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存储 3d 空间中点的信息

发布于 2024-07-22 05:21:39 字数 522 浏览 5 评论 0原文

我正在用 Python 编写一些代码（到目前为止只是为了好玩），它将在 3d 空间中的每个点上存储一些数据。我基本上是在寻找一个存储任意对象的 3d 矩阵对象，它允许我进行一些高级选择，例如：

获取 x=1,y=2,z=3 的点。
获取 y=2 的所有点。
获取位置 x=1,y=2,z=3 3 个单位内的所有点。
获取所有点，其中 point.getType() == "Foo"

在上述所有内容中，我需要最终得到某种输出，该输出将为我提供空间中的原始位置以及该点存储的数据。

显然 numpy 可以做我想做的事，但它似乎针对科学计算进行了高度优化，并且到目前为止我还没有弄清楚如何获取我想要的数据。

有更好的选择吗？还是我应该继续用头撞墙？ :)

编辑：前三个答案让我意识到我应该包括更多信息：我不担心性能，这纯粹是一个概念验证，我更喜欢干净的代码而不是良好的性能。我还将获得给定 3d 空间中每个点的数据，所以我猜备用矩阵不好？

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数理化全能战士 2024-07-29 05:21:39

这是另一种常见方法

class Point( object ):
    def __init__( self, x, y, z, data ):
        self.x, self.y, self.z = x, y, z
        self.data = data
    def distFrom( self, x, y, z )
        return math.sqrt( (self.x-x)**2 + (self.y-y)**2 + (self.z-z)**2 )

database = [ Point(x,y,z,data), Point(x,y,z,data), ... ]

让我们看看您的用例。

求 x=1,y=2,z=3 的点。

[ p for p in database if (p.x, p.y, p.z) == ( 1, 2, 3 ) ]

获取 y=2 的所有点。

[ p for p in database if p.y == 2 ]

获取位置 x=1,y=2,z=3 3 个单位内的所有点。

[ p for p in database if p.distFrom( 1, 2, 3 ) <= 3.0 ]

获取 point.getType() == "Foo" 的所有点

[ p for p in database if type(p.data) == Foo ]

Here's another common approach

class Point( object ):
    def __init__( self, x, y, z, data ):
        self.x, self.y, self.z = x, y, z
        self.data = data
    def distFrom( self, x, y, z )
        return math.sqrt( (self.x-x)**2 + (self.y-y)**2 + (self.z-z)**2 )

database = [ Point(x,y,z,data), Point(x,y,z,data), ... ]

Let's look at your use cases.

Get the point where x=1,y=2,z=3.

[ p for p in database if (p.x, p.y, p.z) == ( 1, 2, 3 ) ]

Getting all points where y=2.

[ p for p in database if p.y == 2 ]

Getting all points within 3 units of position x=1,y=2,z=3.

[ p for p in database if p.distFrom( 1, 2, 3 ) <= 3.0 ]

Getting all points where point.getType() == "Foo"

[ p for p in database if type(p.data) == Foo ]

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我恋#小黄人 2024-07-29 05:21:39

好吧...如果您希望真正填充该空间，那么您可能最好使用密集的类似矩阵的结构，基本上体素。

如果您不希望填充它，请研究一些更优化的东西。我首先会看看八叉树，它经常用于类似的事情。

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呢古 2024-07-29 05:21:39

numpy 的优点之一是它的速度非常快，
例如，计算 8000x8000 邻接矩阵的 pagerank 需要几毫秒。尽管 numpy.ndarray 只接受数字，但您可以将数字/id 对象映射存储在外部哈希表（即字典）中（这又是一个高度优化的数据结构）。

切片就像 python 中的列表切片一样简单：

>>> from numpy import arange

>>> the_matrix = arange(64).reshape(4, 4, 4)
>>> print the_matrix[0][1][2]
    6
>>> print the_matrix[0][1]
    [4 5 6 7]
>>> print the_matrix[0]
    [[ 0  1  2  3]
    [ 4  5  6  7]
    [ 8  9 10 11]
    [12 13 14 15]]

如果您将一些所需的函数（距离）包装在一些核心矩阵和 id-object-mapping 哈希周围，您可以让您的应用程序在短时间内运行。

祝你好运！

One advantage of numpy is that it is blazingly fast,
e.g. calculating the pagerank of a 8000x8000 adjacency matrix takes milliseconds. Even though numpy.ndarray will only accept numbers, you can store number/id-object mappings in an external hash-table i.e. dictionary (which in again is a highly optimized datastructure).

The slicing would be as easy as list slicing in python:

>>> from numpy import arange

>>> the_matrix = arange(64).reshape(4, 4, 4)
>>> print the_matrix[0][1][2]
    6
>>> print the_matrix[0][1]
    [4 5 6 7]
>>> print the_matrix[0]
    [[ 0  1  2  3]
    [ 4  5  6  7]
    [ 8  9 10 11]
    [12 13 14 15]]

If you wrap some of your desired functions (distances) around some core matrix and a id-object-mapping hash, you could have your application running within a short period of time.

Good luck!

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挽手叙旧 2024-07-29 05:21:39

这是一种可能有效的方法。

每个点都是一个 4 元组（x、y、z、数据），您的数据库如下所示：

database = [ (x,y,z,data), (x,y,z,data), ... ]

让我们看看您的用例。

求 x=1,y=2,z=3 的点。

[ (x,y,z,data) for x,y,z,data in database if (x,y,z) == (1,2,3) ]

获取 y=2 的所有点。

[ (x,y,z,data) for x,y,z,data in database if y == 2 ]

获取位置 x=1,y=2,z=3 3 个单位内的所有点。

[ (x,y,z,data) for x,y,z,data in database if math.sqrt((x-1)**2+(y-2)**2+(z-3)**2)<=3.0 ]

获取 point.getType() == "Foo" 的所有点

[ (x,y,z,data) for x,y,z,data in database if type(data) == Foo ]

Here's an approach that may work.

Each point is a 4-tuple (x,y,z,data), and your database looks like this:

database = [ (x,y,z,data), (x,y,z,data), ... ]

Let's look at your use cases.

Get the point where x=1,y=2,z=3.

[ (x,y,z,data) for x,y,z,data in database if (x,y,z) == (1,2,3) ]

Getting all points where y=2.

[ (x,y,z,data) for x,y,z,data in database if y == 2 ]

Getting all points within 3 units of position x=1,y=2,z=3.

[ (x,y,z,data) for x,y,z,data in database if math.sqrt((x-1)**2+(y-2)**2+(z-3)**2)<=3.0 ]

Getting all points where point.getType() == "Foo"

[ (x,y,z,data) for x,y,z,data in database if type(data) == Foo ]

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晨光如昨 2024-07-29 05:21:39

您可以使用 numpy 中的切片来执行前两个查询：

a = numpy.zeros((4, 4, 4))
a[1, 2, 3] # The point at x=1,y=2,z=3
a[:, 2, :] # All points where y=2

对于第三个查询，如果您的意思是“获取半径为 3 且以 x=1,y=2,z=3 为中心的球体内的所有点”，则必须编写一个自定义函数来做到这一点；如果您想要一个立方体，您可以继续进行切片，例如：

a[1:3, 1:3, 1:3] # The 2x2x2 array sliced from the center of 'a'

对于第四个查询，如果数组中存储的唯一数据是单元格类型，您可以将其编码为整数：

FOO = 1
BAR = 2
a = numpy.zeros((4, 4, 4), dtype="i")
a[1, 2, 3] = FOO
a[3, 2, 1] = BAR
def filter(a, type_code):
    coords = []
    for z in range(4):
        for y in range(4):
            for x in range(4):
                if a[x, y, z] == type_code:
                    coords.append((x, y, z))
    return coords
filter(a, FOO) # => [(1, 2, 3)]

numpy 看起来是执行您想要的操作的好工具，因为数组在内存中会更小，可以在 C 中轻松访问（或者更好，cython ！）和扩展切片语法将避免您编写代码。

You can do the first 2 queries with slicing in numpy :

a = numpy.zeros((4, 4, 4))
a[1, 2, 3] # The point at x=1,y=2,z=3
a[:, 2, :] # All points where y=2

For the third one if you mean "getting all points within a sphere of radius 3 and centered at x=1,y=2,z=3", you will have to write a custom function to do that ; if you want a cube you can proceed with slicing, e.g.:

a[1:3, 1:3, 1:3] # The 2x2x2 array sliced from the center of 'a'

For the fourth query if the only data stored in your array is the cells type, you could encode it as integers:

FOO = 1
BAR = 2
a = numpy.zeros((4, 4, 4), dtype="i")
a[1, 2, 3] = FOO
a[3, 2, 1] = BAR
def filter(a, type_code):
    coords = []
    for z in range(4):
        for y in range(4):
            for x in range(4):
                if a[x, y, z] == type_code:
                    coords.append((x, y, z))
    return coords
filter(a, FOO) # => [(1, 2, 3)]

numpy looks like the good tool for doing what you want, as the arrays will be smaller in memory, easily accessible in C (or even better, cython !) and extended slicing syntax will avoid you writing code.

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别念他 2024-07-29 05:21:39

如果您想要使用标准库获得相对简单的解决方案，则使用以 x,y,z 元组作为键的字典是另一种解决方案。

import math

#use indexing to get point at (1,2,3): points[(1,2,3)]
get_points(points, x=None, y=None, x=None):
    """returns dict of all points with given x,y and/or z values.  Call using keywords (eg get_points(points, x=3)"""
    filteredPoints = points.items()
    if x:
        filteredPoints = [p for p in filteredPoints if p[0][0] == x]
    if y:
        filteredPoints = [p for p in filteredPoints if p[0][1] == y]
    if z:
        filteredPoints = [p for p in filteredPoints if p[0][0] == x]
    return dict(filteredPoints)

get_point_with_type(points, type_):
    """returns dict of points with point.getType() == type_"""
    filteredPoints = points.items()
    return dict((position,data) for position,data in points.iterItems() if data.getType == type_)

get_points_in_radius(points,x,y,z,r):
    """Returns a dict of points within radius r of point (x,y,z)"""
    def get_dist(x1,y1,z1,x2,y2,z3):
        return math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2))
    return dict((position,data) for position,data in points.iterItems() if get_dist(x,y,z, *position) <= r))

由于 python 引用，您可以更改返回的字典中的“点”，并且原始点也会更改（我认为）。

Using a dictionary with x,y,z tuples as keys is another solution, if you want a relatively simple solution with the standard library.

import math

#use indexing to get point at (1,2,3): points[(1,2,3)]
get_points(points, x=None, y=None, x=None):
    """returns dict of all points with given x,y and/or z values.  Call using keywords (eg get_points(points, x=3)"""
    filteredPoints = points.items()
    if x:
        filteredPoints = [p for p in filteredPoints if p[0][0] == x]
    if y:
        filteredPoints = [p for p in filteredPoints if p[0][1] == y]
    if z:
        filteredPoints = [p for p in filteredPoints if p[0][0] == x]
    return dict(filteredPoints)

get_point_with_type(points, type_):
    """returns dict of points with point.getType() == type_"""
    filteredPoints = points.items()
    return dict((position,data) for position,data in points.iterItems() if data.getType == type_)

get_points_in_radius(points,x,y,z,r):
    """Returns a dict of points within radius r of point (x,y,z)"""
    def get_dist(x1,y1,z1,x2,y2,z3):
        return math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2))
    return dict((position,data) for position,data in points.iterItems() if get_dist(x,y,z, *position) <= r))

And due to python referencing, you can alter "points" in the returned dictionaries, and have the original points change as well (I think).

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