在python中，有没有比for循环和if语句更快的方法来查找到另一个点的最近点？

Question

有没有一种更快的方法（在Python中，使用CPU）来完成与下面的函数相同的事情？我使用了For循环和if语句，想知道是否有更快的方法？目前，每100个邮政编码运行此功能大约需要1分钟，而我大约需要70000个邮政编码

使用的两个数据帧是：

postcode_df，其中包含71092行和列：

• 邮政编码，例如“BL4 7PD”
• 纬度，例如53.577653
• 经度，例如-2.434136

例如

postcode_df = pd.DataFrame({"Postcode":["SK12 2LH", "SK7 6LQ"],
                                    "Latitude":[53.362549, 53.373812],
                                    "Longitude":[-2.061329, -2.120956]})

air，其中包含421行和列：

• 管参考，例如“ABC01”
• 纬度，例如53.55108
• 经度，例如-2.396236

例如

air = pd.DataFrame({"TubeRef":["Stkprt35", "Stkprt07", "Stkprt33"],
                                    "Latitude":[53.365085, 53.379502, 53.407510],
                                    "Longitude":[-2.0763, -2.120777, -2.145632]})

函数循环使用postcode_df中的每个邮政编码，对于每个邮政编码，循环使用每个TubeRef并计算（使用geopy）它们之间的距离，并使用到邮政编码的最短距离保存TubeRef

输出dfpostcode_nearest_tube_refs包含每个邮政编码最近的管，并包含列：

• 邮政编码，例如“BL4 7PD”
• 最近的空气管，例如“ABC01
• 到空气管的距离KM，例如1.035848

# define function to get nearest air quality monitoring tube per postcode
def get_nearest_tubes(constituency_list):
    
    postcodes = []
    nearest_tubes = []
    distances_to_tubes = []
    
    for postcode in postcode_df["Postcode"]:
            closest_tube = ""
            shortest_dist = 500

            postcode_lat = postcode_df.loc[postcode_df["Postcode"]==postcode, "Latitude"]
            postcode_long = postcode_df.loc[postcode_df["Postcode"]==postcode, "Longitude"]
            postcode_coord = (float(postcode_lat), float(postcode_long))


            for tuberef in air["TubeRef"]:
                tube_lat = air.loc[air["TubeRef"]==tuberef, "Latitude"]
                tube_long = air.loc[air["TubeRef"]==tuberef, "Longitude"]
                tube_coord = (float(tube_lat), float(tube_long))

                # calculate distance between postcode and tube
                dist_to_tube = geopy.distance.distance(postcode_coord, tube_coord).km
                if dist_to_tube < shortest_dist:
                    shortest_dist = dist_to_tube
                    closest_tube = str(tuberef)

            # save postcode's tuberef with shortest distance
            postcodes.append(str(postcode))
            nearest_tubes.append(str(closest_tube))
            distances_to_tubes.append(shortest_dist)
            
    # create dataframe of the postcodes, nearest tuberefs and distance
    postcode_nearest_tube_refs = pd.DataFrame({"Postcode":postcodes, 
                                          "Nearest Air Tube":nearest_tubes, 
                                          "Distance to Air Tube KM": distances_to_tubes})

    return postcode_nearest_tube_refs

我正在使用的库包括：

import numpy as np
import pandas as pd
# !pip install geopy
import geopy.distance

Answer 1

这里是一个工作示例，以秒为单位（<；10）

导入库

import pandas as pd
import numpy as np
from sklearn.neighbors import BallTree
import uuid

我生成一些随机数据，这也需要一秒钟，但至少我们有一些实际的数据

np_rand_post = 5 * np.random.random((72000,2))
np_rand_post = np_rand_post + np.array((53.577653, -2.434136))

并将UUID用于伪造邮政编码

postcode_df = pd.DataFrame( np_rand_post , columns=['lat', 'long'])
postcode_df['postcode'] = [uuid.uuid4().hex[:6] for _ in range(72000)]
postcode_df.head()

我们对空气也是这样

np_rand = 5 * np.random.random((500,2))
np_rand = np_rand + np.array((53.55108, -2.396236))

再次使用uuid作为伪参考

tube_df = pd.DataFrame( np_rand , columns=['lat', 'long'])
tube_df['ref'] = [uuid.uuid4().hex[:5] for _ in range(500)]
tube_df.head()

将gps值提取为numpy

postcode_gps = postcode_df[["lat", "long"]].values
air_gps = tube_df[["lat", "long"]].values

创建一个棒球树

postal_radians =  np.radians(postcode_gps)
air_radians = np.radians(air_gps)

tree = BallTree(air_radians, leaf_size=15, metric='haversine')

查询最接近的第一个

distance, index = tree.query(postal_radians, k=1)

请注意，距离不是以公里为单位，需要先进行转换

earth_radius = 6371000

distance_in_meters = distance * earth_radius
distance_in_meters

例如，使用tube_df.ref[ index[:,0] ]获取ref

Answer 2

可以使用numpy计算集合a中任意点到集合B中任意点的距离矩阵，然后只取集合a中对应于最小距离的点

import numpy as np
import pandas as pd

dfA = pd.DataFrame({'lat':np.random.uniform(0, 30, 3), 'lon':np.random.uniform(0, 30, 3), 'id':[1,2,3]})
dfB = pd.DataFrame({'lat':np.random.uniform(0, 30, 3), 'lon':np.random.uniform(0, 30, 3), 'id':['a', 'b', 'c']})
lat1 = dfA.lat.values.reshape(-1, 1)
lat2 = dfB.lat.values.reshape(1, -1)
lon1 = dfA.lon.values.reshape(-1, 1)
lon2 = dfB.lon.values.reshape(1, -1)
dists = np.sqrt((lat1 - lat2)**2 + (lon1-lon2)**2)
for id1, id2 in zip (dfB.id, dfA.id.iloc[np.argmin(dists, axis=1)]):
    print(f'the closest point in dfA to {id1} is {id2}')