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I would like to find the intersection between (eq1, eq2) and (eq1, eq3) and show that point with the dotted line on each axis. This code does not give me the exact point but just an approximation. I do not understand where am I doing mistake.
import matplotlib.pyplot as plt
import numpy as np
f = []
h = []
j = []
point = []
for x in range(25):
eq1 = x * 185 * 3
eq2 = 11930 - (12502 / 6) + (x * 185) / 6
eq3 = 11930 - (12502 / 3) + (x * 185) / 6
point.append(x)
f.append(eq1)
h.append(eq2)
j.append(eq3)
plt.plot(point, f)
plt.plot(point, h)
plt.plot(point, j)
plt.legend(loc='lower right', fontsize=10)
idx1 = np.argwhere(np.diff(np.sign(np.array(f) - np.array(h)))).flatten()
idx2 = idx = np.argwhere(np.diff(np.sign(np.array(f) - np.array(j)))).flatten()
plt.plot(np.array(point)[idx1+1], np.array(h)[idx1+1], 'ro')
plt.plot(np.array(point)[idx2+1], np.array(j)[idx2+1], 'ro')
plt.show()
Several issues here:
Firstly, your code is unnecessarily long. Make use of NumPy arrays to simplify things. Since NumPy is a dependency of matplotlib, you are not overkilling by importing NumPy.
You need to make a very dense mesh of points between 0 and 25 to get more accurate intersection points. Use linspace with 1000 points for example.
As you can see, with arrays, you don't need to use for loop, neither you need to initialize empty lists and then append values one by one.
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(0, 25, 1000)
f = x * 185 * 3
h = 11930 - (12502 / 6) + (x * 185) / 6
j = 11930 - (12502 / 3) + (x * 185) / 6
plt.plot(x, f, label='f')
plt.plot(x, h, label='h')
plt.plot(x, j, label='j')
plt.legend(loc='lower right', fontsize=12)
idx1 = np.argwhere(np.diff(np.sign(np.array(f) - np.array(h)))).flatten()
idx2 = idx = np.argwhere(np.diff(np.sign(np.array(f) - np.array(j)))).flatten()
plt.plot(x[idx1+1], h[idx1+1], 'ro')
plt.plot(x[idx2+1], j[idx2+1], 'ro')
plt.vlines(x[idx1+1], 0, h[idx1+1], linestyle='--')
plt.vlines(x[idx2+1], 0, j[idx2+1], linestyle='--')
plt.hlines(h[idx1+1], 0, x[idx1+1], linestyle='--')
plt.hlines(j[idx2+1], 0, x[idx2+1], linestyle='--')
plt.xlim(0, None)
plt.ylim(0, None)
plt.show()
I have plotted curve created by a list with several values. How to find out the x-coordinate that correspond with y-coordinate 0.04400918? This value is not exactly included in the list that describes the curve. Thank you very much.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D # 3d graph
from mpl_toolkits.mplot3d import proj3d # 3d graph
import matplotlib.pylab as pl
fig=pl.figure()
ax = Axes3D(fig)
x=[0.02554897, 0.02587839, 0.02623991, 0.02663096, 0.02704882, 0.02749103, 0.02795535, 0.02844018, 0.02894404, 0.02946527, 0.03000235]
y=[0.04739086, 0.0460989, 0.04481555, 0.04354088, 0.04227474, 0.04101689, 0.03976702, 0.03852497, 0.03729052, 0.0360633, 0.03484293]
z=[1.05764017e-18, 1.57788964e-18, 2.00281370e-18, 2.40500994e-18, 2.80239565e-18, 3.19420769e-18, 3.58001701e-18, 3.96024361e-18, 4.33484911e-18, 4.70364652e-18, 5.06672528e-18]
y_point=0.04400918
ax.plot3D(x,y,z)
plt.show()
Here is a specific resolution for your problem.
Some works have already been done for solving line-plane equation. This topic explains how to solve it. Even better, this snippet implements a solution.
For now, we only need to adapt it to our problem.
The first step is to find all the time the line is crossing the plan. To do that, we will iterate over the y dataset and collect all consecutive values when y_point is between them:
lines = []
for i in range(len(y) - 1):
if y[i] >= y_point and y_point >= y[i+1]:
lines.append([[x[i], y[i], z[i]], [x[i+1], y[i+1], z[i+1]]])
Then, for all of these lines, we will solve the intersection equation with the plane. We will use the function provided in sources above.
Finally, we will plot the results
Full code:
# Modules
import numpy as np
import matplotlib.pyplot as plt
# Data
x = [0.02554897, 0.02587839, 0.02623991, 0.02663096, 0.02704882, 0.02749103, 0.02795535, 0.02844018, 0.02894404, 0.02946527, 0.03000235]
y = [0.04739086, 0.0460989, 0.04481555, 0.04354088, 0.04227474, 0.04101689, 0.03976702, 0.03852497, 0.03729052, 0.0360633, 0.03484293]
z = [1.05764017e-18, 1.57788964e-18, 2.00281370e-18, 2.40500994e-18, 2.80239565e-18, 3.19420769e-18, 3.58001701e-18, 3.96024361e-18, 4.33484911e-18, 4.70364652e-18, 5.06672528e-18]
y_point = 0.04400918
# Source: https://rosettacode.org/wiki/Find_the_intersection_of_a_line_with_a_plane#Python
# Resolve intersection
def LinePlaneCollision(planeNormal, planePoint, rayDirection, rayPoint, epsilon=1e-6):
ndotu = planeNormal.dot(rayDirection)
if abs(ndotu) < epsilon:
raise RuntimeError("no intersection or line is within plane")
w = rayPoint - planePoint
si = -planeNormal.dot(w) / ndotu
Psi = w + si * rayDirection + planePoint
return Psi
# For all line, apply the solving process
def solveAllPoints(lines, y_point):
collision_points = []
for line in lines:
# Define plane
planeNormal = np.array([0, 1, 0]) # Plane normal (e.g. y vector)
planePoint = np.array([0, y_point, 0]) # Any point on the plane
# Define ray
rayDirection = line[1] - line[0] # Line direction
rayPoint = line[0] # Any point of the line
# Append point
collision_points.append(LinePlaneCollision(planeNormal, planePoint, rayDirection, rayPoint))
return collision_points
# Find all consecutive Y points crossing the plane.
# This function is only working for the given problem (intersection of the line
# with 1 plan defined by a normal vector = [0,1,0])
def getCrossingLines(y_point, x, y, z):
lines = []
for i in range(len(y) - 1):
if y[i] >= y_point and y_point >= y[i+1]:
lines.append([[x[i], y[i], z[i]], [x[i+1], y[i+1], z[i+1]]])
return np.array(lines)
# Get coordinates for drawing our plane
# Related topic: https://stackoverflow.com/questions/53115276/matplotlib-how-to-draw-a-vertical-plane-in-3d-figure
def getXYZPlane(x, y, z):
xs = np.linspace(min(x), max(x), 100)
zs = np.linspace(min(z), max(z), 100)
X, Z = np.meshgrid(xs, zs)
Y = np.array([y_point for _ in X])
return X, Y, Z
# Create plot
plt3d = plt.figure().gca(projection='3d')
ax = plt.gca()
# Draw data line
ax.plot3D(x,y,z)
# Plot plan
X, Y, Z = getXYZPlane(x, y, z)
ax.plot_surface(X, Y, Z)
# Draw crossing points (lines-planes)
lines = getCrossingLines(y_point, x, y , z)
for pt in solveAllPoints(lines, y_point):
ax.scatter(pt[0], pt[1], pt[2], color='green')
plt.show()
Output
I have a variable (P) which is a function of angle (theta):
In this equation the K is a constant, theta_p is equal to zero and I is the modified Bessel function of the first kind (order 0) which is defined as:
Now, I want to plot the P versus theta for different values of constant K. First I calculated the parameter I and then plug it into the first equation to calculate P for different angles theta. I mapped it into a Cartesian coordinate by putting :
x = P*cos(theta)
y = P*sin(theta)
Here is my python implementation using matplotlib and scipy when the constant k=2.0:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
def integrand(x, a, k):
return a*np.exp(k*np.cos(x))
theta = (np.arange(0, 362, 2))
theta_p = 0.0
X = []
Y = []
for i in range(len(theta)):
a = (1 / np.pi)
k = 2.0
Bessel = quad(integrand, 0, np.pi, args=(a, k))
I = list(Bessel)[0]
P = (1 / (np.pi * I)) * np.exp(k * np.cos(2 * (theta[i]*np.pi/180. - theta_p)))
x = P*np.cos(theta[i]*np.pi/180.)
y = P*np.sin(theta[i]*np.pi/180.)
X.append(x)
Y.append(y)
plt.plot(X,Y, linestyle='-', linewidth=3, color='red')
axes = plt.gca()
plt.show()
I should get a set of graphs like the below figure for different K values:
(Note that the distributions were plotted on a circle of unit 1 to ease visualization)
However it seems like the graphs produced by the above code are not similar to the above figure.
Any idea what is the issue with the above implementation?
Thanks in advance for your help.
Here is how it looks like (for k=2):
The reference for these formulas are the equation 5 and 6 that you could find here
You had a mistake in your formula.
Your formula gives the delta of your function above a unit circle. So in your function to get the plot you want, simply add 1 to it.
Here is what you want, with some tidied up python. ...note you can do the whole calculation of the 'P' values as a numpy vector line, you don't need to loop over the indicies. ...also you can just do a polar plot directly in matplotlib - you don't need to transform it into cartesian.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
theta = np.arange(0, 2*np.pi+0.1, 2*np.pi/100)
def integrand(x, a, k):
return a*np.exp(k*np.cos(x))
for k in np.arange(0, 5, 0.5):
a = (1 / np.pi)
Bessel = quad(integrand, 0, np.pi, args=(a, k))
I = Bessel[0]
P = 1 + (1/(np.pi * I)) * np.exp(k * np.cos(2 * theta))
plt.polar(theta, P)
plt.show()
I'm trying to create animation about how the value of a quantum bit would be changed by computation called X-gate, in form of an arrow.
Here are the codes that I wrote.
#Import libraries
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import animation
import numpy as np
#Accept input (theta, phi) from a user
print("Put angle theta and phi, 0≤ theta ≤180, 0≤ phi ≤360")
theta = input("theta:")
phi = input("phi:")
theta = np.radians(float(theta))
phi = np.radians(float(phi))
#Calculate x,y,z coordinates
X = np.sin(theta) * np.cos(phi)
Y = np.sin(theta) * np.sin(phi)
Z = np.cos(theta)
#Adjusting the length of an arrow
length = np.sqrt(X**2 + Y**2 + Z**2)
if length > 1:
X = X/length
Y = Y/length
Z = Z/length
# Figure of the animation
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.set_aspect("equal")
u, v = np.mgrid[0:2*np.pi:20j, 0:np.pi:10j]
x = np.cos(u)*np.sin(v)
y = np.sin(u)*np.sin(v)
z = np.cos(v)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
ax.plot_wireframe(x,y,z, color="black")
# Calculate x,y,z coordinates in the process of the change
length = 9
xgate_theta = np.linspace(theta,theta+np.pi,length)
xgate_phi = np.linspace(phi,phi,length)
#Array of x,y,z coordinates
xgate= []
# Only x coordinates
xgate_x = []
# Only y coordinates
xgate_y = []
# Only z coordinates
xgate_z = []
for i in range(length):
xgate_x.append(X)
xgate_z.append(np.cos(xgate_theta[i]))
xgate_y.append(np.sqrt(1-np.sqrt(xgate_x[i]**2+xgate_z[i]**2))*(-1))
for j in range(length):
xgate.append(plt.quiver(0,0,0,xgate_x[j],xgate_y[j],xgate_z[j],color="red"))
ani = animation.ArtistAnimation(fig,xgate,interval=1000)
plt.show()
Then, I got the following error.
Traceback (most recent call last):
File "/Users/makotonakai/anaconda3/lib/python3.6/site- packages/matplotlib/cbook/__init__.py", line 388, in process
proxy(*args, **kwargs)
File "/Users/makotonakai/anaconda3/lib/python3.6/site-packages/matplotlib/cbook/__init__.py", line 228, in __call__
return mtd(*args, **kwargs)
File "/Users/makotonakai/anaconda3/lib/python3.6/site-packages/matplotlib/animation.py", line 1026, in _start
self._init_draw()
File "/Users/makotonakai/anaconda3/lib/python3.6/site-packages/matplotlib/animation.py", line 1556, in _init_draw
for artist in f:
TypeError: 'Line3DCollection' object is not iterable
I cannot tell what line causes this error by just looking this error message. Can anybody tell me how I can fix this error?
How do I generate a trapezoidal wave in Python?
I looked into the modules such as SciPy and NumPy, but in vain. Is there a module such as the scipy.signal.gaussian which returns an array of values representing the Gaussian function wave?
I generated this using the trapezoidal kernel of Astropy,
Trapezoid1DKernel(30,slope=1.0)
. I want to implement this in Python without using Astropy.
While the width and the slope are sufficient to define a triangular signal, you would need a third parameter for a trapezoidal signal: the amplitude.
Using those three parameters, you can easily adjust the scipy.signal.sawtooth function to give you a trapeziodal shape by truncating and offsetting the triangular shaped function.
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
def trapzoid_signal(t, width=2., slope=1., amp=1., offs=0):
a = slope*width*signal.sawtooth(2*np.pi*t/width, width=0.5)/4.
a[a>amp/2.] = amp/2.
a[a<-amp/2.] = -amp/2.
return a + amp/2. + offs
t = np.linspace(0, 6, 501)
plt.plot(t,trapzoid_signal(t, width=2, slope=2, amp=1.), label="width=2, slope=2, amp=1")
plt.plot(t,trapzoid_signal(t, width=4, slope=1, amp=0.6), label="width=4, slope=1, amp=0.6")
plt.legend( loc=(0.25,1.015))
plt.show()
Note that you may also like to define a phase, depeding on the use case.
In order to define a single pulse, you might want to modify the function a bit and supply an array which ranges over [0,width].
from scipy import signal
import matplotlib.pyplot as plt
import numpy as np
def trapzoid_signal(t, width=2., slope=1., amp=1., offs=0):
a = slope*width*signal.sawtooth(2*np.pi*t/width, width=0.5)/4.
a += slope*width/4.
a[a>amp] = amp
return a + offs
for w,s,a in zip([2,5], [2,1], [1,0.6]):
t = np.linspace(0, w, 501)
l = "width={}, slope={}, amp={}".format(w,s,a)
plt.plot(t,trapzoid_signal(t, width=w, slope=s, amp=a), label=l)
plt.legend( loc="upper right")
plt.show()
From the SciPy website it looks like this isn't included (they currently have sawtooth and square, but not trapezoid). As a generalised version of the C example the following will do what you want,
import numpy as np
import matplotlib.pyplot as plt
def trapezoidalWave(xin, width=1., slope=1.):
x = xin%(4*width)
if (x <= width):
# Ascending line
return x*slope;
elif (x <= 2.*width):
# Top horizontal line
return width*slope
elif (x <= 3.*width):
# Descending line
return 3.*width*slope - x*slope
elif (x <= 4*width):
# Bottom horizontal line
return 0.
x = np.linspace(0.,20,1000)
for i in x:
plt.plot(i, trapezoidalWave(i), 'k.')
plt.plot(i, trapezoidalWave(i, 1.5, 2.), 'r.')
plt.show()
which looks like,
This can be done more elegantly with Heaviside functions which allow you to use NumPy arrays,
import numpy as np
import matplotlib.pyplot as plt
def H(x):
return 0.5 * (np.sign(x) + 1)
def trapWave(xin, width=1., slope=1.):
x = xin%(4*width)
y = ((H(x)-H(x-width))*x*slope +
(H(x-width)-H(x-2.*width))*width*slope +
(H(x-2.*width)-H(x-3.*width))*(3.*width*slope - x*slope))
return y
x = np.linspace(0.,20,1000)
plt.plot(x, trapWave(x))
plt.plot(x, trapWave(x, 1.5, 2.))
plt.show()
For this example, the Heaviside version is about 20 times faster!
The below example shows how to do that to get points and show scope.
Equation based on reply: Equation for trapezoidal wave equation
import math
import numpy as np
import matplotlib.pyplot as plt
def get_wave_point(x, a, m, l, c):
# Equation from: https://stackoverflow.com/questions/11041498/equation-for-trapezoidal-wave-equation
# a/pi(arcsin(sin((pi/m)x+l))+arccos(cos((pi/m)x+l)))-a/2+c
# a is the amplitude
# m is the period
# l is the horizontal transition
# c is the vertical transition
point = a/math.pi*(math.asin(math.sin((math.pi/m)*x+l))+math.acos(math.cos((math.pi/m)*x+l)))-a/2+c
return point
print('Testing wave')
x = np.linspace(0., 10, 1000)
listofpoints = []
for i in x:
plt.plot(i, get_wave_point(i, 5, 2, 50, 20), 'k.')
listofpoints.append(get_wave_point(i, 5, 2, 50, 20))
print('List of points : {} '.format(listofpoints))
plt.show()
The whole credit goes to #ImportanceOfBeingErnest . I am just revising some edits to his code which just made my day.
from scipy import signal
import matplotlib.pyplot as plt
from matplotlib import style
import numpy as np
def trapzoid_signal(t, width=2., slope=1., amp=1., offs=0):
a = slope*width*signal.sawtooth(2*np.pi*t/width, width=0.5)/4.
a += slope*width/4.
a[a>amp] = amp
return a + offs
for w,s,a in zip([32],[1],[0.0322]):
t = np.linspace(0, w, 34)
plt.plot(t,trapzoid_signal(t, width=w, slope=s, amp=a))
plt.show()
The result:
I'll throw a very late hat into this ring, namely, a function using only numpy that produces a single (symmetric) trapezoid at a desired location, with all the usual parameters. Also posted here
import numpy as np
def trapezoid(x, center=0, slope=1, width=1, height=1, offset=0):
"""
For given array x, returns a (symmetric) trapezoid with plateau at y=h (or -h if
slope is negative), centered at center value of "x".
Note: Negative widths and heights just converted to 0
Parameters
----------
x : array_like
array of x values at which the trapezoid should be evaluated
center : float
x coordinate of the center of the (symmetric) trapezoid
slope : float
slope of the sides of the trapezoid
width : float
width of the plateau of the trapezoid
height : float
(positive) vertical distance between the base and plateau of the trapezoid
offset : array_like
vertical shift (either single value or the same shape as x) to add to y before returning
Returns
-------
y : array_like
y value(s) of trapezoid with above parameters, evaluated at x
"""
# ---------- input checking ----------
if width < 0: width = 0
if height < 0: height = 0
x = np.asarray(x)
slope_negative = slope < 0
slope = np.abs(slope) # Do all calculations with positive slope, invert at end if necessary
# ---------- Calculation ----------
y = np.zeros_like(x)
mask_left = x - center < -width/2.0
mask_right = x - center > width/2.0
y[mask_left] = slope*(x[mask_left] - center + width/2.0)
y[mask_right] = -slope*(x[mask_right] - center - width/2.0)
y += height # Shift plateau up to y=h
y[y < 0] = 0 # cut off below zero (so that trapezoid flattens off at "offset")
if slope_negative: y = -y # invert non-plateau
return y + offset
Which outputs something like
import matplotlib.pyplot as plt
plt.style.use("seaborn-colorblind")
x = np.linspace(-5,5,1000)
for i in range(1,4):
plt.plot(x,trapezoid(x, center=0, slope=1, width=i, height=i, offset = 0), label=f"width = height = {i}\nslope=1")
plt.plot(x,trapezoid(x, center=0, slope=-1, width=2.5, height=1, offset = 0), label=f"width = height = 1.5,\nslope=-1")
plt.ylim((-2.5,3.5))
plt.legend(frameon=False, loc='lower center', ncol=2)
Example output: