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I am trying to find a slope of a line in 3D space. The solution for plotting such line is given in this post
Here's the given code from the link above:
import numpy as np
pts = np.add.accumulate(np.random.random((10,3)))
x,y,z = pts.T
# this will find the slope and x-intercept of a plane
# parallel to the y-axis that best fits the data
A_xz = np.vstack((x, np.ones(len(x)))).T
m_xz, c_xz = np.linalg.lstsq(A_xz, z)[0]
# again for a plane parallel to the x-axis
A_yz = np.vstack((y, np.ones(len(y)))).T
m_yz, c_yz = np.linalg.lstsq(A_yz, z)[0]
# the intersection of those two planes and
# the function for the line would be:
# z = m_yz * y + c_yz
# z = m_xz * x + c_xz
# or:
def lin(z):
x = (z - c_xz)/m_xz
y = (z - c_yz)/m_yz
return x,y
#verifying:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = Axes3D(fig)
zz = np.linspace(0,5)
xx,yy = lin(zz)
ax.scatter(x, y, z)
ax.plot(xx,yy,zz)
plt.savefig('test.png')
plt.show()
Math-wise I know how to find intersections of two planes and the slope of a given line but I am having trouble putting it in code.
How can I find the slope of resulting regression line using this solution?
The "slope" of a 3D line is generally taken to be slopes of the line "projected" onto the x, y and z planes. See the second answer to this question
If this is what you intended then it's easy enough to calculate these; this modified version of your code below does this into the sx, sy and sz variables:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from math import pow, sqrt
pts = np.add.accumulate(np.random.random((10,3)))
x, y, z = pts.T
# plane parallel to the y-axis
A_xz = np.vstack((x, np.ones(len(x)))).T
m_xz, c_xz = np.linalg.lstsq(A_xz, z, rcond=None)[0]
# plane parallel to the x-axis
A_yz = np.vstack((y, np.ones(len(y)))).T
m_yz, c_yz = np.linalg.lstsq(A_yz, z, rcond=None)[0]
# the intersection of those two planes and
# the function for the line would be:
# z = m_yz * y + c_yz
# z = m_xz * x + c_xz
# or:
def lin(z):
x = (z - c_xz)/m_xz
y = (z - c_yz)/m_yz
return x,y
# get 2 points on the intersection line
za = z[0]
zb = z[len(z) - 1]
xa, ya = lin(za)
xb, yb = lin(zb)
# get distance between points
len = sqrt(pow(xb - xa, 2) + pow(yb - ya, 2) + pow(zb - za, 2))
# get slopes (projections onto x, y and z planes)
sx = (xb - xa) / len # x slope
sy = (yb - ya) / len # y slope
sz = (zb - za) / len # z slope
# integrity check - the sum of squares of slopes should equal 1.0
# print (pow(sx, 2) + pow(sy, 2) + pow(sz, 2))
fig = plt.figure()
ax = Axes3D(fig)
ax.set_xlabel("x, slope: %.4f" %sx, color='blue')
ax.set_ylabel("y, slope: %.4f" %sy, color='blue')
ax.set_zlabel("z, slope: %.4f" %sz, color='blue')
ax.scatter(x, y, z)
ax.plot([xa], [ya], [za], markerfacecolor='k', markeredgecolor='k', marker = 'o')
ax.plot([xb], [yb], [zb], markerfacecolor='k', markeredgecolor='k', marker = 'o')
ax.plot([xa, xb], [ya, yb], [za, zb], color = 'r')
plt.show()
The output graph below shows the line in question, which is just drawn between the 2 extreme xyz points.
I hope this may help
I am trying to plot a bivariate gaussian distribution using matplotlib. I want to do this using the xy coordinates of two scatter points (Group A), (Group B).
I want to adjust the distribution by adjusting the COV matrix to account for each Groups velocity and their distance to an additional xy coordinate used as a reference point.
I've calculated the distance of each groups xy coordinate to that of the reference point. The distance is expressed as a radius, labelled [GrA_Rad],[GrB_Rad].
So the further they are away from the reference point the greater the radius. I've also calculated velocity labelled [GrA_Vel],[GrB_Vel]. The direction of each group is expressed as the orientation. This is labelled [GrA_Rotation],[GrB_Rotation]
Question on how I want the distribution to be adjusted for velocity and distance (radius):
I'm hoping to use SVD. Specifically, if I have the rotation angle of each scatter, this provides the direction. The velocity can be used to describe a scaling matrix [GrA_Scaling],[GrB_Scaling]. So this scaling matrix can be used to expand the radius in the x-direction and contract the radius in the y-direction. This expresses the COV matrix.
Finally, the distribution mean value is found by translating the groups location (x,y) by half the velocity.
Put simply: the radius is applied to each group's scatter point. The COV matrix is adjusted by the radius and velocity. So using the scaling matrix to expand the radius in x-direction and contract in y-direction. The direction is measured from the rotation angle. Then determine the distribution mean value by translating the groups location (x,y) by half the velocity.
Below is the df of these variables
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.animation as animation
d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})
df = pd.DataFrame(data = d)
I've made an animated plot of each xy coordinate.
GrA_X = [10,12,17,16,16,14,12,8]
GrA_Y = [10,12,13,7,6,7,8,8]
GrB_X = [5,8,13,16,19,15,13,5]
GrB_Y = [6,15,12,10,8,9,10,8]
Item_X = [6,8,14,18,13,11,16,15]
Item_Y = [10,12,8,12,15,12,10,8]
scatter_GrA = ax.scatter(GrA_X, GrA_Y)
scatter_GrB = ax.scatter(GrB_X, GrB_Y)
scatter_Item = ax.scatter(Item_X, Item_Y)
def animate(i) :
scatter_GrA.set_offsets([[GrA_X[0+i], GrA_Y[0+i]]])
scatter_GrB.set_offsets([[GrB_X[0+i], GrB_Y[0+i]]])
scatter_Item.set_offsets([[Item_X[0+i], Item_Y[0+i]]])
ani = animation.FuncAnimation(fig, animate, np.arange(0,9),
interval = 1000, blit = False)
Update
The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:
Aside from the styling, I think this matches what the OP described.
Here's the code that was used to produce the above plot:
dfake = ({
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})
dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()
and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts
def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])
r = rot(theta)
return r # cov # r.T
def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.
x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees
returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))
# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)
# displace xy by half the velocity
x,y = rot(theta) # (velocity/2, 0) + (x, y)
# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)
# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)
return X, Y, PDF
def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)
if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()
# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)
if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)
# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) # (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')
# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)
# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()
# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)
# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])
# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')
# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')
return fig, ax
def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)
ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)
return ret
def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)
# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)
Old answer -2
I've adjusted my answer to match the example the OP posted:
Here's the code that produced the above image:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts
def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])
r = rot(theta)
return r # cov # r.T
def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad
d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})
df = pd.DataFrame(data=d)
limitpad = .5
clevels = 5
cflevels = 50
xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)
X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))
fig = plt.figure(figsize=(10,6))
ax = plt.gca()
Zs = []
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)
Zrows = []
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]
cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)
Zs.append(np.sum(Zrows, axis=0))
# plot the reference points
# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]
# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()
# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)
# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])
ax.set_aspect('equal', 'box')
Old answer -1
It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts
def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])
r = rot(theta)
return r # cov # r.T
d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})
df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())
X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)
fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)
for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]
cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
ax.contour(X, Y, Z)
ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')
This outputs:
I am trying to draw a spherical harmonics for my college project. The following formula I want to depict,
Y = cos(theta)
for that, I wrote this code
import numpy as np
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
def sph2cart(r, phi, tta):
''' r is from 0 to infinity '''
''' phi is from 0 to 2*pi '''
''' tta is from 0 to pi '''
x = r* np.sin(tta)* np.cos(phi)
y = r* np.sin(tta)* np.sin(phi)
z = r* np.cos(tta)
return x, y, z
# phi running from 0 to pi and tta from 0 to pi
phi = np.linspace(0, 2* np.pi, 25)
tta = np.linspace(0, np.pi, 25)
# meshgrid to generate points
phi, tta = np.meshgrid(phi, tta)
# THIS IS THE FUNCTION
Y = np.cos(tta)
# finally all things in cartesian co-ordinate system
# Note that "Y" is acting as "r"
x, y, z = sph2cart( Y, phi, tta)
# plotting :-
fig = plt.figure()
ax = fig.add_subplot( 111 , projection='3d')
ax.plot_surface(x, y, z, linewidth = 0.5, edgecolors = 'k')
And, get the sphere as a result. Which is not correct, because actual result is dumbbell like shape. See the second row of this image,
https://upload.wikimedia.org/wikipedia/commons/thumb/6/62/Spherical_Harmonics.png/1024px-Spherical_Harmonics.png
The picture in the Wikipedia article Spherical harmonics is obtained by using the absolute value of a spherical harmonic as the r coordinate, and then coloring the surface according to the sign of the harmonic. Here is an approximation.
x, y, z = sph2cart(np.abs(Y), phi, tta)
fig = plt.figure()
ax = fig.add_subplot( 111 , projection='3d')
from matplotlib import cm
ax.set_aspect('equal')
ax.plot_surface(x, y, z, linewidth = 0.5, facecolors = cm.jet(Y), edgecolors = 'k')
When you use Y itself as r, the two hemispheres (positive Y and negative Y) end up mapped onto the same half of the above surface.
The Y you are passing to the function needs to be an absolute value to make it r, else z = cos(theta)^2 is always positive. If r is to be the radius then this what you should be doing.
x, y, z = sph2cart(np.abs(Y), phi, tta)
Win 7, x64, Python 2.7
I'm trying to rotate a square that is initially in the xz plane so that its normal aligns with a given 3D vector. Also I am translating the square to the start of the vector but that isnt a problem.
The path I have taken is as follows,
1) Find the axis of rotation via the cross product of the given vector & the square's normal, a unit vector in the y direction in this case.
2) Find the angle of rotation via the dot product of the given vector and the square's normal.
3) Build appropriate rotation matrix.
4) Apply rotation matrix to the vertices of the square.
5) Translate to the start of the given vector.
The code..
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import math
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
na = np.array
def rotation_matrix(axis, theta):
"""
Return the rotation matrix associated with counterclockwise rotation about
the given axis by theta radians.
"""
axis = np.asarray(axis)
axis = axis/math.sqrt(np.dot(axis, axis))
a = math.cos(theta/2.0)
b, c, d = -axis*math.sin(theta/2.0)
aa, bb, cc, dd = a*a, b*b, c*c, d*d
bc, ad, ac, ab, bd, cd = b*c, a*d, a*c, a*b, b*d, c*d
return np.array([[aa+bb-cc-dd, 2*(bc+ad), 2*(bd-ac)],
[2*(bc-ad), aa+cc-bb-dd, 2*(cd+ab)],
[2*(bd+ac), 2*(cd-ab), aa+dd-bb-cc]])
edgeLen = 4.0 # length of square side
pos = na([2.0,2.0,2.0]) # starting point of vector
dirc = na([6.0,6.0,6.0]) # direction of vector
Ux = na([1.0,0.0,0.0]) # unit basis vectors
Uy = na([0.0,1.0,0.0])
Uz = na([0.0,0.0,1.0])
x = pos[0]
y = pos[1]
z = pos[2]
# corner vertices of square in xz plane
verts = na([[edgeLen/2.0, 0, edgeLen/2.0],
[edgeLen/2.0, 0, -edgeLen/2.0],
[-edgeLen/2.0, 0, -edgeLen/2.0],
[-edgeLen/2.0, 0, edgeLen/2.0]])
# For axis & angle of rotation
dirMag = np.linalg.norm(dirc)
axR = np.cross(dirc, Uy)
theta = np.arccos((np.dot(dirc, Uy) / dirMag))
Rax = rotation_matrix(axR, theta) # rotation matrix
# rotate vertices
rotVerts = na([0,0,0])
for v in verts:
temp = np.dot(Rax, v)
temp = na([temp[0]+x, temp[1]+y, temp[2]+z])
rotVerts = np.vstack((rotVerts, temp))
rotVerts = np.delete(rotVerts, rotVerts[0], axis=0)
# plot
# oringinal square
ax.scatter(verts[:,0], verts[:,1], verts[:,2], s=10, c='r', marker='o')
ax.plot([verts[0,0], verts[1,0]], [verts[0,1], verts[1,1]], [verts[0,2], verts[1,2]], color='g', linewidth=1.0)
ax.plot([verts[1,0], verts[2,0]], [verts[1,1], verts[2,1]], [verts[1,2], verts[2,2]], color='g', linewidth=1.0)
ax.plot([verts[2,0], verts[3,0]], [verts[2,1], verts[3,1]], [verts[2,2], verts[3,2]], color='g', linewidth=1.0)
ax.plot([verts[0,0], verts[3,0]], [verts[0,1], verts[3,1]], [verts[0,2], verts[3,2]], color='g', linewidth=1.0)
# rotated & translated square
ax.scatter(rotVerts[:,0], rotVerts[:,1], rotVerts[:,2], s=10, c='b', marker='o')
ax.plot([rotVerts[0,0], rotVerts[1,0]], [rotVerts[0,1], rotVerts[1,1]], [rotVerts[0,2], rotVerts[1,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[1,0], rotVerts[2,0]], [rotVerts[1,1], rotVerts[2,1]], [rotVerts[1,2], rotVerts[2,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[2,0], rotVerts[3,0]], [rotVerts[2,1], rotVerts[3,1]], [rotVerts[2,2], rotVerts[3,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[0,0], rotVerts[3,0]], [rotVerts[0,1], rotVerts[3,1]], [rotVerts[0,2], rotVerts[3,2]], color='b', linewidth=1.0)
# vector
ax.plot([pos[0], pos[0]+dirc[0]], [pos[1], pos[1]+dirc[1]], [pos[1], pos[1]+dirc[1]], color='r', linewidth=1.0)
ax.set_xlabel('X axis')
ax.set_ylabel('Y axis')
ax.set_zlabel('Z axis')
This gives the following output..
The green square is the original in the xz plane, the blue square the transformed square & the given vector is in red.
As you can see its well off. After many hours pouring through similar questions & replies, I am still none the wiser as to why this does not work.
So what am I missing here?
EDIT: After pouring over the Euler Angles link, given by El Dude in the comments below, I tried the following....
Defined the square in yz plane of a static frame of reference xyz with basis vectors Ux, Uy & Uz
Used a direction vector 'dirVec' as the normal for the plane I want to rotate my square into.
I decided to use the x-convention and the ZXZ rotation matrix as discribed in Euler angles link.
Steps I have taken,
1) Create a rotated frame with Tx, Ty & Tz as basis vectors;
Tx = dirVec
Ty = Tx cross Uz (Tx not allowed to parallel to Uz)
Tz = Ty cross Tx
2) Defined a Node Line, a vector along the intersection of the planes UxUy & TxTy by taking the cross product of Uz & Tz
3) Defined the Euler angles as per the definitions in the above link
4) Defined the ZXZ rotation matrix as per the above link
5) Applied rotation matrix to coordinates of square's vertices
It doesn't work, something odd is happening, no matter what the value of 'dirVec' alpha always comes out as 0.
Is there something obvious going on that I'm just missing?
Here's the amended code...
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import math
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
na = np.array
def rotation_ZXZ(alpha=0.0, beta=0.0, gamma=0.0):
"""
Return ZXZ rotaion matrix
"""
a = alpha
b = beta
g = gamma
ca = np.cos(a)
cb = np.cos(b)
cg = np.cos(g)
sa = np.sin(a)
sb = np.sin(b)
sg = np.sin(g)
return np.array([[(ca*cg-cb*sa*sg), (-ca*sg-cb*cg*sa), sa*sb],
[(cg*sa+ca*cb*sg), (ca*cb*cg-sa*sg), -ca*sb],
[sb*sg, cg*sb, cb]])
def rotated_axes(vector=[0,1,0]):
"""
Return unit basis vectors for rotated frame
"""
vx = np.asarray(vector) / np.linalg.norm(vector)
if vx[1] != 0 or vx[2] != 0:
U = na([1.0, 0.0, 0.0])
else:
U = na([0.0, 1.0, 0.0])
vz = np.cross(vx, U)
vz = vz / np.linalg.norm(vz)
vy = np.cross(vx, vz)
vy = vy / np.linalg.norm(vy)
vx = bv(vx[0], vx[1], vx[2])
vy = bv(vy[0], vy[1], vy[2])
vz = bv(vz[0], vz[1], vz[2])
return vx, vy, vz
def angle_btw_vectors(v1=[1,0,0], v2=[0,1,0]):
"""
Return the angle, in radians, between 2 vectors
"""
v1 = np.asarray(v1)
v2 = np.asarray(v2)
mags = np.linalg.norm(v1) * np.linalg.norm(v2)
return np.arccos(np.dot(v1, v2) / mags)
edgeLen = 4.0 # length of square side
dirVec = na([4,4,4]) # direction of given vector
pos = na([0.0, 0.0, 0.0]) # starting point of given vector
x = pos[0]
y = pos[1]
z = pos[2]
Ux = na([1,0,0]) # Unit basis vectors for static frame
Uy = na([0,1,0])
Uz = na([0,0,1])
Tx, Ty, Tz = rotated_axes(dirVec) # Unit basis vectors for rotated frame
# where Tx = dirVec / |dirVec|
nodeLine = np.cross(Uz, Tz) # Node line - xy intersect XY
alpha = angle_btw_vectors(Ux, nodeLine) #Euler angles
beta = angle_btw_vectors(Uz, Tz)
gamma = angle_btw_vectors(nodeLine, Tx)
Rzxz = rotation_ZXZ(alpha, beta, gamma) # Rotation matrix
print '--------------------------------------'
print 'Tx: ', Tx
print 'Ty: ', Ty
print 'Tz: ', Tz
print 'Node line: ', nodeLine
print 'Tx.dirVec: ', np.dot(Tx, (dirVec / np.linalg.norm(dirVec)))
print 'Ty.dirVec: ', np.dot(Ty, dirVec)
print 'Tz.dirVec: ', np.dot(Tz, dirVec)
print '(Node Line).Tx: ', np.dot(Tx, nodeLine)
print 'alpha: ', alpha * 180 / np.pi
print 'beta: ', beta * 180 / np.pi
print 'gamma: ', gamma * 180 / np.pi
#print 'Rzxz: ', Rxzx
# corner vertices of square in yz plane
verts = na([[0, edgeLen/2.0, edgeLen/2.0],
[0, edgeLen/2.0, -edgeLen/2.0],
[0, -edgeLen/2.0, -edgeLen/2.0],
[0, -edgeLen/2.0, edgeLen/2.0]])
rotVerts = na([0,0,0])
for v in verts:
temp = np.dot(Rzxz, v)
temp = na([temp[0]+x, temp[1]+y, temp[2]+z])
rotVerts = np.vstack((rotVerts, temp))
rotVerts = np.delete(rotVerts, rotVerts[0], axis=0)
# plot
# oringinal square
ax.scatter(verts[:,0], verts[:,1], verts[:,2], s=10, c='g', marker='o')
ax.plot([verts[0,0], verts[1,0]], [verts[0,1], verts[1,1]], [verts[0,2], verts[1,2]], color='g', linewidth=1.0)
ax.plot([verts[1,0], verts[2,0]], [verts[1,1], verts[2,1]], [verts[1,2], verts[2,2]], color='g', linewidth=1.0)
ax.plot([verts[2,0], verts[3,0]], [verts[2,1], verts[3,1]], [verts[2,2], verts[3,2]], color='g', linewidth=1.0)
ax.plot([verts[0,0], verts[3,0]], [verts[0,1], verts[3,1]], [verts[0,2], verts[3,2]], color='g', linewidth=1.0)
# rotated & translated square
ax.scatter(rotVerts[:,0], rotVerts[:,1], rotVerts[:,2], s=10, c='b', marker='o')
ax.plot([rotVerts[0,0], rotVerts[1,0]], [rotVerts[0,1], rotVerts[1,1]], [rotVerts[0,2], rotVerts[1,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[1,0], rotVerts[2,0]], [rotVerts[1,1], rotVerts[2,1]], [rotVerts[1,2], rotVerts[2,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[2,0], rotVerts[3,0]], [rotVerts[2,1], rotVerts[3,1]], [rotVerts[2,2], rotVerts[3,2]], color='b', linewidth=1.0)
ax.plot([rotVerts[0,0], rotVerts[3,0]], [rotVerts[0,1], rotVerts[3,1]], [rotVerts[0,2], rotVerts[3,2]], color='b', linewidth=1.0)
# Rotated reference coordinate system
ax.plot([pos[0], pos[0]+Tx[0]], [pos[1], pos[1]+Tx[1]], [pos[2], pos[2]+Tx[2]], color='r', linewidth=1.0)
ax.plot([pos[0], pos[0]+Ty[0]], [pos[1], pos[1]+Ty[1]], [pos[1], pos[2]+Ty[2]], color='b', linewidth=1.0)
ax.plot([pos[0], pos[0]+Tz[0]], [pos[1], pos[1]+Tz[1]], [pos[1], pos[2]+Tz[2]], color='g', linewidth=1.0)
ax.set_xlabel('X axis')
ax.set_ylabel('Y axis')
ax.set_zlabel('Z axis')
Here's a solution that I came up with - it should work, although there wasn't a whole lot of testing. The solution is somewhat more general, as it would work for any 2D object of any orientation, the only thing you have to adjust are the vertices stored in obj (this could be done better but here I just created a list of points by hand).
Note, that I defined mObj as the "center" of the object - this does not change the functionality but is the anchor point of the normal vector that is displayed.
Here's some explanation for the math:
What we need to do is to find the right rotation axis and angle, such that we only need one matrix multiplication (in principle you could use the Euler angles which would be an equivalent solution). The angle is easy, since it is given by the dot-product:
dot(a, b) = |a| |b| * cos(theta)
where theta is the angle between the vector a and b. To find the rotation axis, we can use the normal vector of the plane spanned by a and b, i.e. use the cross product and normalize it:
rotAxis = cross(a, b) / |cross(a, b)|
Note that this vector is orthogonal to a and b, hence the axis we are looking for.
Hope this helps.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
def rotateVector3D(v, theta, axis):
""" Takes a three-dimensional vector v and rotates it by the angle theta around the specified axis.
"""
return np.dot(rotationMatrix3D(theta, axis), v)
def rotationMatrix3D(theta, axis):
""" Return the rotation matrix associated with counterclockwise rotation about
the given axis by theta radians.
"""
axis = np.asarray(axis) / np.sqrt(np.dot(axis, axis))
a = np.cos(theta/2.0)
b, c, d = -axis*np.sin(theta/2.0)
aa, bb, cc, dd = a**2, b**2, c**2, d**2
bc, ad, ac, ab, bd, cd = b*c, a*d, a*c, a*b, b*d, c*d
return np.array([[aa+bb-cc-dd, 2*(bc+ad), 2*(bd-ac)],
[2*(bc-ad), aa+cc-bb-dd, 2*(cd+ab)],
[2*(bd+ac), 2*(cd-ab), aa+dd-bb-cc]])
def drawObject(ax, pts, color="red"):
""" Draws an object on a specified 3D axis with points and lines between consecutive points.
"""
map(lambda pt: ax.scatter(*pt, s=10, color=color), pts)
for k in range(len(pts)-1):
x, y, z = zip(*pts[k:k+2])
ax.plot(x, y, z, color=color, linewidth=1.0)
x, y, z = zip(*[pts[-1],pts[0]])
ax.plot(x, y, z, color=color, linewidth=1.0)
def normalVector(obj):
""" Takes a set of points, assumed to be flat, and returns a normal vector with unit length.
"""
n = np.cross(np.array(obj[1])-np.array(obj[0]), np.array(obj[2])-np.array(obj[0]))
return n/np.sqrt(np.dot(n,n))
# Set the original object (can be any set of points)
obj = [(2, 0, 2), (2, 0, 4), (4, 0, 4), (4, 0, 2)]
mObj = (3, 0, 3)
nVecObj = normalVector(obj)
# Given vector.
vec = (6, 6, 6)
# Find rotation axis and angle.
rotAxis = normalVector([(0,0,0), nVecObj, vec])
angle = np.arccos(np.dot(nVecObj, vec) / (np.sqrt(np.dot(vec, vec)) * np.sqrt(np.dot(nVecObj, nVecObj))))
print "Rotation angle: {:.2f} degrees".format(angle/np.pi*180)
# Generate the rotated object.
rotObj = map(lambda pt: rotateVector3D(pt, angle, rotAxis), obj)
mRotObj = rotateVector3D(mObj, angle, rotAxis)
nVecRotObj = normalVector(rotObj)
# Set up Plot.
fig = plt.figure()
fig.set_size_inches(18,18)
ax = fig.add_subplot(111, projection='3d')
# Draw.
drawObject(ax, [[0,0,0], np.array(vec)/np.sqrt(np.dot(vec,vec))], color="gray")
drawObject(ax, [mObj, mObj+nVecObj], color="red")
drawObject(ax, obj, color="red")
drawObject(ax, [mRotObj, mRotObj + nVecRotObj], color="green")
drawObject(ax, rotObj, color="green")
# Plot cosmetics.
ax.set_xlabel('X axis')
ax.set_ylabel('Y axis')
ax.set_zlabel('Z axis')
# Check if the given vector and the normal of the rotated object are parallel (cross product should be zero).
print np.round(np.sum(np.cross(vec, nVecRotObj)**2), 5)
I have some data that consist of unevenly sampled 2D spatial locations, where each x, y coordinate has an associated phase value theta between 0 and 2pi. I'd like to be able to interpolate the theta values onto a regular x, y grid. The data is degenerate in the sense that the same (or very nearby) x, y locations may be associated with multiple phase values, and vice versa for values of theta, so this is strictly speaking a smoothing problem rather than straight interpolation.
I've briefly experimented with scipy's radial basis functions, but these give nasty edge effects because of the discontinuity in the theta values from 2pi --> 0.
Here's a toy example (the real spatial distribution of phases is a lot messier):
import numpy as np
from matplotlib import pyplot as plt
from matplotlib import colorbar
from matplotlib.colors import Normalize
from scipy import interpolate
# randomly sampled spatial locations
x, y = np.random.uniform(-1, 1, size=(2, 1000))
# theta varies smoothly with location apart from the singularity at 0, 0
z = np.arctan2(x, y) % (2 * np.pi)
# smooth with a simple linear RBF
rbf = interpolate.Rbf(x, y, z, function='linear', smooth=0.1)
# resample on a finer grid
xi, yi = np.mgrid[-1:1:100j, -1:1:100j].reshape(2, -1)
zi = rbf(xi, yi) % (2 * np.pi)
# plotting
fig, ax = plt.subplots(1, 1, subplot_kw={'aspect': 'equal'})
ax.hold(True)
norm = Normalize(0, 2 * np.pi)
im = ax.imshow(zi.reshape(100, 100).T, extent=(-1, 1, -1, 1),
origin='lower', cmap=plt.cm.hsv, norm=norm)
sc = ax.scatter(x, y, s=30, c=z, cmap=im.cmap, norm=norm)
cax, kw = colorbar.make_axes_gridspec(ax)
cb = plt.colorbar(im, cax=cax, **kw)
ax.set_xlabel(r'$X_0$', fontsize='x-large')
ax.set_ylabel(r'$Y_0$', fontsize='x-large')
cb.set_ticks(np.arange(0, 2.1*np.pi, np.pi/2.))
cb.set_ticklabels([r'$0$', r'$\frac{\pi}{2}$', r'$\pi$',
r'$\frac{3\pi}{2}$', r'$2\pi$'])
cb.set_label(r'$\theta$', fontsize='x-large')
cb.ax.tick_params(labelsize='x-large')
plt.show()
What would be a good way to go about interpolating angular quantities like this? Does scipy have any built in interpolation method that will deal with angles nicely, or will I have to write my own?
I feel pretty stupid now!
The answer was very simple - this answer on MathOverflow clued me in. There is no problem with discontinuity provided that I convert from a polar coordinate space to a Cartesian one, then interpolate the x and y components of the vector independently:
x, y = np.random.uniform(-1, 1, size=(2, 1000))
z = np.arctan2(y, x) % (2*np.pi)
# convert from polar --> cartesian
u, v = np.cos(z), np.sin(z)
# interpolate x and y components separately
rbf_u = interpolate.Rbf(x, y, u, function='linear', smooth=0.1)
rbf_v = interpolate.Rbf(x, y, v, function='linear', smooth=0.1)
xi, yi = np.mgrid[-1:1:100j, -1:1:100j].reshape(2, -1)
ui = rbf_u(xi, yi)
vi = rbf_v(xi, yi)
# convert from cartesian --> polar
zi = np.arctan2(ui, vi) % (2*np.pi)
It would be nice performance-wise if there was a way to avoid performing two separate interpolations on the x and y components, but I don't really see a way around this.
Expanding on the accepted answer, this can be done in a single pass by using complex numbers to store the coordinates:
x, y = np.random.uniform(-1, 1, size=(2, 1000))
z = np.arctan2(y, x) % (2*np.pi)
# convert to cartesian coordinates on the complex plane
u = np.sin(z) + np.cos(z) * 1j
# interpolate x and y components separately
rbf_u = sp.interpolate.Rbf(x, y, u, function='linear', smooth=0.1)
xi, yi = np.mgrid[-1:1:100j, -1:1:100j].reshape(2, -1)
ui = rbf_u(xi, yi)
# convert from cartesian --> polar
zi = np.angle(ui) % (2*np.pi)