我想要生成一个坐标已经旋转的网格。我必须在双循环中进行旋转,我确信有更好的方法来矢量化它。代码如下:
# Define the range for x and y in the unrotated matrix
xspan = linspace(-2*pi, 2*pi, 101)
yspan = linspace(-2*pi, 2*pi, 101)
# Generate a meshgrid and rotate it by RotRad radians.
def DoRotation(xspan, yspan, RotRad=0):
# Clockwise, 2D rotation matrix
RotMatrix = np.array([ [np.cos(RotRad), np.sin(RotRad)],
[-np.sin(RotRad), np.cos(RotRad)]])
print RotMatrix
# This makes two 2D arrays which are the x and y coordinates for each point.
x, y = meshgrid(xspan,yspan)
# After rotating, I'll have another two 2D arrays with the same shapes.
xrot = zeros(x.shape)
yrot = zeros(y.shape)
# Dot the rotation matrix against each coordinate from the meshgrids.
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
# I BELIEVE THERE IS A BETTER WAY THAN THIS DOUBLE LOOP!!!
for i in range(len(xspan)):
for j in range(len(yspan)):
xrot[i,j], yrot[i,j] = dot(RotMatrix, array([x[i,j], y[i,j]]))
# Now the matrix is rotated
return xrot, yrot
# Pick some arbitrary function and plot it (no rotation)
x, y = DoRotation(xspan, yspan, 0)
z = sin(x)+cos(y)
imshow(z)
# And now with 0.3 radian rotation so you can see that it works.
x, y = DoRotation(xspan, yspan, 0.3)
z = sin(x)+cos(y)
figure()
imshow(z)
在两个网格物体上写一个双循环似乎很愚蠢。其中一个向导是否知道如何对其进行矢量化?
答案 0 :(得分:4)
也许我误解了这个问题,但我通常只是......
import numpy as np
pi = np.pi
x = np.linspace(-2.*pi, 2.*pi, 1001)
y = x.copy()
X, Y = np.meshgrid(x, y)
Xr = np.cos(rot)*X + np.sin(rot)*Y # "cloclwise"
Yr = -np.sin(rot)*X + np.cos(rot)*Y
z = np.sin(Xr) + np.cos(Yr)
~100ms也
答案 1 :(得分:2)
np.einsum
)非常快。 1001x1001我得到97毫秒。
def DoRotation(xspan, yspan, RotRad=0):
"""Generate a meshgrid and rotate it by RotRad radians."""
# Clockwise, 2D rotation matrix
RotMatrix = np.array([[np.cos(RotRad), np.sin(RotRad)],
[-np.sin(RotRad), np.cos(RotRad)]])
x, y = np.meshgrid(xspan, yspan)
return np.einsum('ji, mni -> jmn', RotMatrix, np.dstack([x, y]))
答案 2 :(得分:1)
你可以摆脱那些带有reshaping
&的两个嵌套循环。 flattening with np.ravel
并保持矩阵乘以np.dot
,如此 -
mult = np.dot( RotMatrix, np.array([x.ravel(),y.ravel()]) )
xrot = mult[0,:].reshape(xrot.shape)
yrot = mult[1,:].reshape(yrot.shape)