Question

我有一组用户的纬度和经度以及一组办公地点纬度经度。

我必须找到与所有用户之间的平均距离最小的办公地点。

在python中执行此操作的有效方法是什么？我有3k用户和40k办公地点......

例如：

输入：用户1（x1，y1）
用户2（x2，y2）
办公室1（x3，y3）
办公室2（x4，y4）

然后我必须弄清楚所有用户的平均距离最小的办公地点。

Office 1距离用户1 200米，距离用户2米400米。所有用户的平均距离= 300米

Office 2距用户1 100米，距离用户2米2.距所有用户平均距离= 150米

Office 2是首选位置。

Answer 1

以下是使用django的geodjango部分的示例。您可以使用shapely使用pyproj执行相同操作。（安装这些可能有点痛苦，但是一旦你完成了所有设置，这种工作就非常简单了。）

from django.contrib.gis.geos import Point, MultiPoint

WGS84_SRID = 4326
office1 = Point(x1, y1, srid=WGS84_SRID )
office2 = Point(x2, y1, srid=WGS84_SRID )

# load user locations
user_locations = []
with open("user_locations.csv", "r") as in_f:
    # assuming wgs84 decimal degrees 
    # one location per line in format, 'lon, lat'
    for line in in_f:
        x, y = [float(i.strip()) for i in line.split(",")]
        user_locations.append(Point(x, y, srid=WGS84_SRID ))

# get points in a meters projection
GOOGLE_MAPS_SRID = 3857
office1_meters = office1.transform(GOOGLE_MAPS_SRID, clone=True)
office2_meters = office2.transform(GOOGLE_MAPS_SRID, clone=True)
user_locations_meters = [user_loc.transform(GOOGLE_MAPS_SRID, clone=True) for user_loc in user_locations]

# centroid method
mp = MultiPoint(user_locations, srid=4326)
centroid_distance_from_office1 = mp.centroid.distance(office1_meters)
centroid_distance_from_office2 = mp.centroid.distance(office1_meters)

print "Centroid Location: {}".format(mp.centroid.ewkt)
print("centroid_distance_from_office1: {}m".format(centroid_distance_from_office1)
print("centroid_distance_from_office2: {}m".format(centroid_distance_from_office2)

# average distance method
total_user_locations = float(len(user_locations))
office1_user_avg_distance = sum( user_loc.distance(office1_meters) for user_loc in user_locations_meters)/total_user_locations 
office2_user_avg_distance = sum( user_loc.distance(office2_meters) for user_loc in user_locations_meters)/total_user_locations 

print "avg user distance OFFICE-1: {}".format(office1_user_avg_distance)
print "avg user distance OFFICE-2: {}".format(office2_user_avg_distance)

Answer 2

主要是代码，在http://en.wikipedia.org/wiki/Geometric_median#Computation中实现算法并给出一个基于一组随机点的使用示例。

注意：这是针对平面中的点，因为我无法确定如何对两个球面坐标进行求和...因此，您必须预先将球面坐标与平面投影进行映射，但这一点已经是触及了之前的回答。

<强>码

from math import sqrt
from random import seed, uniform
from operator import add
seed(100)

class Point():
    """Very basic point class, supporting "+", scalar "/" and distances."""
    def __init__(self, x, y):
        self.x = x
        self.y = y
    def __repr__(self):
        return "("+repr(self.x)+","+repr(self.y)+")"
    def __add__(self, P):
        return Point(self.x+P.x, self.y+P.y)
    def __div__(self, scalar):
        return Point(self.x/float(scalar), self.y/float(scalar))
    def delta(self, P):
        dx = self.x - P.x
        dy = self.y - P.y
        return sqrt(dx*dx+dy*dy)

def iterate(GM,points):
    "Simple implementation of http://en.wikipedia.org/wiki/Geometric_median#Computation"
    # distances from the tentative GM
    distances = [GM.delta(p) for p in points]
    normalized_positions = [p/d for p,d in zip(points,distances)]
    normalization_factor = sum(1.0/d for d in distances)
    new_median = reduce(add, normalized_positions)/normalization_factor
    return new_median

# The "clients"
nclients = 10
points = [Point(uniform(-3,3),uniform(-3,3)) for i in range(nclients)]

# Centroid of clients and total of distances
centroid = reduce(add,points)/nclients
print "Position of centroid:",centroid
print "Sum of distances from centroid:",
print reduce(add,[centroid.delta(p) for p in points])


print
print "Compute the Geometric Median using random starting points:"
nstart = 10
for P0 in [Point(uniform(-5,5),uniform(-5,5)) for i in range(nstart)]:
    p0 = P0
    for i in range(10):
        P0 = iterate(P0, points)
    print p0,"-->",P0

print
print "Sum of distances from last Geometric Median:",
print reduce(add,[P0.delta(p) for p in points])

<强>输出

Position of centroid: (-0.4647467432024398,0.08675910209912471)
Sum of distances from centroid: 22.846445119

Compute the Geometric Median using random starting points:
(1.2632163919279735,4.633157837008632) --> (-0.8739691868669638,-0.019827884361901298)
(-2.8916600791314986,4.561006461166512) --> (-0.8929310891388812,-0.025857080003665663)
(0.5539966580106901,4.011520429873922) --> (-0.8764828849474395,-0.020607834485528134)
(3.1801819335743033,-3.395781900250662) --> (-0.8550062003820846,-0.014134334529992666)
(1.48542908120573,-3.7590671941155627) --> (-0.8687797019011291,-0.018241177226221747)
(-4.943549141082007,-1.044838193982506) --> (-0.9066276248482427,-0.030440865315529194)
(2.73500702168781,0.6615770729288597) --> (-0.8231318436739281,-0.005320464433689587)
(-3.073593440129266,3.411747144619733) --> (-0.8952513352350909,-0.026600471220747438)
(4.137768422492282,-2.6277493707729596) --> (-0.8471586848200597,-0.011875801531868494)
(-0.5180751681772549,1.377998063140823) --> (-0.8849056106235963,-0.02326386487180884)

Sum of distances from last Geometric Median: 22.7019120091

我自己的评论

在这种情况下，位置（质心与GM）完全不同，但结果相似。当你在一个点（一个城市）周围进行某种聚类，或者某些特征说一条线（一条道路）等时，我希望在位置和平均距离上存在显着差异

最终，人们可以使用numpy加快速度，由于时间资源有限，我已避免执行numpy：）

Python：最小平均距离

2 个答案: