这是我的第一个 geodatframe :
!pip install geopandas
import pandas as pd
import geopandas
city1 = [{'City':"Buenos Aires","Country":"Argentina","Latitude":-34.58,"Longitude":-58.66},
{'City':"Brasilia","Country":"Brazil","Latitude":-15.78 ,"Longitude":-70.66},
{'City':"Santiago","Country":"Chile ","Latitude":-33.45 ,"Longitude":-70.66 }]
city2 = [{'City':"Bogota","Country":"Colombia ","Latitude":4.60 ,"Longitude":-74.08},
{'City':"Caracas","Country":"Venezuela","Latitude":10.48 ,"Longitude":-66.86}]
city1df = pd.DataFrame(city1)
city2df = pd.DataFrame(city2)
gcity1df = geopandas.GeoDataFrame(
city1df, geometry=geopandas.points_from_xy(city1df.Longitude, city1df.Latitude))
gcity2df = geopandas.GeoDataFrame(
city2df, geometry=geopandas.points_from_xy(city2df.Longitude, city2df.Latitude))
城市1
City Country Latitude Longitude geometry
0 Buenos Aires Argentina -34.58 -58.66 POINT (-58.66000 -34.58000)
1 Brasilia Brazil -15.78 -47.91 POINT (-47.91000 -15.78000)
2 Santiago Chile -33.45 -70.66 POINT (-70.66000 -33.45000)
和我的第二个地理数据框:
城市 2 :
City Country Latitude Longitude geometry
1 Bogota Colombia 4.60 -74.08 POINT (-74.08000 4.60000)
2 Caracas Venezuela 10.48 -66.86 POINT (-66.86000 10.48000)
我想要从 city1 到 city2 最近的城市的第三个数据框，距离如下:
City Country Latitude Longitude geometry Nearest Distance
0 Buenos Aires Argentina -34.58 -58.66 POINT (-58.66000 -34.58000) Bogota 111 Km
这是我使用 geodjango 和 dict 的实际解决方案(但它太长了):
from django.contrib.gis.geos import GEOSGeometry
result = []
dict_result = {}
for city01 in city1 :
dist = 99999999
pnt = GEOSGeometry('SRID=4326;POINT( '+str(city01["Latitude"])+' '+str(city01['Longitude'])+')')
for city02 in city2:
pnt2 = GEOSGeometry('SRID=4326;POINT('+str(city02['Latitude'])+' '+str(city02['Longitude'])+')')
distance_test = pnt.distance(pnt2) * 100
if distance_test < dist :
dist = distance_test
result.append(dist)
dict_result[city01['City']] = city02['City']
这是我的尝试:
from shapely.ops import nearest_points
# unary union of the gpd2 geomtries
pts3 = gcity2df.geometry.unary_union
def Euclidean_Dist(df1, df2, cols=['x_coord','y_coord']):
return np.linalg.norm(df1[cols].values - df2[cols].values,
axis=1)
def near(point, pts=pts3):
# find the nearest point and return the corresponding Place value
nearest = gcity2df.geometry == nearest_points(point, pts)[1]
return gcity2df[nearest].City
gcity1df['Nearest'] = gcity1df.apply(lambda row: near(row.geometry), axis=1)
gcity1df
这里 :
City Country Latitude Longitude geometry Nearest
0 Buenos Aires Argentina -34.58 -58.66 POINT (-58.66000 -34.58000) Bogota
1 Brasilia Brazil -15.78 -70.66 POINT (-70.66000 -15.78000) Bogota
2 Santiago Chile -33.45 -70.66 POINT (-70.66000 -33.45000) Bogota
问候
最佳答案
首先，我通过交叉连接合并两个数据框。然后，我使用 map 找到两点之间的距离在 python 中。我用 map ，因为大多数时候它比 apply 快得多, itertuples , iterrows等(引用:https://stackoverflow.com/a/52674448/8205554)
最后，我按数据帧分组并获取距离的最小值。
这里是图书馆，
import pandas as pd
import geopandas
import geopy.distance
from math import radians, cos, sin, asin, sqrt
这里是使用的函数，
def dist1(p1, p2):
lon1, lat1, lon2, lat2 = map(radians, [p1.x, p1.y, p2.x, p2.y])
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
return c * 6373
def dist2(p1, p2):
lon1, lat1, lon2, lat2 = map(radians, [p1[0], p1[1], p2[0], p2[1]])
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
return c * 6373
def dist3(p1, p2):
x = p1.y, p1.x
y = p2.y, p2.x
return geopy.distance.geodesic(x, y).km
def dist4(p1, p2):
x = p1[1], p1[0]
y = p2[1], p2[0]
return geopy.distance.geodesic(x, y).km
还有数据，
city1 = [
{
'City': 'Buenos Aires',
'Country': 'Argentina',
'Latitude': -34.58,
'Longitude': -58.66
},
{
'City': 'Brasilia',
'Country': 'Brazil',
'Latitude': -15.78,
'Longitude': -70.66
},
{
'City': 'Santiago',
'Country': 'Chile ',
'Latitude': -33.45,
'Longitude': -70.66
}
]
city2 = [
{
'City': 'Bogota',
'Country': 'Colombia ',
'Latitude': 4.6,
'Longitude': -74.08
},
{
'City': 'Caracas',
'Country': 'Venezuela',
'Latitude': 10.48,
'Longitude': -66.86
}
]
city1df = pd.DataFrame(city1)
city2df = pd.DataFrame(city2)
与
geopandas 交叉连接数据框，gcity1df = geopandas.GeoDataFrame(
city1df,
geometry=geopandas.points_from_xy(city1df.Longitude, city1df.Latitude)
)
gcity2df = geopandas.GeoDataFrame(
city2df,
geometry=geopandas.points_from_xy(city2df.Longitude, city2df.Latitude)
)
# cross join geopandas
gcity1df['key'] = 1
gcity2df['key'] = 1
merged = gcity1df.merge(gcity2df, on='key')
math函数和 geopandas ,# 6.64 ms ± 588 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
# find distance
merged['dist'] = list(map(dist1, merged['geometry_x'], merged['geometry_y']))
mapping = {
'City_x': 'City',
'Country_x': 'Country',
'Latitude_x': 'Latitude',
'Longitude_x': 'Longitude',
'geometry_x': 'geometry',
'City_y': 'Nearest',
'dist': 'Distance'
}
nearest = merged.loc[merged.groupby(['City_x', 'Country_x'])['dist'].idxmin()]
nearest.rename(columns=mapping)[list(mapping.values())]
City Country Latitude Longitude geometry \
2 Brasilia Brazil -15.78 -70.66 POINT (-70.66000 -15.78000)
0 Buenos Aires Argentina -34.58 -58.66 POINT (-58.66000 -34.58000)
4 Santiago Chile -33.45 -70.66 POINT (-70.66000 -33.45000)
Nearest Distance
2 Bogota 2297.922808
0 Bogota 4648.004515
4 Bogota 4247.586882
geopy和 geopandas ,# 9.99 ms ± 764 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
# find distance
merged['dist'] = list(map(dist3, merged['geometry_x'], merged['geometry_y']))
mapping = {
'City_x': 'City',
'Country_x': 'Country',
'Latitude_x': 'Latitude',
'Longitude_x': 'Longitude',
'geometry_x': 'geometry',
'City_y': 'Nearest',
'dist': 'Distance'
}
nearest = merged.loc[merged.groupby(['City_x', 'Country_x'])['dist'].idxmin()]
nearest.rename(columns=mapping)[list(mapping.values())]
City Country Latitude Longitude geometry \
2 Brasilia Brazil -15.78 -70.66 POINT (-70.66000 -15.78000)
0 Buenos Aires Argentina -34.58 -58.66 POINT (-58.66000 -34.58000)
4 Santiago Chile -33.45 -70.66 POINT (-70.66000 -33.45000)
Nearest Distance
2 Bogota 2285.239605
0 Bogota 4628.641817
4 Bogota 4226.710978
如果您想使用
pandas而不是 geopandas ,# cross join pandas
city1df['key'] = 1
city2df['key'] = 1
merged = city1df.merge(city2df, on='key')
与
math职能，# 8.65 ms ± 2.21 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
# find distance
merged['dist'] = list(
map(
dist2,
merged[['Longitude_x', 'Latitude_x']].values,
merged[['Longitude_y', 'Latitude_y']].values
)
)
mapping = {
'City_x': 'City',
'Country_x': 'Country',
'Latitude_x': 'Latitude',
'Longitude_x': 'Longitude',
'City_y': 'Nearest',
'dist': 'Distance'
}
nearest = merged.loc[merged.groupby(['City_x', 'Country_x'])['dist'].idxmin()]
nearest.rename(columns=mapping)[list(mapping.values())]
City Country Latitude Longitude Nearest Distance
2 Brasilia Brazil -15.78 -70.66 Bogota 2297.922808
0 Buenos Aires Argentina -34.58 -58.66 Bogota 4648.004515
4 Santiago Chile -33.45 -70.66 Bogota 4247.586882
与
geopy ,# 9.8 ms ± 807 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%%timeit
# find distance
merged['dist'] = list(
map(
dist4,
merged[['Longitude_x', 'Latitude_x']].values,
merged[['Longitude_y', 'Latitude_y']].values
)
)
mapping = {
'City_x': 'City',
'Country_x': 'Country',
'Latitude_x': 'Latitude',
'Longitude_x': 'Longitude',
'City_y': 'Nearest',
'dist': 'Distance'
}
nearest = merged.loc[merged.groupby(['City_x', 'Country_x'])['dist'].idxmin()]
nearest.rename(columns=mapping)[list(mapping.values())]
City Country Latitude Longitude Nearest Distance
2 Brasilia Brazil -15.78 -70.66 Bogota 2285.239605
0 Buenos Aires Argentina -34.58 -58.66 Bogota 4628.641817
4 Santiago Chile -33.45 -70.66 Bogota 4226.710978
关于python - 使用 Pandas 中的两个地理数据框获取最近的距离，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59920770/