参考链接: Getting distance between two points based on latitude/longitude
我的数据框如下:
read_randomly_generated_lat_lon.head(3)
Lat Lon
43.937845 -97.905537
44.310739 -97.588820
44.914698 -99.003517
答案 0 :(得分:3)
请注意:以下脚本不考虑地球的曲率。 Convert lat/long to XY上有许多文档解释了此问题。
但是,可以大致确定坐标之间的距离。导出是一个系列,可以很容易地与原始concatenated
df
column
一起提供单独的d = ({
'Lat' : [43.937845,44.310739,44.914698],
'Long' : [-97.905537,-97.588820,-99.003517],
})
df = pd.DataFrame(d)
df = df[['Lat','Long']]
point1 = df.iloc[0]
def to_xy(point):
r = 6371000 #radians of the earth (m)
lam,phi = point
cos_phi_0 = np.cos(np.radians(phi))
return (r * np.radians(lam) * cos_phi_0,
r * np.radians(phi))
point1_xy = to_xy(point1)
df['to_xy'] = df.apply(lambda x:
tuple(x.values),
axis=1).map(to_xy)
df['Y'], df['X'] = df.to_xy.str[0], df.to_xy.str[1]
df = df[['X','Y']]
df = df.diff()
dist = np.sqrt(df['X']**2 + df['Y']**2)
#Convert to km
dist = dist/1000
print(dist)
0 NaN
1 41.149537
2 204.640462
来显示相对于坐标的距离。
LocalDate
答案 1 :(得分:2)
您可以使用scikit-learn进行此操作:
import numpy as np
from sklearn.neighbors import DistanceMetric
dfr = df.copy()
dfr.Lat = np.radians(df.Lat)
dfr.Lon = np.radians(df.Lon)
hs = DistanceMetric.get_metric("haversine")
(hs.pairwise(dfr)*6371) # Earth radius in km
输出:
array([[ 0. , 48.56264446, 139.2836099 ],
[ 48.56264446, 0. , 130.57312786],
[139.2836099 , 130.57312786, 0. ]])
请注意,输出是一个方矩阵,其中元素(i,j)是第i行与第j行之间的距离
这似乎比使用自定义haversine
函数的scipy的pdist更快