Related
I am really confused by the function pywt.cwt, as I've not been able to get it to work. The function seems to integrate instead of differentiating. I would like to work it as the following: Example CWT, but my graph looks like this: My CWT. The idea is to integrate the raw signal (av) with cumtrapz, then differentiate with a gaussian CWT (=> S1), and then once more differentiate with gaussian CWT (=> S2).
As you can see in the pictures, the bottom peaks of the red line should line up in the valleys, but the land under the top peaks for me, and the green line should move 1/4th period to the left but moves to the right... Which makes me think it integrates for some reason.
I currently have no idea what causes this... Does anyone happen to know what is going on?
Thanks in advance!
#Get data from pandas
av = dfRange['y']
#remove gravity & turns av right way up
av = av - dfRange['y'].mean()
av = av * -1
#Filter
[b,a] = signal.butter(4, [0.9/(55.2/2), 20/(55.2/2)], 'bandpass')
av = signal.filtfilt(b,a, av)
#Integrate and differentiate av => S1
integrated_av = integrate.cumtrapz(av)
[CWT_av1, frequency1] = pywt.cwt(integrated_av, 8.8 , 'gaus1', 1/55.2)
CWT_av1 = CWT_av1[0]
CWT_av1 = CWT_av1 * 0.05
#differentiate S1 => S2
[CWT_av2, frequency2] = pywt.cwt(CWT_av1, 8.8 , 'gaus1', 1/55.2)
CWT_av2 = CWT_av2[0]
CWT_av2 = CWT_av2 * 0.8
#Find Peaks
inv_CWT_av1 = CWT_av1 * -1
av1_min, _ = signal.find_peaks(inv_CWT_av1)
av2_max, _ = signal.find_peaks(CWT_av2)
#Plot
plt.style.use('seaborn')
plt.figure(figsize=(25, 7), dpi = 300)
plt.plot_date(dfRange['recorded_naive'], av, linestyle = 'solid', marker = None, color = 'steelblue')
plt.plot_date(dfRange['recorded_naive'][:-1], CWT_av1[:], linestyle = 'solid', marker = None, color = 'red')
plt.plot(dfRange['recorded_naive'].iloc[av1_min], CWT_av1[av1_min], "ob", color = 'red')
plt.plot_date(dfRange['recorded_naive'][:-1], CWT_av2[:], linestyle = 'solid', marker = None, color = 'green')
plt.plot(dfRange['recorded_naive'].iloc[av2_max], CWT_av2[av2_max], "ob", color = 'green')
plt.gcf().autofmt_xdate()
plt.show()
I'm not sure this is your answer, but an observation from playing with pywt...
From the documentation the wavelets are basically given by the differentials of a Gaussian but there is an order dependent normalisation constant.
Plotting the differentials of a Guassian against the wavelets (extracted by putting in an impulse response) gives the following:
The interesting observation is that the order dependent normalisation constant sometimes seems to include a '-1'. In particular, it does for the first order gaus1.
So, my question is, could you actually have differentiation as you expect, but also multiplication by -1?
Code for the graph:
import numpy as np
import matplotlib.pyplot as plt
import pywt
dt = 0.01
t = dt * np.arange(100)
# Calculate the differentials of a gaussian by quadrature:
# start with the gaussian y = exp(-(x - x_0) ^ 2 / dt)
ctr = t[len(t) // 2]
gaus = np.exp(-np.power(t - ctr, 2)/dt)
gaus_quad = [np.gradient(gaus, dt)]
for i in range(7):
gaus_quad.append(np.gradient(gaus_quad[-1], dt))
# Extract the wavelets using the impulse half way through the dataset
y = np.zeros(len(t))
y[len(t) // 2] = 1
gaus_cwt = list()
for i in range(1, 9):
cwt, cwt_f = pywt.cwt(y, 10, f'gaus{i}', dt)
gaus_cwt.append(cwt[0])
fig, axs = plt.subplots(4, 2)
for i, ax in enumerate(axs.flatten()):
ax.plot(t, gaus_cwt[i] / np.max(np.abs(gaus_cwt[i])))
ax.plot(t, gaus_quad[i] / np.max(np.abs(gaus_quad[i])))
ax.set_title(f'gaus {i+1}', x=0.2, y=1.0, pad=-14)
ax.axhline(0, c='k')
ax.set_xticks([])
ax.set_yticks([])
I am trying to fit a smoothing B-spline to some data and I found this very helpful post on here. However, I not only need the spline, but also its derivatives, so I tried to add the following code to the example:
tck_der = interpolate.splder(tck, n=1)
x_der, y_der, z_der = interpolate.splev(u_fine, tck_der)
For some reason this does not seem to work due to some data type issues. I get the following traceback:
Traceback (most recent call last):
File "interpolate_point_trace.py", line 31, in spline_example
tck_der = interpolate.splder(tck, n=1)
File "/home/user/anaconda3/lib/python3.7/site-packages/scipy/interpolate/fitpack.py", line 657, in splder
return _impl.splder(tck, n)
File "/home/user/anaconda3/lib/python3.7/site-packages/scipy/interpolate/_fitpack_impl.py", line 1206, in splder
sh = (slice(None),) + ((None,)*len(c.shape[1:]))
AttributeError: 'list' object has no attribute 'shape'
The reason for this seems to be that the second argument of the tck tuple contains a list of numpy arrays. I thought turning the input data to be a numpy array as well would help, but it does not change the data types of tck.
Does this behavior reflect an error in scipy, or is the input malformed?
I tried manually turning the list into an array:
tck[1] = np.array(tck[1])
but this (which didn't surprise me) also gave an error:
ValueError: operands could not be broadcast together with shapes (0,8) (7,1)
Any ideas of what the problem could be? I have used scipy before and on 1D splines the splder function works just fine, so I assume it has something to do with the spline being a line in 3D.
------- edit --------
Here is a minimum working example:
import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate
from mpl_toolkits.mplot3d import Axes3D
total_rad = 10
z_factor = 3
noise = 0.1
num_true_pts = 200
s_true = np.linspace(0, total_rad, num_true_pts)
x_true = np.cos(s_true)
y_true = np.sin(s_true)
z_true = s_true / z_factor
num_sample_pts = 80
s_sample = np.linspace(0, total_rad, num_sample_pts)
x_sample = np.cos(s_sample) + noise * np.random.randn(num_sample_pts)
y_sample = np.sin(s_sample) + noise * np.random.randn(num_sample_pts)
z_sample = s_sample / z_factor + noise * np.random.randn(num_sample_pts)
tck, u = interpolate.splprep([x_sample, y_sample, z_sample], s=2)
x_knots, y_knots, z_knots = interpolate.splev(tck[0], tck)
u_fine = np.linspace(0, 1, num_true_pts)
x_fine, y_fine, z_fine = interpolate.splev(u_fine, tck)
# this is the part of the code I inserted: the line under this causes the crash
tck_der = interpolate.splder(tck, n=1)
x_der, y_der, z_der = interpolate.splev(u_fine, tck_der)
# end of the inserted code
fig2 = plt.figure(2)
ax3d = fig2.add_subplot(111, projection='3d')
ax3d.plot(x_true, y_true, z_true, 'b')
ax3d.plot(x_sample, y_sample, z_sample, 'r*')
ax3d.plot(x_knots, y_knots, z_knots, 'go')
ax3d.plot(x_fine, y_fine, z_fine, 'g')
fig2.show()
plt.show()
Stumbled into the same problem...
I circumvented the error by using interpolate.splder(tck, n=1) and instead used interpolate.splev(spline_ev, tck, der=1) which returns the derivatives at the points spline_ev (see Scipy Doku).
If you need the spline I think you can then use interpolate.splprep() again.
In total something like:
import numpy as np
from scipy import interpolate
import matplotlib.pyplot as plt
points = np.random.rand(10,2) * 10
(tck, u), fp, ier, msg = interpolate.splprep(points.T, s=0, k=3, full_output=True)
spline_ev = np.linspace(0.0, 1.0, 100, endpoint=True)
spline_points = interpolate.splev(spline_ev, tck)
# Calculate derivative
spline_der_points = interpolate.splev(spline_ev, tck, der=1)
spline_der = interpolate.splprep(spline_der_points.T, s=0, k=3, full_output=True)
# Plot the data and derivative
fig = plt.figure()
plt.plot(points[:,0], points[:,1], '.-', label="points")
plt.plot(spline_points[0], spline_points[1], '.-', label="tck")
plt.plot(spline_der_points[0], spline_der_points[1], '.-', label="tck_der")
# Show tangent
plt.arrow(spline_points[0][23]-spline_der_points[0][23], spline_points[1][23]-spline_der_points[1][23], 2.0*spline_der_points[0][23], 2.0*spline_der_points[1][23])
plt.legend()
plt.show()
EDIT:
I also opened an Issue on Github and according to ev-br the usage of interpolate.splprep is depreciated and one should use make_interp_spline / BSpline instead.
As noted in other answers, splprep output is incompatible with splder, but is compatible with splev. And the latter can evaluate the derivatives.
However, for interpolation, there is an alternative approach, which avoids splprep altogether. I'm basically copying a reply on the SciPy issue tracker (https://github.com/scipy/scipy/issues/10389):
Here's an example of replicating the splprep outputs. First let's make sense out of the splprep output:
# start with the OP example
import numpy as np
from scipy import interpolate
points = np.random.rand(10,2) * 10
(tck, u), fp, ier, msg = interpolate.splprep(points.T, s=0, k=3, full_output=True)
# check the meaning of the `u` array: evaluation of the spline at `u`
# gives back the original points (up to a list/transpose)
xy = interpolate.splev(u, tck)
xy = np.asarray(xy)
np.allclose(xy.T, points)
Next, let's replicate it without splprep. First, build the u array: the curve is represented parametrically, and u is essentially an approximation for the arc length. Other parametrizations are possible, but here let's stick to what splprep does. Translating the pseudocode from the doc page, https://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.splprep.html
vv = np.sum((points[1:, :] - points[:-1, :])**2, axis=1)
vv = np.sqrt(vv).cumsum()
vv/= vv[-1]
vv = np.r_[0, vv]
# check:
np.allclose(u, vv)
Now, interpolate along the parametric curve: points vs vv:
spl = interpolate.make_interp_spline(vv, points)
# check spl.t vs knots from splPrep
spl.t - tck[0]
The result, spl, is a BSpline object which you can evaluate, differentiate etc in a usual way:
np.allclose(points, spl(vv))
# differentiate
spl_derivative = spl.derivative(vv)
Explanation:
I have two numpy arrays: dataX and dataY, and I am trying to filter each array to reduce the noise. The image shown below shows the actual input data (blue dots) and an example of what I want it to be like(red dots). I do not need the filtered data to be as perfect as in the example but I do want it to be as straight as possible. I have provided sample data in the code.
What I have tried:
Firstly, you can see that the data isn't 'continuous', so I first divided them into individual 'segments' ( 4 of them in this example), and then applied a filter to each 'segment'. Someone suggested that I use a Savitzky-Golay filter. The full, run-able code is below:
import scipy as sc
import scipy.signal
import numpy as np
import matplotlib.pyplot as plt
# Sample Data
ydata = np.array([1,0,1,2,1,2,1,0,1,1,2,2,0,0,1,0,1,0,1,2,7,6,8,6,8,6,6,8,6,6,8,6,6,7,6,5,5,6,6, 10,11,12,13,12,11,10,10,11,10,12,11,10,10,10,10,12,12,10,10,17,16,15,17,16, 17,16,18,19,18,17,16,16,16,16,16,15,16])
xdata = np.array([1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32,33, 1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32])
# Used a diff array to find where there is a big change in Y.
# If there's a big change in Y, then there must be a change of 'segment'.
diffy = np.diff(ydata)
# Create empty numpy arrays to append values into
filteredX = np.array([])
filteredY = np.array([])
# Chose 3 to be the value indicating the change in Y
index = np.where(diffy >3)
# Loop through the array
start = 0
for i in range (0, (index[0].size +1) ):
# Check if last segment is reached
if i == index[0].size:
print xdata[start:]
partSize = xdata[start:].size
# Window length must be an odd integer
if partSize % 2 == 0:
partSize = partSize - 1
filteredDataX = sc.signal.savgol_filter(xdata[start:], partSize, 3)
filteredDataY = sc.signal.savgol_filter(ydata[start:], partSize, 3)
filteredX = np.append(filteredX, filteredDataX)
filteredY = np.append(filteredY, filteredDataY)
else:
print xdata[start:index[0][i]]
partSize = xdata[start:index[0][i]].size
if partSize % 2 == 0:
partSize = partSize - 1
filteredDataX = sc.signal.savgol_filter(xdata[start:index[0][i]], partSize, 3)
filteredDataY = sc.signal.savgol_filter(ydata[start:index[0][i]], partSize, 3)
start = index[0][i]
filteredX = np.append(filteredX, filteredDataX)
filteredY = np.append(filteredY, filteredDataY)
# Plots
plt.plot(xdata,ydata, 'bo', label = 'Input Data')
plt.plot(filteredX, filteredY, 'ro', label = 'Filtered Data')
plt.xlabel('X')
plt.ylabel('Y')
plt.title('Result')
plt.legend()
plt.show()
This is my result:
When each point is connected, the result looks as follows.
I have played around with the order, but it seems like a third order gave the best result.
I have also tried these filters, among a few others:
scipy.signal.medfilt
scipy.ndimage.filters.uniform_filter1d
But so far none of the filters I have tried were close to what I really wanted. What is the best way to filter data such as this? Looking forward to your help.
One way to get something looking close to your ideal would be clustering + linear regression.
Note that you have to provide the number of clusters and I also cheated a bit in scaling up y before clustering.
import numpy as np
from scipy import cluster, stats
ydata = np.array([1,0,1,2,1,2,1,0,1,1,2,2,0,0,1,0,1,0,1,2,7,6,8,6,8,6,6,8,6,6,8,6,6,7,6,5,5,6,6, 10,11,12,13,12,11,10,10,11,10,12,11,10,10,10,10,12,12,10,10,17,16,15,17,16, 17,16,18,19,18,17,16,16,16,16,16,15,16])
xdata = np.array([1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32,33, 1,2,3,1,5,4,7,8,6,10,11,12,13,10,12,13,17,16,19,18,21,19,23,21,25,20,26,27,28,26,26,26,29,30,30,29,30,32])
def split_to_lines(x, y, k):
yo = np.empty_like(y, dtype=float)
# get the cluster centers and the labels for each point
centers, map_ = cluster.vq.kmeans2(np.array((x, y * 2)).T.astype(float), k)
# for each cluster, use the labels to select the points belonging to
# the cluster and do a linear regression
for i in range(k):
slope, interc, *_ = stats.linregress(x[map_==i], y[map_==i])
# use the regression parameters to construct y values on the
# best fit line
yo[map_==i] = x[map_==i] * slope + interc
return yo
import pylab
pylab.plot(xdata, ydata, 'or')
pylab.plot(xdata, split_to_lines(xdata, ydata, 4), 'ob')
pylab.show()
I am trying to replicate some graphs from 3 equations.
My equation are shown in my python code below
The first two equations are plotted on the 1st graph
The first equation and the 3rd equation are plotted on the 2nd graph
The last two equations are plotted on the 3rd graph
My parameters are:
k1 = a range of values, k2=1, n1=3, n2=3, K1=0.3, K2=0.3, b1=0, b2=0.05
i'm unsure about how to plot this on a graph. any help would be great. thanks
note that i am a python beginner and have basic knowledge of coding
here's my code so far:
import numpy as np
import matplotlib.pyplot as plt
rateinact= k1*X
rateact12= k2*(b2*(K2**n2/(K2**n2 + X**n2)) + (X**n2/(K2**n2 + X**n2))*(1-X)
Rateinact2 = k1*(B1*((X**n1)/(K1**n1+X**n1))+((K1**n1)/(K1**n1+X**n1)))*(X)
#define parameters
k1 = []
k2=1
n1=3
n2=3
K1=0.3
K2=0.3
b1=0
b2=0.05
#graph 1
plt.plot(inact, 'r')
plt.plot(act12, 'b')
plt.ylabel('Rate of A inactivation or activation')
plt.xlabel('Fraction of activated A)
plt.show()
The trick you seem to be missing is how to make a range of values. This can be easily done with np.linspace since you're already using numpy.
x = np.linspace(0,1,100)
y = x**2 + 3*x + 4
plt.plot(x,y,'g-')
Or, for your specific example, with the syntax errors removed (I had to guess a little on what some values were):
-- EDIT --
I see now a little better what you wanted. Here's a quick demo of how you might accomplish this (I'm using subfigures instead of separate figures, but the idea is the same):
import numpy as np
import matplotlib.pyplot as plt
k1 = np.linspace(0,1,5)
k2=1
n1=3
n2=3
K1=0.3
K2=0.3
beta1=0
beta2=0.05
x = np.linspace(0,1)
def r_inact_1loop(x,k1=1):
return k1*x
def r_activation(x,k1=1):
return k2*(beta2*K2**n2/(K2**n2+x**n2) + x**n2/(K2**n2+x**n2))*(1-x)
def r_inact_2loop(x,k1=1):
return k1*(beta1*x**n1/(K1**n1+x**n1) + K1**n1/(K1**n1+x**n1))*x
#define parameters
#graph 1
fig = plt.figure(figsize=(16,8))
ax1 = fig.add_subplot(1,3,1)
ax2 = fig.add_subplot(1,3,2)
ax3 = fig.add_subplot(1,3,3)
ax1.set_xlabel('Fraction of activated A')
ax2.set_xlabel('Fraction of activated A')
ax3.set_xlabel('Fraction of activated A')
ax1.set_ylabel('Rate of A inactivation or activation')
ax2.set_ylabel('Rate of A inactivation or activation')
ax3.set_ylabel('Rate of A inactivation or activation')
for k in k1:
ax1.plot(x,r_inact_1loop(x,k),label='in1 k1={:.1f}'.format(k))
ax1.plot(x,r_activation(x,k),label='ac k1={:.1f}'.format(k))
ax1.legend(loc='best')
ax2.plot(x,r_inact_1loop(x,k),label='in1 k1={:.1f}'.format(k))
ax2.plot(x,r_inact_2loop(x,k),label='in2 k1={:.1f}'.format(k))
ax1.set_xlabel('Fraction of activated A')
ax2.legend(loc='best')
ax3.plot(x,r_activation(x,k),label='ac k1={:.1f}'.format(k))
ax3.plot(x,r_inact_2loop(x,k),label='in2 k1={:.1f}'.format(k))
ax3.legend(loc='best')
plt.show()
--ADDENDUM--
If you want the cool shading of the plots you were looking at, here's the code for that
ax2.fill_between(x,r_inact_1loop(x,k1[0]),r_inact_1loop(x,k1[-1]))
ax3.fill_between(x,r_inact_2loop(x,k1[0]),r_inact_2loop(x,k1[-1]))
I have about 50,000 data points in 3D on which I have run scipy.spatial.Delaunay from the new scipy (I'm using 0.10) which gives me a very useful triangulation.
Based on: http://en.wikipedia.org/wiki/Delaunay_triangulation (section "Relationship with the Voronoi diagram")
...I was wondering if there is an easy way to get to the "dual graph" of this triangulation, which is the Voronoi Tesselation.
Any clues? My searching around on this seems to show no pre-built in scipy functions, which I find almost strange!
Thanks,
Edward
The adjacency information can be found in the neighbors attribute of the Delaunay object. Unfortunately, the code does not expose the circumcenters to the user at the moment, so you'll have to recompute those yourself.
Also, the Voronoi edges that extend to infinity are not directly obtained in this way. It's still probably possible, but needs some more thinking.
import numpy as np
from scipy.spatial import Delaunay
points = np.random.rand(30, 2)
tri = Delaunay(points)
p = tri.points[tri.vertices]
# Triangle vertices
A = p[:,0,:].T
B = p[:,1,:].T
C = p[:,2,:].T
# See http://en.wikipedia.org/wiki/Circumscribed_circle#Circumscribed_circles_of_triangles
# The following is just a direct transcription of the formula there
a = A - C
b = B - C
def dot2(u, v):
return u[0]*v[0] + u[1]*v[1]
def cross2(u, v, w):
"""u x (v x w)"""
return dot2(u, w)*v - dot2(u, v)*w
def ncross2(u, v):
"""|| u x v ||^2"""
return sq2(u)*sq2(v) - dot2(u, v)**2
def sq2(u):
return dot2(u, u)
cc = cross2(sq2(a) * b - sq2(b) * a, a, b) / (2*ncross2(a, b)) + C
# Grab the Voronoi edges
vc = cc[:,tri.neighbors]
vc[:,tri.neighbors == -1] = np.nan # edges at infinity, plotting those would need more work...
lines = []
lines.extend(zip(cc.T, vc[:,:,0].T))
lines.extend(zip(cc.T, vc[:,:,1].T))
lines.extend(zip(cc.T, vc[:,:,2].T))
# Plot it
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
lines = LineCollection(lines, edgecolor='k')
plt.hold(1)
plt.plot(points[:,0], points[:,1], '.')
plt.plot(cc[0], cc[1], '*')
plt.gca().add_collection(lines)
plt.axis('equal')
plt.xlim(-0.1, 1.1)
plt.ylim(-0.1, 1.1)
plt.show()
As I spent a considerable amount of time on this, I'd like to share my solution on how to get the Voronoi polygons instead of just the edges.
The code is at https://gist.github.com/letmaik/8803860 and extends on the solution of tauran.
First, I changed the code to give me vertices and (pairs of) indices (=edges) separately, as many calculations can be simplified when working on indices instead of point coordinates.
Then, in the voronoi_cell_lines method I determine which edges belong to which cells. For that I use the proposed solution of Alink from a related question. That is, for each edge find the two nearest input points (=cells) and create a mapping from that.
The last step is to create the actual polygons (see voronoi_polygons method). First, the outer cells which have dangling edges need to be closed. This is as simple as looking through all edges and checking which ones have only one neighboring edge. There can be either zero or two such edges. In case of two, I then connect these by introducing an additional edge.
Finally, the unordered edges in each cell need to be put into the right order to derive a polygon from them.
The usage is:
P = np.random.random((100,2))
fig = plt.figure(figsize=(4.5,4.5))
axes = plt.subplot(1,1,1)
plt.axis([-0.05,1.05,-0.05,1.05])
vertices, lineIndices = voronoi(P)
cells = voronoi_cell_lines(P, vertices, lineIndices)
polys = voronoi_polygons(cells)
for pIdx, polyIndices in polys.items():
poly = vertices[np.asarray(polyIndices)]
p = matplotlib.patches.Polygon(poly, facecolor=np.random.rand(3,1))
axes.add_patch(p)
X,Y = P[:,0],P[:,1]
plt.scatter(X, Y, marker='.', zorder=2)
plt.axis([-0.05,1.05,-0.05,1.05])
plt.show()
which outputs:
The code is probably not suitable for large numbers of input points and can be improved in some areas. Nevertheless, it may be helpful to others who have similar problems.
I came across the same problem and built a solution out of pv.'s answer and other code snippets I found across the web. The solution returns a complete Voronoi diagram, including the outer lines where no triangle neighbours are present.
#!/usr/bin/env python
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
from scipy.spatial import Delaunay
def voronoi(P):
delauny = Delaunay(P)
triangles = delauny.points[delauny.vertices]
lines = []
# Triangle vertices
A = triangles[:, 0]
B = triangles[:, 1]
C = triangles[:, 2]
lines.extend(zip(A, B))
lines.extend(zip(B, C))
lines.extend(zip(C, A))
lines = matplotlib.collections.LineCollection(lines, color='r')
plt.gca().add_collection(lines)
circum_centers = np.array([triangle_csc(tri) for tri in triangles])
segments = []
for i, triangle in enumerate(triangles):
circum_center = circum_centers[i]
for j, neighbor in enumerate(delauny.neighbors[i]):
if neighbor != -1:
segments.append((circum_center, circum_centers[neighbor]))
else:
ps = triangle[(j+1)%3] - triangle[(j-1)%3]
ps = np.array((ps[1], -ps[0]))
middle = (triangle[(j+1)%3] + triangle[(j-1)%3]) * 0.5
di = middle - triangle[j]
ps /= np.linalg.norm(ps)
di /= np.linalg.norm(di)
if np.dot(di, ps) < 0.0:
ps *= -1000.0
else:
ps *= 1000.0
segments.append((circum_center, circum_center + ps))
return segments
def triangle_csc(pts):
rows, cols = pts.shape
A = np.bmat([[2 * np.dot(pts, pts.T), np.ones((rows, 1))],
[np.ones((1, rows)), np.zeros((1, 1))]])
b = np.hstack((np.sum(pts * pts, axis=1), np.ones((1))))
x = np.linalg.solve(A,b)
bary_coords = x[:-1]
return np.sum(pts * np.tile(bary_coords.reshape((pts.shape[0], 1)), (1, pts.shape[1])), axis=0)
if __name__ == '__main__':
P = np.random.random((300,2))
X,Y = P[:,0],P[:,1]
fig = plt.figure(figsize=(4.5,4.5))
axes = plt.subplot(1,1,1)
plt.scatter(X, Y, marker='.')
plt.axis([-0.05,1.05,-0.05,1.05])
segments = voronoi(P)
lines = matplotlib.collections.LineCollection(segments, color='k')
axes.add_collection(lines)
plt.axis([-0.05,1.05,-0.05,1.05])
plt.show()
Black lines = Voronoi diagram, Red lines = Delauny triangles
I do not know of a function to do this, but it does not seem like an overly complicated task.
The Voronoi graph is the junction of the circumcircles, as described in the wikipedia article.
So you could start with a function that finds the center of the circumcircles of a triangle, which is basic mathematics (http://en.wikipedia.org/wiki/Circumscribed_circle).
Then, just join centers of adjacent triangles.