Can this matrix math be done faster?
我正在使用Python以透视的方式渲染3D点.速度很重要，因为它将在以后的某个时间点直接转换为帧速率.
我试图使用NumPy函数，但我无法清除两个令人讨厌的列表理解.我的程序90%的运行时是在列表理解期间，这是有意义的，因为它们包含了所有的数学知识，所以如果可能的话，我想找到一个更快的方法.
- 第一个列表理解发生在
pos的时候-它做了一个 和矩阵乘法vert_array - 第二个数
persp将x值和y值相乘 每一行都基于该特定行的z值.
可以用NumPy中的某些东西替换这些列表理解吗？我读到了有关numpy.einsum和numpy.from函数的内容，但我很难理解它们是否与我的问题相关.
Here is the function that does the main rendering calculations:
我想让pos和persp更快:
import time
from random import randint
import numpy as np
def render_all_verts(vert_array):
"""
:param vert_array: a 2-dimensional numpy array of float32 values and
size 3 x n, formatted as follows, where each row represents one
vertex's coordinates in world-space coordinates:
[[vert_x_1, vert_y_1, vert_z_1],
[vert_x_2, vert_y_2, vert_z_2],
...
[vert_x_n, vert_y, vert_z]]
:return: a 2-dimensional numpy array of the same data type, size
and format as vert_array, but in screen-space coordinates
"""
# Unit Vector is a 9 element, 2D array that represents the rotation matrix
# for the camera after some rotation (there's no rotation in this example)
unit_vec = np.array([[1, 0, 0],
[0, 1, 0],
[0, 0, 1]], dtype='float32')
# Raw Shift is a 3 element, 1D array that represents the position
# vector (x, y, z) of the camera in world-space coordinates
shift = np.array([0, 0, 10], dtype='float32')
# PURPOSE: This converts vert_array, with its coordinates relative
# to the world-space axes and origin, into coordinates relative
# to camera-space axes and origin (at the camera).
# MATH DESCRIPTION: For every row, raw_shift is added, then matrix
# multiplication is performed with that sum (1x3) and unit_array (3x3).
pos = np.array([np.matmul(unit_vec, row + shift) for row in vert_array], dtype='float32')
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
# This is a constant used to solve for the perspective
focus = 5
# PURPOSE: This calculation does the math to change the vertex coordinates,
# which are relative to the camera, into a representation of how they'll
# appear on screen in perspective. The x and y values are scaled based on
# the z value (distance from the camera)
# MATH DESCRIPTION: Each row's first two columns are multiplied
# by a scalar, which is derived from that row's third column value.
persp = np.array([np.multiply(row, np.array([focus / abs(row[2]), focus / abs(row[2]), 1])) for row in pos])
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
return persp
我写了这段代码到时间render_all_verts，并生成一个随机顶点坐标列表，以重复运行它.
# TESTING RENDERING SPEED
start_time = time.time()
# The next few lines make an array, similar to the 3D points I'd be rendering.
# It contains n vertices with random coordinate values from -m to m
n = 1000
m = 50
example_vertices = np.array([(randint(-m, m), randint(-m, m), randint(-m, m)) for i in range(n)])
# This empty array, which is the same shape as example_vertices. The results are saved here.
rendered_verts = np.empty(example_vertices.shape)
print('Example vertices:\n', example_vertices)
# This loop will render the example vertices many times
render_times = 2000
for i in range(render_times):
rendered_verts = render_all_verts(example_vertices)
print('\n\nLast calculated render of vertices:\n', rendered_verts)
print(f'\n\nThis program took an array of {n} vertices with randomized coordinate')
print(f'values between {-m} and {m} and rendered them {render_times} times.')
print(f'--- {time.time() - start_time} seconds ---')
最后，下面是终端输出的一个实例:
C:\...\simplified_demo.py
Example vertices:
[[-45 4 -43]
[ 42 27 28]
[-33 24 -18]
...
[ -5 48 5]
[-17 -17 29]
[ -5 -46 -24]]
C:\...\simplified_demo.py:45: RuntimeWarning: divide by zero encountered in divide
persp = np.array([np.multiply(row, np.array([focus / abs(row[2]), focus / abs(row[2]), 1]))
Last calculated render of vertices:
[[ -6.81818182 0.60606061 -33. ]
[ 5.52631579 3.55263158 38. ]
[-20.625 15. -8. ]
...
[ -1.66666667 16. 15. ]
[ -2.17948718 -2.17948718 39. ]
[ -1.78571429 -16.42857143 -14. ]]
This program took an array of 1000 vertices with randomized coordinate
values between -50 and 50 and rendered them 2000 times.
--- 15.910243272781372 seconds ---
Process finished with exit code 0
附注:NumPy目前似乎可以很好地处理被零除数和溢出值，所以我并不担心运行警告.我重新访问了我的文件路径...
附注:是的，我知道我可以只使用OpenGL或任何其他现有的渲染引擎，它们已经处理了所有这些数学运算，但我更感兴趣的是重新发明这个轮子.对我来说，这主要是一个学习Python和NumPy的实验.