Dapatkan jarak terdekat dengan dua geodataframe di panda

Ini adalah geodatframe pertama saya:

!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))

City1

           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)

dan geodataframe kedua saya: City2:

         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)

Saya ingin frame data ketiga dengan kota terdekat dari city1 ke city2 dengan jarak seperti:

           City    Country  Latitude  Longitude                     geometry    Nearest    Distance
0  Buenos Aires  Argentina    -34.58     -58.66  POINT (-58.66000 -34.58000)    Bogota    111 Km

Inilah solusi aktual saya menggunakan geodjango dan dict (tapi terlalu lama):

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']

Inilah beberapa percobaan saya:

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

di sini:

    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

Salam

Pertama, saya menggabungkan dua frame data dengan cross join. Dan kemudian, saya menemukan jarak antara dua titik menggunakan mappython. Saya menggunakan map, karena sebagian besar waktu itu jauh lebih cepat daripada apply, itertuples, iterrowsdll (Referensi: https://stackoverflow.com/a/52674448/8205554 )

Terakhir, saya mengelompokkan berdasarkan frame data dan mengambil nilai jarak minimum.

Inilah perpustakaan,

import pandas as pd
import geopandas
import geopy.distance
from math import radians, cos, sin, asin, sqrt

Berikut adalah fungsi yang digunakan,

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

Dan data,

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)

Gabung silang dengan geopandasbingkai data,

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')

mathfungsi dan 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 

geopydan 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 

Jika Anda ingin menggunakan pandasbukan geopandas,

# cross join pandas
city1df['key'] = 1
city2df['key'] = 1
merged = city1df.merge(city2df, on='key')

Dengan mathfungsi,

# 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

Dengan 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

Saya pikir cukup sulit untuk menemukan solusi dengan kompleksitas waktu yang lebih baik daripada O (m · n) , di mana m dan n adalah ukuran city1dancity2 . Menjaga perbandingan jarak (satu-satunya operasi O (m · n)) sederhana, dan mengambil keuntungan dari operasi vektor yang disediakan oleh numpy dan panda, kecepatan seharusnya tidak menjadi masalah untuk ukuran input yang masuk akal.

Idenya adalah bahwa, untuk membandingkan jarak pada bola, Anda dapat membandingkan jarak antara titik-titik dalam 3D. Kota terdekat adalah juga yang paling dekat lewat melalui bola. Selain itu, Anda biasanya mengambil akar kuadrat untuk menghitung jarak, tetapi jika Anda hanya perlu membandingkannya, Anda dapat menghindari akar kuadrat.

from geopy.distance import distance as dist
import numpy as np
import pandas as pd

def find_closest(lat1, lng1, lat2, lng2):
    def x_y_z_of_lat_lng_on_unit_sphere(lat, lng):
        rad_lat, rad_lng = np.radians(lat), np.radians(lng)
        sin_lat, sin_lng = np.sin(rad_lat), np.sin(rad_lng)
        cos_lat, cos_lng = np.cos(rad_lat), np.cos(rad_lng)
        return cos_lat * cos_lng, cos_lat * sin_lng, sin_lat
    x1, y1, z1 = x_y_z_of_lat_lng_on_unit_sphere(lat1, lng1)
    x2, y2, z2 = x_y_z_of_lat_lng_on_unit_sphere(lat2, lng2)
    return pd.Series(map(lambda x, y, z:
                         ((x2-x)**2 + (y2-y)**2 + (z2-z)**2).idxmin(),
                         x1, y1, z1))

city1 = [{"City":"Tokyo",    "Ctry":"JP", "Latitude": 35.68972, "Longitude": 139.69222},
         {"City":"Pretoria", "Ctry":"ZA", "Latitude":-25.71667, "Longitude": 28.28333},
         {"City":"London",   "Ctry":"GB", "Latitude": 51.50722, "Longitude": -0.12574}]
city2 = [{"City":"Seattle",  "Ctry":"US", "Latitude": 47.60972, "Longitude":-122.33306},
         {"City":"Auckland", "Ctry":"NZ", "Latitude":-36.84446, "Longitude": 174.76364}]
city1df = pd.DataFrame(city1)
city2df = pd.DataFrame(city2)

closest = find_closest(city1df.Latitude, city1df.Longitude, city2df.Latitude, city2df.Longitude)

resultdf = city1df.join(city2df, on=closest, rsuffix='2')
km = pd.Series(map(lambda latlng1, latlng2: round(dist(latlng1, latlng2).km),
                   resultdf[['Latitude',  'Longitude' ]].to_numpy(),
                   resultdf[['Latitude2', 'Longitude2']].to_numpy()))
resultdf['Distance'] = km
print(resultdf.to_string())
#        City Ctry  Latitude  Longitude     City2 Ctry2  Latitude2  Longitude2  Distance
# 0     Tokyo   JP  35.68972  139.69222   Seattle    US   47.60972  -122.33306      7715
# 1  Pretoria   ZA -25.71667   28.28333  Auckland    NZ  -36.84446   174.76364     12245
# 2    London   GB  51.50722   -0.12574   Seattle    US   47.60972  -122.33306      7723

Perhatikan bahwa setiap solusi yang menggunakan lintang dan bujur seolah-olah mereka adalah koordinat Cartesius salah, karena bergerak ke kutub meridian (garis bujur yang sama) semakin dekat satu sama lain.

Solusi ini mungkin bukan cara tercepat untuk menyelesaikan masalah Anda, tetapi saya yakin ini akan berhasil.

#New dataframe is basicly a copy of first but with more columns
gcity3df = gcity1df.copy()
gcity3df["Nearest"] = None
gcity3df["Distance"] = None

#For each city (row in gcity3df) we will calculate the nearest city from gcity2df and 
fill the Nones with results

for index, row in gcity3df.iterrows():
    #Setting neareast and distance to None, 
    #we will be filling those variables with results

    nearest = None
    distance = None
    for df2index, df2row in gcity2df.iterrows():
        d = row.geometry.distance(df2row.geometry)
        #If df2index city is closer than previous ones, replace nearest with it
        if distance is None or d < distance:
            distance = d
            nearest = df2row.City 
    #In the end we appends the closest city to gdf
    gcity3df.at[index, "Nearest"] = nearest
    gcity3df.at[index, "Distance"] = distance

Jika Anda perlu bekerja pada meter dan bukan derajat, Anda selalu dapat memproyeksi ulang layer Anda (itu juga akan menghapus kesalahan yang dimaksudkan Walter). Anda dapat melakukannya dengan gcity3df = gcity3df.to_crs({'init': 'epsg:XXXX'})XXXX sebagai kode epsg untuk CR yang digunakan di wilayah dunia Anda.