Question

我正在处理大量的3D点，每个点都存储在numpy数组中的x，y，z值。对于背景，点将始终落在固定半径的圆柱内，并且高度=点的最大z值。我的目标是将边界圆柱（或更容易的列）分成例如。 1米高的地层，然后计算每个单元格内的点数在每个层上叠加规则网格（例如1米×1米）。

从概念上讲，操作与覆盖栅格和计算与每个像素相交的点相同。细胞网格可以形成正方形或圆盘，无所谓。

经过大量的搜索和阅读，我目前的想法是使用numpy.linspace和numpy.meshgrid的某种组合来生成存储在数组中的每个单元格的顶点，并针对每个点测试每个单元格以查看它是否是'在'。这似乎效率低下，尤其是在处理数千个点时。

numpy / scipy套件似乎很适合这个问题，但我还没有找到解决方案。任何建议将不胜感激。我已经包含了一些示例点和一些代码来可视化数据。

# Setup
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# Load in X,Y,Z values from a sub-sample of 10 points for testing
# XY Values are scaled to a reasonable point of origin
z_vals = np.array([3.08,4.46,0.27,2.40,0.48,0.21,0.31,3.28,4.09,1.75])
x_vals = np.array([22.88,20.00,20.36,24.11,40.48,29.08,36.02,29.14,32.20,18.96])
y_vals = np.array([31.31,25.04,31.86,41.81,38.23,31.57,42.65,18.09,35.78,31.78])

# This plot is instructive to visualize the problem
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x_vals, y_vals, z_vals, c='b', marker='o')
plt.show()

Answer 1

我不确定我完全理解你在寻找什么，但由于每个“细胞”似乎都有一个1米的方向，所以你不能：

可能使用floor函数将所有值舍入为整数（栅格化数据）;
使用以下内容从这些整数坐标创建一个双向投影更方便：

(64**2)*x + (64)*y + z # assuming all values are in [0,63]

如果您想在以后更轻松地关注身高，可以将z放在开头
计算每个“细胞”的直方图（numpy / scipy或numpy的几个函数可以做到）;
如果需要还原双射（即知道计数后知道每个单元格的“真实”坐标）

也许我不太了解，但万一它有帮助...

Answer 2

Thanks @Baruchel. It turns out the n-dimensional histograms suggested by @DilithiumMatrix provides a fairly simple solution to the problem I posted. After some reading, here is my current solution for anyone else that faces a similar problem. As this is my first Python/Numpy effort any improvements/suggestions, especially regarding performance, would be welcome. Thanks.

# Setup
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# Load in X,Y,Z values from a sub-sample of 10 points for testing
# XY Values are scaled to a reasonable point of origin
z_vals = np.array([3.08,4.46,0.27,2.40,0.48,0.21,0.31,3.28,4.09,1.75])
x_vals = np.array([22.88,20.00,20.36,24.11,40.48,29.08,36.02,29.14,32.20,18.96])
y_vals = np.array([31.31,25.04,31.86,41.81,38.23,31.57,42.65,18.09,35.78,31.78])

# Updated code below
# Variables needed for 2D,3D histograms
xmax, ymax, zmax = int(x_vals.max())+1, int(y_vals.max())+1, int(z_vals.max())+1
xmin, ymin, zmin = int(x_vals.min()), int(y_vals.min()), int(z_vals.min())
xrange, yrange, zrange = xmax-xmin, ymax-ymin, zmax-zmin
xedges = np.linspace(xmin, xmax, (xrange + 1), dtype=int)
yedges = np.linspace(ymin, ymax, (yrange + 1), dtype=int)
zedges = np.linspace(zmin, zmax, (zrange + 1), dtype=int)

# Make the 2D histogram
h2d, xedges, yedges = np.histogram2d(x_vals, y_vals, bins=(xedges, yedges))
assert np.count_nonzero(h2d) == len(x_vals), "Unclassified points in the array"
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
plt.imshow(h2d.transpose(), extent=extent,  interpolation='none', origin='low')
# Transpose and origin must be used to make the array line up when using imshow, unsure why
# Plot settings, not sure yet on matplotlib update/override objects
plt.grid(b=True, which='both')
plt.xticks(xedges)
plt.yticks(yedges)
plt.xlabel('X-Axis')
plt.ylabel('Y-Axis')
plt.plot(x_vals, y_vals, 'ro')
plt.show()

# 3-dimensional histogram with 1 x 1 x 1 m bins. Produces point counts in each 1m3 cell.
xyzstack = np.stack([x_vals,y_vals,z_vals], axis=1)
h3d, Hedges = np.histogramdd(xyzstack, bins=(xedges, yedges, zedges))
assert np.count_nonzero(h3d) == len(x_vals), "Unclassified points in the array"
h3d.shape  # Shape of the array should be same as the edge dimensions
testzbin = np.sum(np.logical_and(z_vals >= 1, z_vals < 2))  # Slice to test with
np.sum(h3d[:,:,1]) == testzbin  # Test num points in second bins
np.sum(h3d, axis=2)  # Sum of all vertical points above each x,y 'pixel'
# only in this example the h2d and np.sum(h3d,axis=2) arrays will match as no z bins have >1 points

# Remaining issue - how to get a r x c count of empty z bins.
# i.e. for each 'pixel'  how many z bins contained no points?
# Possible solution is to reshape to use logical operators
count2d = h3d.reshape(xrange * yrange, zrange)  # Maintain dimensions per num 3D cells defined
zerobins = (count2d == 0).sum(1)
zerobins.shape
# Get back to x,y grid with counts - ready for output as image with counts=pixel digital number
bincount_pixels = zerobins.reshape(xrange,yrange)
# Appears to work, perhaps there is a way without reshapeing?

PS if you are facing a similar problem scikit patch extraction looks like another possible solution.

使用numpy数组计算常规网格单元格内的点数

2 个答案: