计算两个列表的 haversine

Question

我目前正在尝试计算我在 Geopandas 数据框中的 (lat/long) 坐标的路线距离。我对这个包裹还很陌生，但基本上我有几个点组成了一条路线，我想做的就是找到这条路线的总实际距离。我可以用两个固定点来做到这一点，我欠用户@steve clark 的帮助：

# Start
lon1 = 41.592181
lat1 = -87.638856
# End
lat2 = -86.754688
lon2 = 41.877575

def haversine(lat1, lon1, lat2, lon2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a))
    r = 6371 # Radius of earth in kilometers
    print('Distance from beginning to end of route in km: ',round((c * r), 2),'\n')

我被两件事卡住了，我目前正在四处寻找是否可以计算与 Geopandas point() 对象的距离，但老实说，我找到的例子要么与我的无关问题，或者超出我的理解范围（目前）。

我可以将我的 gpd 中的纬度和经度列拉入列表，但我无法循环应用它

LatList = geo_data['latitude'].tolist()
LonList = geo_data['longitude'].tolist()

我尝试将迭代的内容追加到一个新列表中并对距离求和，但我最终得到一个具有相同值的列表追加了 2,850 次。感谢任何帮助或指导！

编辑：根据要求，这是失败的代码

distance = []

    for i, j in zip(LatList, LonList):

        dlat = i - i+1
        dlon = j - j+1

        a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
        c = 2 * asin(sqrt(a))
        r = 6371 # Radius of earth in kilometers

        distance.append(round((c * r), 2))

    print(distance)

Answer 1

您需要调整i、i+1、j和j+1的定义，否则循环将无法执行您想要的操作。

distance = []

LatLonList = list(zip(LatList, LonList))

    # notice that if you do "for n in len(LatLonList)", the loop will fail in the last element
    for n in len(LatLonList) -1:

        dlat = LatLonList[n][0] - LatLonList[n+1][0]  # this is for i
        dlon = LatLonList[n][1] - LatLonList[n+1][1]  # this is for j

        a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
        c = 2 * asin(sqrt(a))
        r = 6371 # Radius of earth in kilometers

        distance.append(round((c * r), 2))

    print(distance)

Answer 2

以geopandas参考中的数据为例：

import pandas as pd
import geopandas
from shapely.geometry import Point

df = pd.DataFrame(
    {'City': ['Buenos Aires', 'Brasilia', 'Santiago', 'Bogota', 'Caracas'],
     'Country': ['Argentina', 'Brazil', 'Chile', 'Colombia', 'Venezuela'],
     'Latitude': [-34.58, -15.78, -33.45, 4.60, 10.48],
     'Longitude': [-58.66, -47.91, -70.66, -74.08, -66.86]})
df['Coordinates'] = list(zip(df.Longitude, df.Latitude))
df['Coordinates'] = df['Coordinates'].apply(Point)
gdf = geopandas.GeoDataFrame(df, geometry='Coordinates')

以两点为输入的距离可以写成：

def haversine(point1, point2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    lon1, lat1 = point1.bounds[0], point1.bounds[1]
    lon2, lat2 = point2.bounds[0], point2.bounds[1]

    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a))
    r = 6371 # Radius of earth in kilometers
    # print('Distance from beginning to end of route in km: ',round((c * r), 2),'\n')
    return c * r

和使用pandas.DataFrame.apply的计算：

gdf['Coordinates'].apply(lambda x: gdf['Coordinates'].apply(lambda y: haversine(x, y)))

编辑：只计算矩阵的一半

gdf[['Coordinates']].apply(lambda x: gdf.loc[:x.name, 'Coordinates'].apply(lambda y: haversine(x['Coordinates'], y)), axis=1)

Answer 3

从、

中找到此代码

def calculate_distance(positions):
    results = []
    for i in range(1, len(positions)):
        loc1 = positions[i - 1]
        loc2 = positions[i]

        lat1 = loc1[0]
        lng1 = loc1[1]

        lat2 = loc2[0]
        lng2 = loc2[1]

        degreesToRadians = (math.pi / 180)
        latrad1 = lat1 * degreesToRadians
        latrad2 = lat2 * degreesToRadians
        dlat = (lat2 - lat1) * degreesToRadians
        dlng = (lng2 - lng1) * degreesToRadians

        a = math.sin(dlat / 2) * math.sin(dlat / 2) + math.cos(latrad1) * \
        math.cos(latrad2) * math.sin(dlng / 2) * math.sin(dlng / 2)
        c = 2 * math.atan2(math.sqrt(a), math.sqrt(1 - a))
        r = 6371000

        results.append(r * c)

    return (sum(results) / 1000)  # Converting from m to km

Answer 4

还有一个计算Haversine距离的包 https://pypi.org/project/haversine/ 它带有一个 numpy 版本，可以加快列表的计算速度。

计算两个列表的 haversine

computing haversine over two lists

python

pandas

geopandas