Python 求解测量点列表（x，y）中点之间距离的函数

我正在写一个函数路由。此函数具有一个强制参数points，该参数接受点列表。如果依次访问给定列表中的每个点，则函数必须返回行驶的总距离。除强制参数外，该函数还有两个可选参数：

循环：获取一个布尔值，该值指示路线的终点是否等于其起点（True）或（False）；此参数的默认值为False

距离：采用距离函数，用于计算给定路线中两个连续点之间的总距离；如果没有向该参数传递显式值，则必须使用欧几里德距离

问题：任何人都知道上一个定义route（）如何解决这种情况：

route([(41.79, 13.59), (41.68, 14.65), (21.16, -4.79)], distance=lambda p1, p2: abs(p1[0] + p2[0]))

正确答案：146.31

我提到的代码的一部分：

 if cycle == False and distance is λ(p1, p2): abs(p1[0] + p2[0]):

            l = list()
            count = 0

            for items in range(len(points)-1):
                a = points[items]
                b = points[items+1]
                d = euclidean(a[0], b[0])
                l.append(d)
                count += 1

            return sum(l)

在这一部分中，我被第一条规则和更进一步的规则卡住了

工作正常的完整代码（上述部分除外）：

---

我同意邓肯的观点。你有太多的重复。 这里有一个更直接的方法：

euclidean=lambda p1，p2:sqrt（sum（（p1_i-p2_i）**2表示p1_i，p2_i在zip中（p1，p2）））
曼哈顿=lambda p1，p2:sum（在zip（p1，p2）中p1_i，p2_i的abs（p1_i-p2_i）
棋盘=lambda p1，p2:max（用于压缩（p1，p2）中p1_i，p2_i的abs（p1_i-p2_i））
def路线（点，周期=假，度量=欧几里德）：
l=0.0
对于范围内的i（长度（点）-1）：
l+=度量（点[i]，点[i+1]）
如果循环：
l+=度量（点[-1]，点[0]）
返回l

---

可以传递任何度量函数，然后使用它们代替欧几里德度量。

不要比较函数：两个不同的lambda表达式不会与

is

或

=

进行相等比较。只需拨打距离（a，b），消除所有重复。这是一个有用的答案。

  def euclidean(a, b):
    '''
    >>> euclidean((42.36, 56.78), (125.65, 236.47))
    198.05484139500354
    '''

    from math import sqrt

    return sqrt(sum((a - b)**2 for a, b in zip(a, b)))




def manhattan(c, d):
    '''
    >>> manhattan((42.36, 56.78), (125.65, 236.47))
    262.98
    '''

    return sum(abs(c - d) for c, d in zip(c, d))



def chessboard(e, f):
    '''
    >>> chessboard((42.36, 56.78), (125.65, 236.47))
    179.69
    '''

    return max(abs(e - f) for e, f in zip(e, f))



def route(points, cycle=False, distance=None):
    '''
    >>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)])
    21.861273201261746
    >>> route(cycle=True, points=[(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)])
    42.60956710702662
    >>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)], distance=manhattan)
    23.45
    >>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)], cycle=True, distance=manhattan)
    45.42
    '''



    if cycle == False and distance is None: 

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = euclidean(a, b)
            l.append(d)
            count += 1

        return sum(l)


    if cycle == False and distance is euclidean:

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = euclidean(a, b)
            l.append(d)
            count += 1

        return sum(l)


    if cycle == False and distance is λ(p1, p2): abs(p1[0] + p2[0]):

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = euclidean(a[0], b[0])
            l.append(d)
            count += 1

        return sum(l)



    if cycle == True and distance is None:

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = euclidean(a, b)
            l.append(d)
            count += 1

        f = points[0]
        g = points[-1] 
        r = euclidean(g, f)

        k = sum(l) + r

        return k


    if cycle == True and distance is euclidean:

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = euclidean(a, b)
            l.append(d)
            count += 1

        f = points[0]
        g = points[-1] 
        r = euclidean(g, f)

        k = sum(l) + r

        return k



    if cycle is False and distance is manhattan:

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = manhattan(a, b)
            l.append(d)
            count += 1

        return sum(l)


    if cycle is True and distance is manhattan:

        l = list()
        count = 0

        for items in range(len(points)-1):
            a = points[items]
            b = points[items+1]
            d = manhattan(a, b)
            l.append(d)
            count += 1

        f = points[0]
        g = points[-1] 
        r = manhattan(g, f)

        k = sum(l) + r

        return k