Question

我有一个未还原点（2D）的列表，我想计算它们之间的距离之和。因为我的背景是c ++开发人员，所以我会这样做：

import math

class Point:
    def __init__(self, x,y):
        self.x = x
        self.y = y

def distance(P1, P2):
    return math.sqrt((P2.x-P1.x)**2 + (P2.y-P1.y)**2)

points = [Point(rand(1), rand(1)) for i in range(10)]

#this part should be in a nicer way
pathLen = 0
for i in range(1,10):
    pathLen += distance(points[i-1], points[i])

是否有更多的Python方式替换for循环？例如用reduce或类似的方法？

最诚挚的问候！

Answer 1

您可以将生成器表达式与sum，zip和itertools islice结合使用，以避免重复数据：

from itertools import islice
paathLen = sum(distance(x, y) for x, y in zip(points, islice(points, 1, None)))

这里有live example

Answer 2

一些修复，因为C ++方法可能不是最好的解决方法

import math
# you need this import here, python has no rand in the main namespace
from random import random

class Point:
    def __init__(self, x,y):
        self.x = x
        self.y = y
    # there's usually no need to encapsulate variables in Python

def distance(P1, P2):
    # your distance formula was wrong 
    # you were adding positions on each axis instead of subtracting them
    return math.sqrt((P1.x-P2.x)**2 + (P1.y-P2.y)**2)

points = [Point(random(), random()) for i in range(10)]
# use a sum over a list comprehension:
pathLen = sum([distance(points[i-1], points[i]) for i in range(10)])

@Robin Zigmond的zip方法也是一种很好的实现方法，尽管对我来说并不是立即可以在这里使用它。

用Python方式计算点列表之间的距离

2 个答案: