我目前正在尝试使用Python进行一些地理编码.过程如下：我有两个数据框(df1和df2,房屋和学校),具有纬度和经度值,并希望在df2中找到df1中每个观测值的最近邻居.我使用以下代码：

from tqdm import tqdm
import numpy as np
import pandas as pd
import math 

def distance(lat1, long1, lat2, long2):
        R = 6371 # Earth Radius in Km
        dLat = math.radians(lat2 - lat1) # Convert Degrees 2 Radians 
        dLong = math.radians(long2 - long1)
        lat1 = math.radians(lat1)
        lat2 = math.radians(lat2)
        a = math.sin(dLat/2) * math.sin(dLat/2) + math.sin(dLong/2) * math.sin(dLong/2) * math.cos(lat1) * math.cos(lat2)
        c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))
        d = R * c
        return d

dists = []
schools =[]
for index, row1 in tqdm(df1.iterrows()):
    for index, row2 in df2.iterrows():
        dists.append(distance(row1.lat, row1.lng, row2.Latitude, row2.Longitude))
    schools.append(min(dists))
    del dists [:]

df1["school"] = pd.Series(schools)

代码有效,但需要很长时间.使用tqdm,我得到平均速度为每秒df1的2次迭代.作为比较,我使用geonear在STATA中完成了整个任务,并且在df1(950)中所有观察需要1秒.我在geonear的帮助文件中读到他们使用聚类,不计算所有距离,而只计算最接近的距离.但是,在我添加一个集群功能(也可能需要CPU功率)之前,我想知道是否有人看到了一些方法来加快进程的速度(我是python的新手,并且可能有一些低效的代码会降低进程的速度) .或者是否有一个包可以更快地完成这个过程？

如果它比STATA需要更长的时间我会好的,但不会接近7分钟……

先感谢您

解决方法:

你这样做的方式很慢,因为你使用的是O(n²)算法：每一行都会查看每一行. Georgy’s answer虽然引入了矢量化,但并未解决这种基本的低效问题.

我建议将数据点加载到kd-tree：这种数据结构提供了一种快速查找多维度最近邻居的方法.这种树的构造在O(n log n)中,并且查询采用O(log n),因此总时间在O(n log n)中.

如果您的数据已本地化为可以通过平面很好地逼近的地理区域,则投影数据,然后在两个维度中执行查找.否则,如果您的数据是全局分散的,请投影到spherical cartesian coordinates并在那里执行查找.

您可以如何执行此操作的示例如下所示：

#/usr/bin/env python3

import numpy as np
import scipy as sp
import scipy.spatial

Rearth = 6371

#Generate uniformly-distributed lon-lat points on a sphere
#See: http://mathworld.wolfram.com/SpherePointPicking.html
def GenerateUniformSpherical(num):
  #Generate random variates
  pts      = np.random.uniform(low=0, high=1, size=(num,2))
  #Convert to sphere space
  pts[:,0] = 2*np.pi*pts[:,0]          #0-360 degrees
  pts[:,1] = np.arccos(2*pts[:,1]-1)   #0-180 degrees
  #Convert to degrees
  pts = np.degrees(pts)
  #Shift ranges to lon-lat
  pts[:,0] -= 180
  pts[:,1] -= 90
  return pts

def ConvertToXYZ(lonlat):
  theta  = np.radians(lonlat[:,0])+np.pi
  phi    = np.radians(lonlat[:,1])+np.pi/2
  x      = Rearth*np.cos(theta)*np.sin(phi)
  y      = Rearth*np.sin(theta)*np.sin(phi)
  z      = Rearth*np.cos(phi)
  return np.transpose(np.vstack((x,y,z)))

#For each entry in qpts, find the nearest point in the kdtree
def GetNearestNeighbours(qpts,kdtree):
  pts3d        = ConvertToXYZ(qpts)
  #See: https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.spatial.KDTree.query.html#scipy.spatial.KDTree.query
  #p=2 implies Euclidean distance, eps=0 implies no approximation (slower)
  return kdtree.query(pts3d,p=2,eps=0) 

#Generate uniformly-distributed test points on a sphere. Note that you'll want
#to find a way to extract your pandas columns into an array of width=2, height=N
#to match this format.
df1 = GenerateUniformSpherical(10000)
df2 = GenerateUniformSpherical(10000)

#Convert df2 into XYZ coordinates. WARNING! Do not alter df2_3d or kdtree will
#malfunction!
df2_3d = ConvertToXYZ(df2)
#Build a kd-tree from df2_3D
kdtree = sp.spatial.KDTree(df2_3d, leafsize=10) #Stick points in kd-tree for fast look-up

#Return the distance to, and index of, each of df1's nearest neighbour points
distance, indices = GetNearestNeighbours(df1,kdtree)