Python - 地理坐標之間的距離矩陣

Question

我有一個擁有 600 多個地理坐標點的 dataframe 熊貓。 他的摘錄如下：

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
from math import sin, cos, sqrt, atan2, radians

lat_long = pd.DataFrame({'LATITUDE':[-22.98, -22.97, -22.92, -22.87, -22.89], 'LONGITUDE': [-43.19, -43.39, -43.24, -43.28, -43.67]})
lat_long

要手動計算兩點之間的距離，我使用以下代碼：

lat1 = radians(lat_long['LATITUDE'][0])
lon1 = radians(lat_long['LONGITUDE'][0])
lat2 = radians(lat_long['LATITUDE'][1])
lon2 = radians(lat_long['LONGITUDE'][1])

R = 6373.0

dlon = lon2 - lon1
dlat = lat2 - lat1

a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))

distance = R * c

print("Result:", round(distance,4))

我需要做的是創建一個 function 使用上面的公式來計算從所有點到所有點的距離，就像在數組中一樣。 但我很難考慮 function 做什么和存儲點之間的距離。 歡迎任何幫助。 Output 示例（僅用於說明目的，如果我不清楚的話）：

|       |point 0 | point1 | point2 |
|point0 |    0   |    2   |   3    |
|point1 |    2   |    0   |   4    |
|point2 |    3   |    4   |   0    |
        |distance|distance|distance|

Answer 1

您可以使用pdist計算成對距離：

import pandas as pd

import numpy as np
from math import sin, cos, sqrt, atan2, radians

from scipy.spatial.distance import pdist, squareform

lat_long = pd.DataFrame({'LATITUDE': [-22.98, -22.97, -22.92, -22.87, -22.89], 'LONGITUDE': [-43.19, -43.39, -43.24, -43.28, -43.67]})


def dist(x, y):
    """Function to compute the distance between two points x, y"""

    lat1 = radians(x[0])
    lon1 = radians(x[1])
    lat2 = radians(y[0])
    lon2 = radians(y[1])

    R = 6373.0

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat / 2) ** 2 + cos(lat1) * cos(lat2) * sin(dlon / 2) ** 2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))

    distance = R * c

    return round(distance, 4)


distances = pdist(lat_long.values, metric=dist)

points = [f'point_{i}' for i in range(1, len(lat_long) + 1)]

result = pd.DataFrame(squareform(distances), columns=points, index=points)

print(result)

Output

         point_1  point_2  point_3  point_4  point_5
point_1   0.0000  20.5115   8.4123  15.3203  50.1784
point_2  20.5115   0.0000  16.3400  15.8341  30.0319
point_3   8.4123  16.3400   0.0000   6.9086  44.1838
point_4  15.3203  15.8341   6.9086   0.0000  40.0284
point_5  50.1784  30.0319  44.1838  40.0284   0.0000

請注意， squareform從稀疏矩陣轉換為密集矩陣，因此結果存儲在 numpy 數組中。

Answer 2

另一種可能的解決方案是

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np
from math import sin, cos, sqrt, atan2, radians

lat_long = pd.DataFrame({'LATITUDE':[-22.98, -22.97, -22.92, -22.87, -22.89], 'LONGITUDE': [-43.19, -43.39, -43.24, -43.28, -43.67]})
lat_long

test = lat_long.iloc[2:,:]

def distance(city1, city2):
    lat1 = radians(city1['LATITUDE'])
    lon1 = radians(city1['LONGITUDE'])
    lat2 = radians(city2['LATITUDE'])
    lon2 = radians(city2['LONGITUDE'])

    R = 6373.0

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))

    distance = R * c

    return distance

dist = np.zeros([lat_long.shape[0],lat_long.shape[0]])
for i1, city1 in lat_long.iterrows():
    for i2, city2 in lat_long.iloc[i1+1:,:].iterrows():
        dist[i1,i2] = distance(city1, city2)

print(dist)

Output

[[ 0.         20.51149047  8.41230771 15.32026132 50.17836849]
 [ 0.          0.         16.33997119 15.83407186 30.03192954]
 [ 0.          0.          0.          6.90864606 44.18376436]
 [ 0.          0.          0.          0.         40.02842872]
 [ 0.          0.          0.          0.          0.        ]]

距離矩陣的下三角形是空的，因為矩陣是對稱的（ dist[i1,i2]==dist[i2,i1] ）

Answer 3

這似乎快了兩倍：

# imports
import pandas as pd
import numpy as np

# supporting functions
def create_cartestin(df: pd.DataFrame):
    """
    This function returns cartesian of a dataframe with itself.

    df:
        dataframe to combine with itself

    """

    # create artifical id
    df['temp_id'] = [i for i in range(len(df))]

    # create cartesian merging key
    df['temp_key'] = 1
    df_cartesian = df.merge(df, on=['temp_key']).drop(columns=['temp_key'])

    return df_cartesian

def hav(theta):
    return np.sin(theta/2)**2

def give_me_straight_line_distance(
    df: pd.DataFrame,
    lattitude_x: str,
    longitude_x: str,
    lattitude_y: str,
    longitude_y: str
):
    """
    This function calculates distance between coordinates with haversine formula.

    df:
        dataframe containing cartesian product of tables containing points of interest
    lattitude_x:
        name of column containing lattitude of points x (1st set of points of interest)
    longitude_x:
        name of column containing longitude of points x (1st set of points of interest)
    lattitude_y:
        name of column containing lattitude of points y (2nd set of points of interest)
    longitude_y:
        name of column containing longitude of points y (2nd set of points of interest)

    """

    # assumed Earth radius
    r = 6371.009

    coords = df[[lattitude_x,longitude_x,lattitude_y,longitude_y]].values
    coordinates = np.deg2rad(coords)
    lat1 = coordinates[:, 0]
    lng1 = coordinates[:, 1]
    lat2 = coordinates[:, 2]
    lng2 = coordinates[:, 3]
    coslat1 = np.cos(lat1)
    coslat2 = np.cos(lat2)
    t = hav(lat2-lat1) + coslat1[:]*coslat2[:]*hav(lng2-lng1)
    d = 2*r*np.arcsin(np.sqrt(t))

    return d

# THE FUNCTION
def give_me_distance_matrix(df: pd.DataFrame):
    # create cartesian
    df_cartesian = create_cartestin(df)

    # calc distance for each pair of points
    df_cartesian['distance_km'] = \
    give_me_straight_line_distance(
        df = df_cartesian,
        lattitude_x = 'LATITUDE_x',
        longitude_x = 'LONGITUDE_x',
        lattitude_y = 'LATITUDE_y',
        longitude_y = 'LONGITUDE_y'
    )

    # turn into matrix format
    df_cartesian = df_cartesian.set_index(['temp_id_x','temp_id_y'])[['distance_km']].unstack(['temp_id_y'])

    # erasing artifical names
    df_cartesian = df_cartesian.reset_index(drop = True)
    df_cartesian = df_cartesian.T.reset_index(drop = True)

    return df_cartesian

lat_long = pd.DataFrame({'LATITUDE':[-22.98, -22.97, -22.92, -22.87, -22.89]*100, 'LONGITUDE': [-43.19, -43.39, -43.24, -43.28, -43.67]*100})
lat_long.shape
(500, 2)

%%timeit
result = give_me_distance_matrix(lat_long)

244 ms ± 33.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

比：

import pandas as pd
import numpy as np
from math import sin, cos, sqrt, atan2, radians
from scipy.spatial.distance import pdist, squareform

def dist(x, y):
    """Function to compute the distance between two points x, y"""

    lat1 = radians(x[0])
    lon1 = radians(x[1])
    lat2 = radians(y[0])
    lon2 = radians(y[1])

    R = 6373.0

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = sin(dlat / 2) ** 2 + cos(lat1) * cos(lat2) * sin(dlon / 2) ** 2
    c = 2 * atan2(sqrt(a), sqrt(1 - a))

    distance = R * c

    return round(distance, 4)

%%timeit
distances = pdist(lat_long.values, metric=dist)
points = [f'point_{i}' for i in range(1, len(lat_long) + 1)]
result = pd.DataFrame(squareform(distances), columns=points, index=points)

563 ms ± 77.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)