Question

我正试图解决一个奇怪的问题。也许你们知道一些算法可以解决这个问题。

我有货运卡车的数据，想要提取一些数据。假设我有一个从GPS获得的分类点列表。那是那辆卡车的路线：

[
    {
        "lng": "-111.5373066",
        "lat": "40.7231711",
        "time": "1970-01-01T00:00:04Z",
        "elev": "1942.1789265256325"
    },
    {
        "lng": "-111.5372056",
        "lat": "40.7228762",
        "time": "1970-01-01T00:00:07Z",
        "elev": "1942.109892409177"
    }
]

现在，我想得到的是“最快的里程”列表。我会做一个例子：

鉴于要点：

A, B, C, D, E, F

从A点到B点的距离是1英里，货物需要10:32分钟。从B点到D点，我有另一英里，货物需要10分钟等等。所以，我需要一个按时间排序的清单。类似于：

B -> D: 10
A -> B: 10:32
D -> F: 11:02

你知道任何有效的算法让我计算出来吗？

谢谢大家。

PS：我正在使用Python。

修改

我有距离。我知道如何计算它并且有很多帖子可以做到这一点。我需要的是一种算法，按英里标记并从中获得速度。有一个距离函数是没有用的：

results = {}
for point in points:
  aux_points = points.takeWhile(point>n) #This doesn't exist, just trying to be simple
  for aux_point in aux_points:
    d = distance(point, aux_point)
    if d == 1_MILE:
      time_elapsed = time(point, aux_point)
      results[time_elapsed] = (point, aux_point)

我还在做一些非常低效的计算。

Answer 1

如果您有获取位置数据的位置和时间戳，您可以执行以下操作：

def CalculateSpeeds(list_of_points_in_time_order):
  """Calculate a list of (average) speeds for a list of geographic points."""

  points = list_of_points_in_time_order
  segment_start = points[0]
  speed_list = []

  for segment_end in points[1:]:
    dt = ElapsedTime(segment_start, segment_end)
    # If you're looking at skipping points, with a slight risk of degraded data
    # you could do something like "if dt < MIN_ELAPSED_TIME:" and indent
    # the rest of the loop. However, you'd need to then check if the last point 
    # has been accounted for, as it might've been too close to the last considered
    # point.
    d = Distance(segment_start, segment_end)
    speed_list.append(d/dt)
    segment_start = segment_end
  return speed_list

你已经说过（在评论中）你可以为一对做这个，所以你需要做的就是为所有连续对做。

Answer 2

所以，如果你有n个这样的积分，行程中会有n - 1个“腿”。您可以通过以下方式形成该列表：

legs = []
for i in xrange(n - 1):
  legs.append(build_leg(point[i], point[i + 1]))

假设point是点列表，build_leg()是一个接受两个点并计算距离和平均速度的函数。

以上循环将调用build_leg第一个点0和1，然后是1和2，依此类推，直到n - 2和n - 1为最后两个点。

Answer 3

我已经爱上了滑动窗口，在这里可能会有所帮助。与其他答案概念相同，只是略有不同的方法。

from itertools import islice
def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result    
    for elem in it:
        result = result[1:] + (elem,)
        yield result


results = {}
# presort your points in time if necessary
for point_a, point_b in window(points):
    d = distance(point_a, point_b)
    if d == 1_MILE:
        time_elapsed = time(point_a, point_b)
        results[time_elapsed] = (point_a, point_b)

如何从一组GPS点获取“最快里程”列表

3 个答案: