问题描述
是否有更快的方法（在带cpu的Python中）执行与以下函数相同的操作？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个邮政编码大约需要1分钟才能运行此功能，而我大约需要70,000个电话才能通过。
使用的2个数据帧是：
postcode_df，其中包含71,092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：
- TubeRef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]})
该函数在postcode_df中的每个邮政编码中循环，对于每个邮政编码，在每个TubeRef中循环并计算（使用geopy）它们之间的距离，并以距邮政编码最短的距离保存TubeRef。
输出df postcode_nearest_tube_refs包含每个邮政编码最近的试管，并包含列：
- 邮政编码，例如“ BL4 7PD”
- 最近的空气管“ ABC01
- 与空气管KM的距离，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
我正在使用的图书馆是：
import numpy as np
import pandas as pd
# !pip install geopy
import geopy.distance
解决方法
这里是一个有效的示例，耗时数为10秒。
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))
并使用UUID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()
我们在空中做同样的事情
np_rand = 5 * np.random.random((500,2))
np_rand = np_rand + np.array((53.55108,-2.396236))
并再次使用uuid进行伪引用
tube_df = pd.DataFrame( np_rand,'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
创建一棵balltree
postal_radians = np.radians(postcode_gps)
air_radians = np.radians(air_gps)
tree = BallTree(air_radians,leaf_size=15,metric='haversine')
查询最接近的第一位
distance,index = tree.query(postal_radians,k=1)
请注意，距离不是以KM为单位，您需要先进行转换。
earth_radius = 6371000
distance_in_meters = distance * earth_radius
distance_in_meters
例如，使用tube_df.ref[ index[:,0] ]
您可以使用numpy计算A组中任何点到B组中任何点的距离矩阵，然后仅取A组中与最小距离相对应的点。
import numpy as np
import pandas as pd
dfA = pd.DataFrame({'lat':np.random.uniform(0,30,3),'lon':np.random.uniform(0,'id':[1,2,3]})
dfB = pd.DataFrame({'lat':np.random.uniform(0,'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}')