两点层之间的距离矩阵 [英] Distance matrix between two point layers

问题描述

我有两个数组，其中包含点坐标为shapely.geometry.Point，具有不同的大小.

I have two arrays containing point coordinates as shapely.geometry.Point with different sizes.

例如:

[Point(X Y), Point(X Y)...]
[Point(X Y), Point(X Y)...]

我想用距离函数创建这两个数组的叉积".距离函数来自shape.geometry，这是一种简单的几何矢量距离计算.我正在尝试在M:N点之间创建距离矩阵:

I would like to create a "cross product" of these two arrays with a distance function. Distance function is from shapely.geometry, which is a simple geometry vector distance calculation. I am tryibg to create distance matrix between M:N points:

现在我具有此功能:

    source = gpd.read_file(source)
    near = gpd.read_file(near)

    source_list = source.geometry.values.tolist()
    near_list = near.geometry.values.tolist()

    array = np.empty((len(source.ID_SOURCE), len(near.ID_NEAR)))

    for index_source, item_source in enumerate(source_list):
        for index_near, item_near in enumerate(near_list):
            array[index_source, index_near] = item_source.distance(item_near)

    df_matrix = pd.DataFrame(array, index=source.ID_SOURCE, columns = near.ID_NEAR)

这可以很好地完成工作，但是速度很慢.4000 x 4000点大约是100秒(我的数据集要大得多，所以速度是主要问题).如果可能的话，我想避免这种双重循环.我试图在Pandas数据帧中这样做(速度太快了):

Which does the job fine, but is slow. 4000 x 4000 points is around 100 seconds (I have datasets which are way bigger, so speed is main issue). I would like to avoid this double loop if possible. I tried to do in in pandas dataframe as in (which has terrible speed):

for index_source, item_source in source.iterrows():
         for index_near, item_near in near.iterrows():
             df_matrix.at[index_source, index_near] = item_source.geometry.distance(item_near.geometry)

速度更快(但仍然比numpy慢4倍):

A bit faster is (but still 4x slower than numpy):

    for index_source, item_source in enumerate(source_list):
        for index_near, item_near in enumerate(near_list):
             df_matrix.at[index_source, index_near] = item_source.distance(item_near)

有更快的方法吗?我想有，但我不知道如何进行.我也许可以将数据帧分成较小的块，然后将块发送到不同的内核，然后合并结果-这是不得已的方法.如果我们能以某种方式仅将numpy与仅索引魔术结合使用，我可以将其发送到GPU并立即完成.但是double for循环现在是不可以的.我也想不使用除Pandas/Numpy之外的任何其他库.我可以使用SAGA处理及其点距离"模块( http://www.saga-gis.org/saga_tool_doc/2.2.2/shapes_points_3.html )，该死的速度非常快，但我正在寻找仅适用于Python的解决方案.

Is there a faster way to do this? I guess there is, but I have no idea how to proceed. I might be able to chunk the dataframe into smaller pieces and send the chunk onto different core and concat the results - this is the last resort. If somehow we can use numpy only with some indexing only magic, I can send it to GPU and be done with it in no time. But the double for loop is a no no right now. Also I would like to not use any other library than Pandas/Numpy. I can use SAGA processing and its Point distances module (http://www.saga-gis.org/saga_tool_doc/2.2.2/shapes_points_3.html), which is pretty damn fast, but I am looking for Python only solution.

推荐答案

如果可以在单独的向量中获取坐标，则可以尝试以下操作:

If you can get the coordinates in separate vectors, I would try this:

import numpy as np

x = np.asarray([5.6, 2.1, 6.9, 3.1]) # Replace with data
y = np.asarray([7.2, 8.3, 0.5, 4.5]) # Replace with data

x_i = x[:, np.newaxis]
x_j = x[np.newaxis, :]

y_i = y[:, np.newaxis]
y_j = y[np.newaxis, :]

d = (x_i-x_j)**2+(y_i-y_j)**2

np.sqrt(d, out=d)

这篇关于两点层之间的距离矩阵的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！