计算每个场点在等高线内的频率

Question

我在处理二维地理数据。我有一长串的轮廓路径。现在我想确定在我的领域内的每一个点有多少个轮廓（也就是说，我想计算轮廓所代表的特征的空间频率分布）。在

为了说明我想做什么，这里有一个非常天真的实现：

import numpy as np
from shapely.geometry import Polygon, Point

def comp_frequency(paths,lonlat):
    """
    - paths: list of contour paths, made up of (lon,lat) tuples
    - lonlat: array containing the lon/lat coordinates; shape (nx,ny,2)
    """
    frequency = np.zeros(lonlat.shape[:2])
    contours = [Polygon(path) for path in paths]

    # Very naive and accordingly slow implementation
    for (i,j),v in np.ndenumerate(frequency):
        pt = Point(lonlat[i,j,:])
        for contour in contours:
            if contour.contains(pt):
                frequency[i,j] += 1

    return frequency

lon = np.array([
    [-1.10e+1,-7.82+0,-4.52+0,-1.18+0, 2.19e+0,5.59e+0,9.01+0,1.24+1,1.58+1,1.92+1,2.26+1],
    [-1.20e+1,-8.65+0,-5.21+0,-1.71+0, 1.81e+0,5.38e+0,8.97+0,1.25+1,1.61+1,1.96+1,2.32+1],
    [-1.30e+1,-9.53+0,-5.94+0,-2.29+0, 1.41e+0,5.15e+0,8.91+0,1.26+1,1.64+1,2.01+1,2.38+1],
    [-1.41e+1,-1.04+1,-6.74+0,-2.91+0, 9.76e-1,4.90e+0,8.86+0,1.28+1,1.67+1,2.06+1,2.45+1],
    [-1.53e+1,-1.15+1,-7.60+0,-3.58+0, 4.98e-1,4.63e+0,8.80+0,1.29+1,1.71+1,2.12+1,2.53+1],
    [-1.66e+1,-1.26+1,-8.55+0,-4.33+0,-3.00e-2,4.33e+0,8.73+0,1.31+1,1.75+1,2.18+1,2.61+1],
    [-1.81e+1,-1.39+1,-9.60+0,-5.16+0,-6.20e-1,3.99e+0,8.66+0,1.33+1,1.79+1,2.25+1,2.70+1],
    [-1.97e+1,-1.53+1,-1.07+1,-6.10+0,-1.28e+0,3.61e+0,8.57+0,1.35+1,1.84+1,2.33+1,2.81+1],
    [-2.14e+1,-1.69+1,-1.21+1,-7.16+0,-2.05e+0,3.17e+0,8.47+0,1.37+1,1.90+1,2.42+1,2.93+1],
    [-2.35e+1,-1.87+1,-1.36+1,-8.40+0,-2.94e+0,2.66e+0,8.36+0,1.40+1,1.97+1,2.52+1,3.06+1],
    [-2.58e+1,-2.08+1,-1.54+1,-9.86+0,-3.99e+0,2.05e+0,8.22+0,1.44+1,2.05+1,2.65+1,3.22+1]])

lat = np.array([
    [ 29.6, 30.3, 30.9, 31.4, 31.7, 32.0, 32.1, 32.1, 31.9, 31.6, 31.2],
    [ 32.4, 33.2, 33.8, 34.4, 34.7, 35.0, 35.1, 35.1, 34.9, 34.6, 34.2],
    [ 35.3, 36.1, 36.8, 37.3, 37.7, 38.0, 38.1, 38.1, 37.9, 37.6, 37.1],
    [ 38.2, 39.0, 39.7, 40.3, 40.7, 41.0, 41.1, 41.1, 40.9, 40.5, 40.1],
    [ 41.0, 41.9, 42.6, 43.2, 43.7, 44.0, 44.1, 44.0, 43.9, 43.5, 43.0],
    [ 43.9, 44.8, 45.6, 46.2, 46.7, 47.0, 47.1, 47.0, 46.8, 46.5, 45.9],
    [ 46.7, 47.7, 48.5, 49.1, 49.6, 49.9, 50.1, 50.0, 49.8, 49.4, 48.9],
    [ 49.5, 50.5, 51.4, 52.1, 52.6, 52.9, 53.1, 53.0, 52.8, 52.4, 51.8],
    [ 52.3, 53.4, 54.3, 55.0, 55.6, 55.9, 56.1, 56.0, 55.8, 55.3, 54.7],
    [ 55.0, 56.2, 57.1, 57.9, 58.5, 58.9, 59.1, 59.0, 58.8, 58.3, 57.6],
    [ 57.7, 59.0, 60.0, 60.8, 61.5, 61.9, 62.1, 62.0, 61.7, 61.2, 60.5]])

lonlat = np.dstack((lon,lat))

paths = [
    [(-1.71,34.4),(1.81,34.7),(5.15,38.0),(4.9,41.0),(4.63,44.0),(-0.03,46.7),(-4.33,46.2),(-9.6,48.5),(-8.55,45.6),(-3.58,43.2),(-2.91,40.3),(-2.29,37.3),(-1.71,34.4)],
    [(0.976,40.7),(-4.33,46.2),(-0.62,49.6),(3.99,49.9),(4.33,47.0),(4.63,44.0),(0.976,40.7)],
    [(2.9,55.8),(2.37,56.0),(8.47,56.1),(3.17,55.9),(-2.05,55.6),(-1.28,52.6),(-0.62,49.6),(4.33,47.0),(8.8,44.1),(2.29,44.0),(2.71,43.9),(3.18,46.5),(3.25,49.4),(3.33,52.4),(2.9,55.8)],
    [(2.25,35.1),(2.26,38.1),(8.86,41.1),(5.15,38.0),(5.38,35.0),(9.01,32.1),(2.25,35.1)]]

frequency = comp_frequency(paths,lonlat)

当然，这是尽可能低效率地编写的，使用所有显式循环，因此需要花费很长时间。在

我如何才能有效地做到这一点？

编辑：根据要求添加了一些样本数据。请注意，我的实际域比原来的域大150**2（就分辨率而言），因为我通过切片原始数组创建了样本坐标：lon[::150]。在

Answer 1

因此，同时我找到了一个好的解决方案，这要感谢一位同事，他在某个时候实现了类似的东西（基于this blog post）。在

使用shapely的旧的、非常慢的方法

首先，这里是我使用shapely的参考实现，它只是我在开头文章中第一个“天真”方法的一个稍微改进的版本。它很容易理解并且有效，但是非常慢。在

import numpy as np
from shapely.geometry import Polygon, Point

def spatial_contour_frequency_shapely(paths,lon,lat):

    frequency = np.zeros(lon.shape)
    contours = [Polygon(path) for path in paths]

    for (i,j),v in np.ndenumerate(frequency):
        pt = Point([lon[i,j],lat[i,j]])
        for contour in contours:
            if contour.contains(pt):
                frequency[i,j] += 1

    return frequency

新的，非常快速的解决方案使用PIL

我的（几乎）最终解决方案不再使用shapely，而是使用PIL（Python图像库）中的图像处理技术。这个解决方案要快得多，多得多，虽然这种形式只适用于规则的矩形网格（见下面的注释）。在

^{pr2}$

值得注意的是，这两种方法的结果并不完全相同。用PIL方法识别的特征比用shapely方法识别的特征稍大一些，但实际上其中一个并不比另一个好。

时间安排

以下是一些使用简化数据集创建的计时（尽管不是来自开场白的半人工示例数据）：

spatial_contour_frequency/shapely             :   191.8843
spatial_contour_frequency/pil                 :     0.3287
spatial_contour_frequency/pil-tree-inside     :     2.3629
spatial_contour_frequency/pil-regular_grid    :     0.3276

最耗时的步骤是在轮廓点的不规则lon/lat网格上寻找索引。其中最消耗时间的部分是构造cKDTree，这就是为什么我把它移出get_indices。为了将其放在透视图中，pil-tree-inside是在get_indices中创建树的版本。pil-regular-grid与regular_grid=True一起使用，对于我的数据集，这会产生错误的结果，但是给出了一个ide，说明了它在常规网格上的运行速度。在

总的来说，现在我已经设法消除了非规则网格的影响（pilvs.pil-regular-grid），这是我在开始时所希望的！：）

Answer 2

def comp_frequency_lc(paths,lonlat):

    import operator
    frequency = np.zeros(lonlat.shape[:2])
    contours = [Polygon(path) for path in paths]

    for (i,j),v in np.ndenumerate(frequency):
        pt = Point(lonlat[i,j,:])
        [
            operator.setitem(frequency,(i,j),
                    operator.getitem(frequency,(i,j))+1)
            for contour in contours if contour.contains(pt)
         ]

    return frequency

    print(comp_frequency(paths,lonlat))

**result in**:

[[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  1.  0.  0.  1.  2.  2.  2.]
 [ 0.  1.  0.  0.  1.  0.  0.  1.  1.  1.  1.]
 [ 0.  2.  0.  0.  2.  0.  0.  2.  2.  2.  1.]
 [ 0.  2.  0.  0.  1.  0.  0.  1.  1.  1.  2.]
 [ 0.  1.  0.  0.  0.  0.  0.  1.  2.  1.  1.]
 [ 0.  1.  1.  0.  0.  0.  0.  1.  1.  0.  0.]
 [ 0.  1.  1.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]]

以下是计时的结果。。。。其中comp_frequency_lc是指 使用列表理解

^{pr2}$

Answer 3

如果你的输入多边形实际上是轮廓，那么你最好直接使用你的输入网格，而不是计算轮廓并测试其中是否有一个点。在

等高线遵循网格数据的常量值。每个轮廓都是一个多边形，包含输入栅格中大于该值的区域。在

如果您需要知道一个给定的点在里面有多少个轮廓，那么在该点的位置对输入网格进行采样并操作返回的“z”值会更快。如果知道创建轮廓的值，可以直接从中提取它内部的轮廓数。在

例如：

import numpy as np
from scipy.interpolate import RegularGridInterpolator
import matplotlib.pyplot as plt

# One of your input gridded datasets
y, x = np.mgrid[-5:5:100j, -5:5:100j]
z = np.sin(np.hypot(x, y)) + np.hypot(x, y) / 10

contour_values = [-1, -0.5, 0, 0.5, 1, 1.5, 2]

# A point location...
x0, y0 = np.random.normal(0, 2, 2)

# Visualize what's happening...
fig, ax = plt.subplots()
cont = ax.contourf(x, y, z, contour_values, cmap='gist_earth')
ax.plot([x0], [y0], marker='o', ls='none', color='salmon', ms=12)
fig.colorbar(cont)

# Instead of working with whether or not the point intersects the
# contour polygons we generated, we'll turn the problem on its head:

# Sample the grid at the point location
interp = RegularGridInterpolator((x[0,:], y[:,0]), z)
z0 = interp([x0, y0])

# How many contours would the point be inside?
num_inside = sum(z0 > c for c in contour_values)[0]

ax.set(title='Point is inside {} contours'.format(num_inside))
plt.show()