如何计算边界框的中心？ [关闭]

Question

我试图用这些给定点计算边界框的中心

(50.607041876988994, -1.3187316344406208, 52.40735812301099, 1.5737316344406207)

这是我在python中完成的代码

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)))

但我不断收到此错误

line 64, in center_geolocation
    for lat, lon in geolocations:
TypeError: 'float' object is not iterable

谁能解释我做错了什么，或者谁能纠正我可能做的错误，

谢谢你:)

Answer 1

使用shapely 库的解决方案：

from shapely.geometry import box

bounds = (-1.3187316344406208, 50.607041876988994, 1.5737316344406207, 52.40735812301099)
polygon = box(*bounds)

print(polygon.centroid.x, polygon.centroid.y)

bounds定义为（西南经度、西南纬度、东北经度、东北纬度）。

Answer 2

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)))

在您给出的示例中没有浮动错误...。我不喜欢输出 - 但这不是问题。