonic 的方法：

def nearest(point, all_points):
    """Returns the closest point in all_points from the first parameter."""
    distance_from_point = functools.partial(distance, point)
    return min(all_points, key=distance_from_point)

上述解决方案的总体思路是构建偏函数。该部分函数采用单个参数和 returns 到作为参数给定的点的距离。这可以重写为 lambda other_point: distance(point, other_point) 但这样更漂亮。

请注意，如果您使用空的点列表 nearest(point, []) 调用上面的函数，它将引发 ValueError。如有必要，您可以为这种情况添加一个 if。

Answer 2

将 min() 与 key 函数一起使用：

#!/usr/bin/env python3

import math
from functools import partial


def distance_between_points(a, b):
    ax, ay = a
    bx, by = b
    return math.sqrt(pow(ax - bx, 2) + pow(ay - by, 2))


def nearest(point, more_points):
    '''
    Finds the nearest point to the base point
    '''
    distance_to_point = partial(distance_between_points, point)
    return min(more_points, key=distance_to_point)


point = (-4, 3)
my_points = [(10, 6), (1, 7), (6, 3), (1, 9)]
n = nearest(point, my_points)
print(n)

将一对元组与元组对列表进行比较

Comparing a tuple of a pair to a list of tuple pairs

python

tuples

list

python-3.6