我正在尝试计算多组纬度和经度坐标之间的距离。简单来说，我已经找到了许多使用数学或geopy的教程。当我只想找到一个坐标集（或两个唯一位置）之间的距离时，这些教程非常有用。然而，我的目标是扫描具有400k个起点和终点坐标组合的数据集。下面列出了我使用过的代码示例，但是在我的数组大于1个记录时似乎会出现错误。非常感谢任何有用的提示。谢谢。
# starting dataframe is df
lat1 = df.lat1.as_matrix()
long1 = df.long1.as_matrix()
lat2 = df.lat2.as_matrix()
long2 = df.df_long2.as_matrix()
from geopy.distance import vincenty
point1 = (lat1, long1)
point2 = (lat2, long2)
print(vincenty(point1, point2).miles)
一般情况下，假设您有一个包含点的DataFrame列，并且您想要计算它们之间的距离（如果您有单独的列，请首先将它们合并为(lon,lat)元组）。将新列命名为coords。
import pandas as pd
import numpy as np
from geopy.distance import vincenty
# assumes your DataFrame is named df, and its lon and lat columns are named lon and lat. Adjust as needed.
df['coords'] = zip(df.lat, df.lon)
# first, let's create a square DataFrame (think of it as a matrix if you like)
square = pd.DataFrame(
np.zeros(len(df) ** 2).reshape(len(df), len(df)),
index=df.index, columns=df.index)
这个函数会从 df 数据框中使用输入的列名称查找我们的“结束”坐标，然后对输入列中的每一行应用 geopy 的 vincenty() 函数，使用 square.coords 列作为第一个参数。之所以能够工作，是因为该函数在应用时按列从右向左处理。
def get_distance(col):
end = df.ix[col.name]['coords']
return df['coords'].apply(vincenty, args=(end,), ellipsoid='WGS-84')
.T)，因为我们将使用loc[]方法来检索距离，该方法引用索引标签、行标签。然而，我们上面的内部应用函数会用检索到的值填充列。distances = square.apply(get_distance, axis=1).T
geopy 值（如果我没有记错）以公里为单位返回，因此你可能需要使用 .meters, .miles 等将其转换为所需的单位。def units(input_instance):
return input_instance.meters
distances_meters = distances.applymap(units)
loc[row_index, column_index]来索引您的距离矩阵。您应该能够相对容易地进行适应。您可能需要调整get_distance函数中apply调用以确保将正确的值传递给great_circle。Pandas的apply文档可能会有所帮助，特别是关于使用args传递位置参数的部分（这需要一个较新的pandas版本才能工作）。(lon, lat)还是(lat, lon)了。我打赌是后者（叹息）。import geopy.distance
# geopy DOES use latlon configuration
df['latlon'] = list(zip(df['lat'], df['lon']))
square = pd.DataFrame(
np.zeros((df.shape[0], df.shape[0])),
index=df.index, columns=df.index
)
# replacing distance.vicenty with distance.distance
def get_distance(col):
end = df.loc[col.name, 'latlon']
return df['latlon'].apply(geopy.distance.distance,
args=(end,),
ellipsoid='WGS-84'
)
distances = square.apply(get_distance, axis=1).T
vincenty需要被替换为distance。而在pandas中分配值的语法现在是：df.loc[col.name, 'coords']。 - BodoB我最近也做了类似的工作，我编写了一个我认为非常容易理解和调整以满足您需求的解决方案，但可能不是最佳/最快的：
它与urschrei发布的内容非常相似：假设您想从Pandas DataFrame中获得每两个连续坐标之间的距离，我们可以编写一个函数来处理每对点作为一条路径的起点和终点，计算距离，然后构建一个新的DataFrame来返回：
import pandas as pd
from geopy import Point, distance
def get_distances(coords: pd.DataFrame,
col_lat='lat',
col_lon='lon',
point_obj=Point) -> pd.DataFrame:
traces = len(coords) -1
distances = [None] * (traces)
for i in range(traces):
start = point_obj((coords.iloc[i][col_lat], coords.iloc[i][col_lon]))
finish = point_obj((coords.iloc[i+1][col_lat], coords.iloc[i+1][col_lon]))
distances[i] = {
'start': start,
'finish': finish,
'path distance': distance.geodesic(start, finish),
}
return pd.DataFrame(distances)
coords = pd.DataFrame({
'lat': [-26.244333, -26.238000, -26.233880, -26.260000, -26.263730],
'lon': [-48.640946, -48.644670, -48.648480, -48.669770, -48.660700],
})
print('-> coords DataFrame:\n', coords)
print('-'*79, end='\n\n')
distances = get_distances(coords)
distances['total distance'] = distances['path distance'].cumsum()
print('-> distances DataFrame:\n', distances)
print('-'*79, end='\n\n')
# Or if you want to use tuple for start/finish coordinates:
print('-> distances DataFrame using tuples:\n', get_distances(coords, point_obj=tuple))
print('-'*79, end='\n\n')
-> coords DataFrame:
lat lon
0 -26.244333 -48.640946
1 -26.238000 -48.644670
2 -26.233880 -48.648480
3 -26.260000 -48.669770
4 -26.263730 -48.660700
-------------------------------------------------------------------------------
-> distances DataFrame:
start finish \
0 26 14m 39.5988s S, 48 38m 27.4056s W 26 14m 16.8s S, 48 38m 40.812s W
1 26 14m 16.8s S, 48 38m 40.812s W 26 14m 1.968s S, 48 38m 54.528s W
2 26 14m 1.968s S, 48 38m 54.528s W 26 15m 36s S, 48 40m 11.172s W
3 26 15m 36s S, 48 40m 11.172s W 26 15m 49.428s S, 48 39m 38.52s W
path distance total distance
0 0.7941932910049856 km 0.7941932910049856 km
1 0.5943709651000332 km 1.3885642561050187 km
2 3.5914909016938505 km 4.980055157798869 km
3 0.9958396130609087 km 5.975894770859778 km
-------------------------------------------------------------------------------
-> distances DataFrame using tuples:
start finish path distance
0 (-26.244333, -48.640946) (-26.238, -48.64467) 0.7941932910049856 km
1 (-26.238, -48.64467) (-26.23388, -48.64848) 0.5943709651000332 km
2 (-26.23388, -48.64848) (-26.26, -48.66977) 3.5914909016938505 km
3 (-26.26, -48.66977) (-26.26373, -48.6607) 0.9958396130609087 km
-------------------------------------------------------------------------------
import pandas as pd
from geopy import Point, distance
def get_distances(coords: pd.DataFrame,
col_lat='lat',
col_lon='lon',
point_obj=Point) -> pd.DataFrame:
traces = len(coords) -1
distances = [None] * (traces)
for i in range(traces):
start = point_obj((coords.iloc[i][col_lat], coords.iloc[i][col_lon]))
finish = point_obj((coords.iloc[i+1][col_lat], coords.iloc[i+1][col_lon]))
distances[i] = {
'start': start,
'finish': finish,
'path distance': distance.geodesic(start, finish),
}
output = pd.DataFrame(distances)
output.to_csv('geopy_output.csv')
return output
我使用相同的代码为50,000多个坐标生成了距离数据。