Question

我想使用GeoDjango或GeoPy计算基于方向和距离的点。

例如，如果我有一个点（-24680.1613,6708860.65389），我想找到一个点1KM North，1KM East，1KM Sourh和1KM west使用Vincenty距离公式。

我能找到的最接近的东西是＆＃34;目的地＆＃34;函数在distance.py（https://code.google.com/p/geopy/source/browse/trunk/geopy/distance.py?r=105）。虽然我无法在任何地方找到这些文件，但我还没弄明白如何使用它。

非常感谢任何帮助。

Answer 1

修改2

好的，有一个带有geopy的开箱即用的解决方案，它没有详细记录：

import geopy
import geopy.distance

# Define starting point.
start = geopy.Point(48.853, 2.349)

# Define a general distance object, initialized with a distance of 1 km.
d = geopy.distance.VincentyDistance(kilometers = 1)

# Use the `destination` method with a bearing of 0 degrees (which is north)
# in order to go from point `start` 1 km to north.
print d.destination(point=start, bearing=0)

输出为48 52m 0.0s N, 2 21m 0.0s E（或Point(48.861992239749355, 2.349, 0.0)）。

90度的轴承对应东方，180度对应南方，依此类推。

旧答案：

一个简单的解决方案是：

def get_new_point():
    # After going 1 km North, 1 km East, 1 km South and 1 km West
    # we are back where we were before.
    return (-24680.1613, 6708860.65389)

但是，我不确定这是否符合您的目的。

好的，说真的，你可以开始使用geopy了。首先，您需要在已知的geopy坐标系中定义起点。乍一看，似乎你不能只是添加＆＃34;到某个方向一定距离。我认为，原因是距离的计算是一个问题，没有简单的逆解。或者我们如何反转https://code.google.com/p/geopy/source/browse/trunk/geopy/distance.py#217中定义的measure函数？

因此，您可能希望采用迭代方法。

如上所述：https://stackoverflow.com/a/9078861/145400您可以计算两个给定点之间的距离：

pt1 = geopy.Point(48.853, 2.349)
pt2 = geopy.Point(52.516, 13.378)
# distance.distance() is the  VincentyDistance by default.
dist = geopy.distance.distance(pt1, pt2).km

向北行驶一公里，你会反复将纬度改为正方向并检查距离。您可以使用简单的迭代求解器自动执行此方法。 SciPy：只需通过http://docs.scipy.org/doc/scipy/reference/optimize.html#root-finding中列出的优化器之一找到geopy.distance.distance().km - 1的根。

我认为很明显，你可以通过改变经度将纬度改为负向，向西和向东转向南方。

我没有这种地理计算的经验，这种迭代方法只有在没有简单的直接方式进入北方时才有意义。一定距离。

修改：我的提案的示例实现：

import geopy
import geopy.distance
import scipy.optimize


def north(startpoint, distance_km):
    """Return target function whose argument is a positive latitude
    change (in degrees) relative to `startpoint`, and that has a root
    for a latitude offset that corresponds to a point that is 
    `distance_km` kilometers away from the start point.
    """
    def target(latitude_positive_offset):
        return geopy.distance.distance(
            startpoint, geopy.Point(
                latitude=startpoint.latitude + latitude_positive_offset,
                longitude=startpoint.longitude)
            ).km - distance_km
    return target


start = geopy.Point(48.853, 2.349)
print "Start: %s" % start

# Find the root of the target function, vary the positve latitude offset between
# 0 and 2 degrees (which is for sure enough for finding a 1 km distance, but must
# be adjusted for larger distances).
latitude_positive_offset = scipy.optimize.bisect(north(start, 1),  0, 2)


# Build Point object for identified point in space.
end = geopy.Point(
    latitude=start.latitude + latitude_positive_offset,
    longitude=start.longitude
    )

print "1 km north: %s" % end

# Make the control.
print "Control distance between both points: %.4f km." % (
     geopy.distance.distance(start, end).km)

输出：

$ python test.py 
Start: 48 51m 0.0s N, 2 21m 0.0s E
1 km north: 48 52m 0.0s N, 2 21m 0.0s E
Control distance between both points: 1.0000 km.

Answer 2

根据Jan-Philip Gehrcke博士的回答，此问题的2020年更新。

VincentyDistance已正式弃用，从未完全精确，有时甚至不准确。

此代码段显示了如何与最新版本一起使用（GeoPy的将来版本-Vincenty将在2.0中弃用）

import geopy
import geopy.distance

# Define starting point.
start = geopy.Point(48.853, 2.349)

# Define a general distance object, initialized with a distance of 1 km.
d = geopy.distance.distance(kilometers=1)

# Use the `destination` method with a bearing of 0 degrees (which is north)
# in order to go from point `start` 1 km to north.
final = d.destination(point=start, bearing=0)

final是一个新的Point对象，该对象在打印时返回48 51m 43.1721s N, 2 20m 56.4s E

如您所见，它比Vincenty更准确，并且应该在极点附近保持更好的准确性。

希望有帮助！

Answer 3

我不得不处理在经度和纬度上添加米数。

这是我做过的事情，受到this source的启发：

import math
from geopy.distance import vincenty

initial_location = '50.966086,5.502027'
lat, lon = (float(i) for i in location.split(','))
r_earth = 6378000
lat_const = 180 / math.pi
lon_const = lat_const / math.cos(lat * math.pi / 180)

# dx = distance in meters on x axes (longitude)
dx = 1000
new_longitude = lon + (dx / r_earth) * lon_const
new_longitude = round(new_longitude, 6)
new_latitude = lat + (dy / r_earth) * lat_const
new_latitude = round(new_latitude, 6)

# dy = distance on y axes (latitude)
new_latitude = lat + (dy / r_earth) * lat_const
new_latitude = round(new_latitude, 6)
new_location = ','.join([str(y_lat), str(x_lon)])

dist_to_location = vincenty(location, new_location).meters

根据距离和方向计算点

3 个答案: