python - 将 numpy 点数组分配给二维方形网格

Question

我要超越 my上一个问题因为速度问题。我有一个点的纬度/经度坐标数组，我想将它们分配给一个索引代码，该索引代码从具有相同大小单元格的二维正方形网格派生。这是一个例子。让我们将第一个包含六个点的坐标(称为 [x y] 对)的数组称为 points:

points = [[ 1.5  1.5]
 [ 1.1  1.1]
 [ 2.2  2.2]
 [ 1.3  1.3]
 [ 3.4  1.4]
 [ 2.   1.5]]

然后我有另一个数组，其中包含 [minx,miny,maxx,maxy] 形式的两个单元格的网格的顶点坐标；我们称它为 bounds:

bounds = [[ 0.  0.  2.  2.]
 [ 2.  2.  3.  3.]]

我想找到哪些点在哪个边界，然后分配一个从 bounds 数组索引派生的代码(在这种情况下，第一个单元格的代码为 0，第二个单元格的代码为 1，依此类推...)。由于单元格是正方形，计算每个点是否在每个单元格中的最简单方法是评估:

x > minx & x < maxx & y > miny & y < maxy

这样生成的数组将显示为:

results = [0 0 1 0 NaN NaN]

其中 NaN 表示该点在单元格外。在我的真实案例中，元素的数量是在 10^4 个单元格中找到 10^6 个点的顺序。有没有一种方法可以使用 numpy 数组快速完成此类操作？

编辑:澄清一下，results 数组预期意味着第一个点在第一个单元格内(bounds 数组的 0 索引)，所以第二个和第一个位于 bounds 数组的第二个单元格内，依此类推...

Answer 1

这是解决您的问题的矢量化方法。它应该会显着加快速度。

import numpy as np
def findCells(points, bounds):
    # make sure points is n by 2 (pool.map might send us 1D arrays)
    points = points.reshape((-1,2))

    # check for each point if all coordinates are in bounds
    # dimension 0 is bound
    # dimension 1 is is point
    allInBounds = (points[:,0] > bounds[:,None,0])
    allInBounds &= (points[:,1] > bounds[:,None,1])
    allInBounds &= (points[:,0] < bounds[:,None,2])
    allInBounds &= (points[:,1] < bounds[:,None,3])


    # now find out the positions of all nonzero (i.e. true) values
    # nz[0] contains the indices along dim 0 (bound)
    # nz[1] contains the indices along dim 1 (point)
    nz = np.nonzero(allInBounds)

    # initialize the result with all nan
    r = np.full(points.shape[0], np.nan)
    # now use nz[1] to index point position and nz[0] to tell which cell the
    # point belongs to
    r[nz[1]] = nz[0]
    return r

def findCellsParallel(points, bounds, chunksize=100):
    import multiprocessing as mp
    from functools import partial

    func = partial(findCells, bounds=bounds)

    # using python3 you could also do 'with mp.Pool() as p:'  
    p = mp.Pool()
    try:
        return np.hstack(p.map(func, points, chunksize))
    finally:
        p.close()

def main():
    nPoints = 1e6
    nBounds = 1e4

    # points = np.array([[ 1.5, 1.5],
    #                    [ 1.1, 1.1],
    #                    [ 2.2, 2.2],
    #                    [ 1.3, 1.3],
    #                    [ 3.4, 1.4],
    #                    [ 2. , 1.5]])

    points = np.random.random([nPoints, 2])

    # bounds = np.array([[0,0,2,2],
    #                    [2,2,3,3]])

    # bounds = np.array([[0,0,1.4,1.4],
    #                    [1.4,1.4,2,2],
    #                    [2,2,3,3]])

    bounds = np.sort(np.random.random([nBounds, 2, 2]), 1).reshape(nBounds, 4)

    r = findCellsParallel(points, bounds)
    print(points[:10])
    for bIdx in np.unique(r[:10]):
        if np.isnan(bIdx):
            continue
        print("{}: {}".format(bIdx, bounds[bIdx]))
    print(r[:10])

if __name__ == "__main__":
    main()

编辑:
用你的数据量尝试它给了我一个 MemoryError。如果您使用 multiprocessing.Pool 及其 map 函数，您可以避免这种情况，甚至可以加快速度，请参阅更新的代码。

结果:

>time python test.py
[[ 0.69083585  0.19840985]
 [ 0.31732711  0.80462512]
 [ 0.30542996  0.08569184]
 [ 0.72582609  0.46687164]
 [ 0.50534322  0.35530554]
 [ 0.93581095  0.36375539]
 [ 0.66226118  0.62573407]
 [ 0.08941219  0.05944215]
 [ 0.43015872  0.95306899]
 [ 0.43171644  0.74393729]]
9935.0: [ 0.31584562  0.18404152  0.98215445  0.83625487]
9963.0: [ 0.00526106  0.017255    0.33177741  0.9894455 ]
9989.0: [ 0.17328876  0.08181912  0.33170444  0.23493507]
9992.0: [ 0.34548987  0.15906761  0.92277442  0.9972481 ]
9993.0: [ 0.12448765  0.5404578   0.33981119  0.906822  ]
9996.0: [ 0.41198261  0.50958195  0.62843379  0.82677092]
9999.0: [ 0.437169    0.17833114  0.91096133  0.70713434]
[ 9999.  9993.  9989.  9999.  9999.  9935.  9999.  9963.  9992.  9996.]

real 0m 24.352s
user 3m  4.919s
sys  0m  1.464s