计算每个场点在轮廓内的频率

Question

我正在处理2D地理数据。我有一长串轮廓路径。现在我想为我的域中的每个点确定它所在的轮廓（即我想计算由轮廓表示的特征的空间频率分布）。

为了说明什么，我想做的事，这里的第一个很天真的实现：

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，这大约是低效写越好，所有的显式循环，并相应地需要永远。

我该如何有效地做到这一点？

编辑：根据要求添加了一些示例数据。请注意，由于我通过对原始数组进行切片来创建样本坐标，所以我的实际域大小为150 ** 2（分辨率方面）：lon[::150]。

Answer 1

所以同时我已经找到了一个很好的解决方案感谢谁的执行在一些点（基于this blog post）类似的一个同事。

老，身材匀称的使用

首先非常缓慢的办法，这里是用我的身材匀称的参考实现，这仅仅是在开封后我的第一个“天真”的做法有点改良版本。这很容易理解和运作，但速度很慢。

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，使用PIL

我（almost-）最终解决方案不使用身材匀称了，而是使用图像处理技术，从PIL（Python图像库）非常快速的解决方案。这种解决方案要快得多，尽管这种形式只适用于常规的矩形网格（见下面的注释）。

import numpy as np 
from PIL import Image, ImageDraw 

def _spatial_contour_frequency_pil(paths,lon,lat,regular_grid=False, 
     method_ind=None): 

    def get_indices(points,lon,lat,tree=None,regular=False): 

     def get_indices_regular(points,lon,lat): 
      lon,lat = lon.T,lat.T 
      def _get_ij(lon,lat,x,y): 
       lon0 = lon[0,0] 
       lat0 = lat[0,0] 
       lon1 = lon[-1,-1] 
       lat1 = lat[-1,-1] 
       nx,ny = lon.shape 
       dx = (lon1-lon0)/nx 
       dy = (lat1-lat0)/ny 
       i = int((x-lon0)/dx) 
       j = int((y-lat0)/dy) 
       return i,j 
      return [_get_ij(lon,lat,x,y) for x,y in points] 

     def get_indices_irregular(points,tree,shape): 

      dist,idx = tree.query(points,k=1) 
      ind = np.column_stack(np.unravel_index(idx,lon.shape)) 
      return [(i,j) for i,j in ind] 

     if regular: 
      return get_indices_regular(points,lon,lat) 
     return get_indices_irregular(points,tree,lon.T.shape) 

    tree = None 
    if not regular_grid: 
     lonlat = np.column_stack((lon.T.ravel(),lat.T.ravel())) 
     tree = sp.spatial.cKDTree(lonlat) 

    frequency = np.zeros(lon.shape) 
    for path in paths: 
     path_ij = get_indices(path,lon,lat,tree=tree,regular=regular_grid) 
     raster_poly = Image.new("L",lon.shape,0) 
     rasterize = ImageDraw.Draw(raster_poly) 
     rasterize.polygon(path_ij,1) 
     mask = sp.fromstring(raster_poly.tobytes(),'b') 
     mask.shape = raster_poly.im.size[1],raster_poly.im.size[0] 
     frequency += mask 

    return frequency 

应当指出的是，这两种方法的结果是不exaclty相同。使用PIL方法确定的特征略大于用匀称方法确定的特征，但实际上一个并不比另一个更好。

计时

下面是具有减小的数据集创建一些定时（不从所述开口后的半人工数据。例如，虽然）：

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长/纬网格上查找索引。其中最耗时的部分是cKDTree的构建，这就是为什么我将它从get_indices中移出的原因。为了对此进行透视，pil-tree-inside是在get_indices内创建树的版本。 pil-regular-grid与regular_grid=True一致，对于我的数据集，这会产生错误的结果，但会给出在规则网格上运行速度有多快的问题。

总体而言，现在我已经设法消除非规则网格（pil与pil-regular-grid）的影响，这是我一开始就希望的！ :)

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.]] 

，这里是来自时序....那里comp_frequency_lc是结果一个 使用列表理解

# 10000 runs 
# comp_frequency 185.10419257196094 
# comp_frequency_lc 124.42639064462102 

Answer 3

如果输入的多边形实际上是轮廓，那么你最好直接与工作你的输入网格比计算轮廓和测试点是否在他们的内部。

轮廓遵循网格数据的恒定值。每个轮廓都是一个多边形，封闭了大于该值的输入网格区域。

如果您需要知道给定点的内部有多少个等值线，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。如果您知道自己创建的轮廓线的值，则可以直接从中提取其内部的轮廓线数量。

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()