python – 需要一个适当的数据结构或索引,以便根据3d点和重要性

pythagoras(point,point2) * max(importance_factor,importance_factor2) * (abs(importance_factor - importance_factor2) + 1)

既然你还没有得到任何答案,我想我至少会提出一些想法.我使用了一个python k-d树模块来快速搜索最近的邻居点：
http://code.google.com/p/python-kdtree/downloads/detail?name=kdtree.py
只要它们的大小相同,它就需要任意点长度.

我不确定你将如何应用“重要性”的权重,但这里只是一个关于如何使用kdtree模块至少让最近的“人”到达给定人集的每个点的头脑风暴：

import numpy
from kdtree import KDTree
from itertools import chain

class PersonPoint(object):

    def __init__(self,person,point,factor):
        self.person = person 
        self.point = point 
        self.factor = factor 

    def __repr__(self):
        return '<%s: %s,%0.2f>' % (self.person,['%0.2f' % p for p in self.point],self.factor) 

    def __iter__(self):
        return self.point

    def __len__(self):
        return len(self.point)

    def __getitem__(self,i):
        return self.point[i]


people = {}
for name in ('bill','john','mary','jenny','phil','george'):
    factors = numpy.random.rand(6)
    points = numpy.random.rand(6,3).tolist()
    people[name] = [PersonPoint(name,p,f) for p,f in zip(points,factors)]

bill_points = people['bill']
others = list(chain(*[people[name] for name in people if name != 'bill']))

tree = KDTree.construct_from_data(others)

for point in bill_points:
    # t=1 means only return the 1 closest.
    # You could set it higher to return more.
    print point,"=>",tree.query(point,t=1)[0]

结果：

<bill: ['0.22','0.64','0.14'],0.07> => 
    <phil: ['0.23','0.54','0.11'],0.90>

<bill: ['0.31','0.87','0.16'],0.88> => 
    <phil: ['0.36','0.80',0.40>

<bill: ['0.34','0.25'],0.65> => 
    <jenny: ['0.29','0.77','0.28'],0.40>

<bill: ['0.24','0.90','0.23'],0.53> => 
    <jenny: ['0.29',0.40>

<bill: ['0.50','0.69','0.06'],0.68> => 
    <phil: ['0.36',0.40>

<bill: ['0.13','0.67','0.93'],0.54> => 
    <jenny: ['0.05','0.62','0.94'],0.84>

我想结果,你可以看看最常见的匹配“人”或者然后考虑权重.或者也许你可以总结结果中的重要因素,然后取最高评分.那样的话,如果玛丽只匹配一次,但有10个因子,而菲尔有3个匹配,但只总计5,玛丽可能更相关？

我知道你有一个更强大的功能来创建一个索引,但它需要遍历你的集合中的每个点.