数以百万计的3D指出：如何找到他们最近的10到指定的点？ [英] Millions of 3D points: How to find the 10 of them closest to a given point?

问题描述

在三维甲点由（X，Y，Z）中所定义。任意两点（X，Y，Z）和（X，Y，Z）之间的距离是d =的Sqrt [（二十）^ 2 +（YY）^ 2 +（ZZ）^ 2]。 现在也有在文件中的万个条目，每个条目是一些点在空间中，在没有特定的顺序。给定任意点（A，B，C）寻找最近的10点到它。你将如何保存万点以及如何将从该数据结构检索这些10点。

A point in 3-d is defined by (x,y,z). Distance d between any two points (X,Y,Z) and (x,y,z) is d= Sqrt[(X-x)^2 + (Y-y)^2 + (Z-z)^2]. Now there are a million entries in a file, each entry is some point in space, in no specific order. Given any point (a,b,c) find the nearest 10 points to it. How would you store the million points and how would you retrieve those 10 points from that data structure.

推荐答案

万点是一个小数目。最直接的方法在这里工作（基于KDTree code是慢（用于查询仅1点））。

Million points is a small number. The most straightforward approach works here (code based on KDTree is slower (for querying only one point)).

#!/usr/bin/env python
import numpy

NDIM = 3 # number of dimensions

# read points into array
a = numpy.fromfile('million_3D_points.txt', sep=' ')
a.shape = a.size / NDIM, NDIM

point = numpy.random.uniform(0, 100, NDIM) # choose random point
print 'point:', point
d = ((a-point)**2).sum(axis=1)  # compute distances
ndx = d.argsort() # indirect sort 

# print 10 nearest points to the chosen one
import pprint
pprint.pprint(zip(a[ndx[:10]], d[ndx[:10]]))

运行：

$ time python nearest.py 
point: [ 69.06310224   2.23409409  50.41979143]
[(array([ 69.,   2.,  50.]), 0.23500677815852947),
 (array([ 69.,   2.,  51.]), 0.39542392750839772),
 (array([ 69.,   3.,  50.]), 0.76681859086988302),
 (array([ 69.,   3.,  50.]), 0.76681859086988302),
 (array([ 69.,   3.,  51.]), 0.9272357402197513),
 (array([ 70.,   2.,  50.]), 1.1088022980015722),
 (array([ 70.,   2.,  51.]), 1.2692194473514404),
 (array([ 70.,   2.,  51.]), 1.2692194473514404),
 (array([ 70.,   3.,  51.]), 1.801031260062794),
 (array([ 69.,   1.,  51.]), 1.8636121147970444)]

real    0m1.122s
user    0m1.010s
sys 0m0.120s

下面是产生百万三维点的脚本：

Here's the script that generates million 3D points:

#!/usr/bin/env python
import random
for _ in xrange(10**6):
    print ' '.join(str(random.randrange(100)) for _ in range(3))

输出：

$ head million_3D_points.txt

18 56 26
19 35 74
47 43 71
82 63 28
43 82 0
34 40 16
75 85 69
88 58 3
0 63 90
81 78 98

您可以使用code键测试更复杂的数据结构和算法（例如，它们是否实际上消耗更少的存储器或更快那么上述简单的方法）。值得一提的是，目前它是一个包含工作code是唯一的答案。

You could use that code to test more complex data structures and algorithms (for example, whether they actually consume less memory or faster then the above simplest approach). It is worth noting that at the moment it is the only answer that contains working code.

#!/usr/bin/env python
import numpy

NDIM = 3 # number of dimensions

# read points into array
a = numpy.fromfile('million_3D_points.txt', sep=' ')
a.shape = a.size / NDIM, NDIM

point =  [ 69.06310224,   2.23409409,  50.41979143] # use the same point as above
print 'point:', point


from scipy.spatial import KDTree

# find 10 nearest points
tree = KDTree(a, leafsize=a.shape[0]+1)
distances, ndx = tree.query([point], k=10)

# print 10 nearest points to the chosen one
print a[ndx]

运行：

$ time python nearest_kdtree.py  

point: [69.063102240000006, 2.2340940900000001, 50.419791429999997]
[[[ 69.   2.  50.]
  [ 69.   2.  51.]
  [ 69.   3.  50.]
  [ 69.   3.  50.]
  [ 69.   3.  51.]
  [ 70.   2.  50.]
  [ 70.   2.  51.]
  [ 70.   2.  51.]
  [ 70.   3.  51.]
  [ 69.   1.  51.]]]

real    0m1.359s
user    0m1.280s
sys 0m0.080s

部分排序的C ++（时间〜1.1秒）

// $ g++ nearest.cc && (time ./a.out < million_3D_points.txt )
#include <algorithm>
#include <iostream>
#include <vector>

#include <boost/lambda/lambda.hpp>  // _1
#include <boost/lambda/bind.hpp>    // bind()
#include <boost/tuple/tuple_io.hpp>

namespace {
  typedef double coord_t;
  typedef boost::tuple<coord_t,coord_t,coord_t> point_t;

  coord_t distance_sq(const point_t& a, const point_t& b) { // or boost::geometry::distance
    coord_t x = a.get<0>() - b.get<0>();
    coord_t y = a.get<1>() - b.get<1>();
    coord_t z = a.get<2>() - b.get<2>();
    return x*x + y*y + z*z;
  }
}

int main() {
  using namespace std;
  using namespace boost::lambda; // _1, _2, bind()

  // read array from stdin
  vector<point_t> points;
  cin.exceptions(ios::badbit); // throw exception on bad input
  while(cin) {
    coord_t x,y,z;
    cin >> x >> y >> z;    
    points.push_back(boost::make_tuple(x,y,z));
  }

  // use point value from previous examples
  point_t point(69.06310224, 2.23409409, 50.41979143);
  cout << "point: " << point << endl;  // 1.14s

  // find 10 nearest points using partial_sort() 
  // Complexity: O(N)*log(m) comparisons (O(N)*log(N) worst case for the implementation)
  const size_t m = 10;
  partial_sort(points.begin(), points.begin() + m, points.end(), 
               bind(less<coord_t>(), // compare by distance to the point
                    bind(distance_sq, _1, point), 
                    bind(distance_sq, _2, point)));
  for_each(points.begin(), points.begin() + m, cout << _1 << "\n"); // 1.16s
}

运行：

g++ -O3 nearest.cc && (time ./a.out < million_3D_points.txt )
point: (69.0631 2.23409 50.4198)
(69 2 50)
(69 2 51)
(69 3 50)
(69 3 50)
(69 3 51)
(70 2 50)
(70 2 51)
(70 2 51)
(70 3 51)
(69 1 51)

real    0m1.152s
user    0m1.140s
sys 0m0.010s

C语言优先级队列++（时间约1.2秒），

#include <algorithm>           // make_heap
#include <functional>          // binary_function<>
#include <iostream>

#include <boost/range.hpp>     // boost::begin(), boost::end()
#include <boost/tr1/tuple.hpp> // get<>, tuple<>, cout <<

namespace {
  typedef double coord_t;
  typedef std::tr1::tuple<coord_t,coord_t,coord_t> point_t;

  // calculate distance (squared) between points `a` & `b`
  coord_t distance_sq(const point_t& a, const point_t& b) { 
    // boost::geometry::distance() squared
    using std::tr1::get;
    coord_t x = get<0>(a) - get<0>(b);
    coord_t y = get<1>(a) - get<1>(b);
    coord_t z = get<2>(a) - get<2>(b);
    return x*x + y*y + z*z;
  }

  // read from input stream `in` to the point `point_out`
  std::istream& getpoint(std::istream& in, point_t& point_out) {    
    using std::tr1::get;
    return (in >> get<0>(point_out) >> get<1>(point_out) >> get<2>(point_out));
  }

  // Adaptable binary predicate that defines whether the first
  // argument is nearer than the second one to given reference point
  template<class T>
  class less_distance : public std::binary_function<T, T, bool> {
    const T& point;
  public:
    less_distance(const T& reference_point) : point(reference_point) {}

    bool operator () (const T& a, const T& b) const {
      return distance_sq(a, point) < distance_sq(b, point);
    } 
  };
}

int main() {
  using namespace std;

  // use point value from previous examples
  point_t point(69.06310224, 2.23409409, 50.41979143);
  cout << "point: " << point << endl;

  const size_t nneighbours = 10; // number of nearest neighbours to find
  point_t points[nneighbours+1];

  // populate `points`
  for (size_t i = 0; getpoint(cin, points[i]) && i < nneighbours; ++i)
    ;

  less_distance<point_t> less_distance_point(point);
  make_heap  (boost::begin(points), boost::end(points), less_distance_point);

  // Complexity: O(N*log(m))
  while(getpoint(cin, points[nneighbours])) {
    // add points[-1] to the heap; O(log(m))
    push_heap(boost::begin(points), boost::end(points), less_distance_point); 
    // remove (move to last position) the most distant from the
    // `point` point; O(log(m))
    pop_heap (boost::begin(points), boost::end(points), less_distance_point);
  }

  // print results
  push_heap  (boost::begin(points), boost::end(points), less_distance_point);
  //   O(m*log(m))
  sort_heap  (boost::begin(points), boost::end(points), less_distance_point);
  for (size_t i = 0; i < nneighbours; ++i) {
    cout << points[i] << ' ' << distance_sq(points[i], point) << '\n';  
  }
}

运行：

$ g++ -O3 nearest.cc && (time ./a.out < million_3D_points.txt )

point: (69.0631 2.23409 50.4198)
(69 2 50) 0.235007
(69 2 51) 0.395424
(69 3 50) 0.766819
(69 3 50) 0.766819
(69 3 51) 0.927236
(70 2 50) 1.1088
(70 2 51) 1.26922
(70 2 51) 1.26922
(70 3 51) 1.80103
(69 1 51) 1.86361

real    0m1.174s
user    0m1.180s
sys 0m0.000s

线性搜索为基础的方法（时间〜1.15秒）

// $ g++ -O3 nearest.cc && (time ./a.out < million_3D_points.txt )
#include <algorithm>           // sort
#include <functional>          // binary_function<>
#include <iostream>

#include <boost/foreach.hpp>
#include <boost/range.hpp>     // begin(), end()
#include <boost/tr1/tuple.hpp> // get<>, tuple<>, cout <<

#define foreach BOOST_FOREACH

namespace {
  typedef double coord_t;
  typedef std::tr1::tuple<coord_t,coord_t,coord_t> point_t;

  // calculate distance (squared) between points `a` & `b`
  coord_t distance_sq(const point_t& a, const point_t& b);

  // read from input stream `in` to the point `point_out`
  std::istream& getpoint(std::istream& in, point_t& point_out);    

  // Adaptable binary predicate that defines whether the first
  // argument is nearer than the second one to given reference point
  class less_distance : public std::binary_function<point_t, point_t, bool> {
    const point_t& point;
  public:
    explicit less_distance(const point_t& reference_point) 
        : point(reference_point) {}
    bool operator () (const point_t& a, const point_t& b) const {
      return distance_sq(a, point) < distance_sq(b, point);
    } 
  };
}

int main() {
  using namespace std;

  // use point value from previous examples
  point_t point(69.06310224, 2.23409409, 50.41979143);
  cout << "point: " << point << endl;
  less_distance nearer(point);

  const size_t nneighbours = 10; // number of nearest neighbours to find
  point_t points[nneighbours];

  // populate `points`
  foreach (point_t& p, points)
    if (! getpoint(cin, p))
      break;

  // Complexity: O(N*m)
  point_t current_point;
  while(cin) {
    getpoint(cin, current_point); //NOTE: `cin` fails after the last
                                  //point, so one can't lift it up to
                                  //the while condition

    // move to the last position the most distant from the
    // `point` point; O(m)
    foreach (point_t& p, points)
      if (nearer(current_point, p)) 
        // found point that is nearer to the `point` 

        //NOTE: could use insert (on sorted sequence) & break instead
        //of swap but in that case it might be better to use
        //heap-based algorithm altogether
        std::swap(current_point, p);
  }

  // print results;  O(m*log(m))
  sort(boost::begin(points), boost::end(points), nearer);
  foreach (point_t p, points)
    cout << p << ' ' << distance_sq(p, point) << '\n';  
}

namespace {
  coord_t distance_sq(const point_t& a, const point_t& b) { 
    // boost::geometry::distance() squared
    using std::tr1::get;
    coord_t x = get<0>(a) - get<0>(b);
    coord_t y = get<1>(a) - get<1>(b);
    coord_t z = get<2>(a) - get<2>(b);
    return x*x + y*y + z*z;
  }

  std::istream& getpoint(std::istream& in, point_t& point_out) {    
    using std::tr1::get;
    return (in >> get<0>(point_out) >> get<1>(point_out) >> get<2>(point_out));
  }
}

测量显示，大部分的时间都花在从文件读取数组，实际计算采取量级的时间更少了。

Measurements shows that most of the time is spent reading array from the file, actual computations take on order of magnitude less time.

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