Python 如何计算边界框的中心？

我试图用这些给定的点计算边界框的中心

(50.607041876988994, -1.3187316344406208, 52.40735812301099, 1.5737316344406207)

Edit Solved：这是我用python编写的代码

from math import cos, sin, atan2, sqrt

    def center_geolocation(geolocations):
        """
        Provide a relatively accurate center lat, lon returned as a list pair, given
        a list of list pairs.
        ex: in: geolocations = ((lat1,lon1), (lat2,lon2),)
            out: (center_lat, center_lon)
        """
        x = 0
        y = 0
        z = 0

        for lat, lon in geolocations:
            lat = float(lat)
            lon = float(lon)
            x += cos(lat) * cos(lon)
            y += cos(lat) * sin(lon)
            z += sin(lat)

        x = float(x / len(geolocations))
        y = float(y / len(geolocations))
        z = float(z / len(geolocations))

        return (atan2(y, x), atan2(z, sqrt(x * x + y * y)))

谢谢：）

---

您给出的示例中没有浮点错误。。。。我不喜欢输出-但这不是问题。

你将什么作为参数传递给函数？我试图传递我计算的边界框的坐标（50.6070418769494，-1.3187316344406208，52.40735812301099，1.5737316344406207）一个问题，它打印出来（-1.4093883217883079，0.6741482698408712）它应该打印51.51608899635712，0.09891956707558282，你知道我可能出了什么问题吗？这个解决方案似乎对度做了所有的计算，我认为它们应该首先转换成弧度，然后在最后转换成度，即

lat=float（lat）*pi/180

（与lon相同）和

atan2（y，x）*180/pi

（与lon相同）。此外，我认为输出是（center_lon，center_lat），与描述的不同。

from math import *

def center_geolocation(geolocations):
    """
    Provide a relatively accurate center lat, lon returned as a list pair, given
    a list of list pairs.
    ex: in: geolocations = ((lat1,lon1), (lat2,lon2),)
        out: (center_lat, center_lon)
"""
x = 0
y = 0
z = 0

for lat, lon in geolocations:
    lat = float(lat)
    lon = float(lon)
    x += cos(lat) * cos(lon)
    y += cos(lat) * sin(lon)
    z += sin(lat)

x = float(x / len(geolocations))
y = float(y / len(geolocations))
z = float(z / len(geolocations))

return (atan2(y, x), atan2(z, sqrt(x * x + y * y)))


center_geolocation( 
 ((50.607041876988994, -1.3187316344406208),   
  (52.40735812301099, 1.5737316344406207)))