来自点列表的成对欧几里得距离

Question

我正在尝试编写一个 Python 函数（不使用模块），它将遍历坐标列表并找到两个后续点之间的欧氏距离（例如，点之间的距离a 和 b、b 和 c、c 和 d 等）。经过几个小时的搜索，我遇到了 this post，我认为它解决了我的问题，所以我写了这个：

myList = [[2, 3], [3,4], [4,5], [5,6], [6,7]]

def distance(pointOne,pointTwo):
    eucDist = ((pointOne[0] - pointTwo[0])**2 + (pointOne[1] - pointTwo[1])**2)**0.5
    return eucDist

def totalDistance(inputPoints):
    dist = []
    for item in inputPoints[1:]:
        coordDist = distance(inputPoints[0],item)
        dist.append(coordDist)
    return sum(dist)

print totalDistance(myList)

但是，这会检索第一个点与其他所有点之间的距离。我一直在试图弄清楚如何为序列中的下一点定义变量，但我对 Python 还很陌生，只是不太了解如何到达那里。我目前正在这样编写 totalDistance 函数：

def totalDistance(inputPoints):
    dist = []
    for item in inputPoints:
        pOne = item
        pTwo = 
        coordDist = distance(pOne,pTwo)
        dist.append(coordDist)
    return sum(dist)

但不知道如何定义 pTwo。

Answer 1

一种方法是：

def totalDistance(inputPoints):
    dist = []
    pTwo = inputPoints[0]
    for item in inputPoints[1:]:
        pOne = pTwo
        pTwo = item
        coordDist = distance(pOne,pTwo)
        dist.append(coordDist)
    return sum(dist)

基本上是记录第一项，从列表的第二项开始迭代。最好交换 pOne 和 pTwo 以便更容易理解，或者更清楚并使用更多 Pythonic 名称：

def totalDistance(input_points):
    dist = []
    this_item = input_points[0]
    for item in input_points[1:]:
        prev_item = this_item
        this_item = item
        coord_dist = distance(prev_item, this_item)
        dist.append(coord_dist)
    return sum(dist)

Answer 2

使用 list comprehension and zip 可以这样做：

代码：

def distance(point_one, point_two):
    return ((point_one[0] - point_two[0]) ** 2 +
            (point_one[1] - point_two[1]) ** 2) ** 0.5

def total_distance(points):
    return sum(distance(p1, p2) for p1, p2 in zip(points, points[1:]))

或者 Python 3（来自评论）使用 map:

def total_distance(points):
    return sum(map(distance, points, points[1:]))

测试代码

my_points = [[2, 3], [3, 4], [4, 5], [5, 6], [6, 7]]
print(total_distance(my_points))

结果：

5.656854249492381

Answer 3

使用 itertools 和 NumPy：

from itertools import tee
import numpy as np

def pairwise(iterable):
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

def total_dist(points):
    return np.sum(np.sqrt(np.sum(np.square(
                  np.diff(tuple(pairwise(points)))), axis=-2)))

total_dist(myList)
# 5.656854249492381

来自点列表的成对欧几里得距离

Pairwise Euclidean distance from a list of points

python

iteration

for-loop

point

python-2.7

代码：

测试代码

结果：