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[英]Iterative euclidean distance calculation between consecutive points (x,y tuples) which belongs to a list of lines
[英]Solving a function that measures distance between points in a list of points (x, y)
我正在写一个函数路由。 这个函数有一个强制参数points,它接受一个点列表。 如果依次访问给定列表中的每个点,则该函数必须返回行进的总距离。 除了强制参数外,该函数还有两个可选参数:
循环:取一个布尔值,指示路线的终点是否等于其起点(真)或不等于(假); 此参数的默认值为 False
distance:采用距离函数,用于计算给定路径中两个连续点之间的总距离; 如果没有明确的值传递给这个参数,则必须使用欧几里德距离
问题:任何人都知道最后一个定义 route() 如何解决这个问题:
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
我引用的部分代码:
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)
在这部分中,我陷入了第一条规则和更远的地方。
工作正常的完整代码(除了上面的部分):
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
我同意邓肯。 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 for p1_i, p2_i in zip(p1, p2)))
manhattan = lambda p1, p2: sum(abs(p1_i - p2_i) for p1_i, p2_i in zip(p1, p2))
chessboard = lambda p1, p2: max(abs(p1_i - p2_i) for p1_i, p2_i in zip(p1, p2))
def route(points, cycle=False, metric=euclidean):
l = 0.0
for i in range(len(points) - 1):
l += metric(points[i], points[i + 1])
if cycle:
l += metric(points[-1], points[0])
return l
可以传递任何度量函数,然后使用它来代替欧几里德度量。
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