I have a function f(x,y) where t is a parameter. I'm trying to plot the function where t = 1 for x and y values ranging from -5 to 5. The plot doesn't render.
import sympy as sp
import numpy as np
import matplotlib.pyplot as plt
%matplotlib notebook
C = sv.CoordSys3D("")
x, y, z = C.base_scalars()
t = sp.symbols("t")
f = sp.sin(2*sp.pi*t)*sp.exp(-(x-3*sp.sin(sp.pi*t))**2 -(y-3*sp.cos(sp.pi*t))**2)
fig = plt.figure(figsize=(6, 6))
ax = fig.add_subplot(projection='3d')
X = np.linspace(-5,5,100)
Y = np.linspace(-5,5,100)
xvals, yvals = np.meshgrid(X,Y)
zvals = sp.lambdify((x,y),f.subs(t,1),"numpy")(xvals,yvals)
ax.plot_surface(xvals,yvals,zvals)
plt.show()
I get an error 'int' object has no attribute 'ndim' which I don't know how to solve.
The problem is that when you execute f.subs(t,1) it returns a number (zero in this case). So, f=0 is the expression that you are going to lambdify. Let's see the function generated by lambdify:
import inspect
print(inspect.getsource(sp.lambdify((x,y),f.subs(t,1),"numpy")))
# def _lambdifygenerated(Dummy_25, Dummy_24):
# return 0
So, no matter the values and shape of xvals and yvals, that numerical function will always return 0, which is an integer number.
However, ax.plot_surface requires zvals to have the same shape as xvals or yval. Luckily, we can easily fix that with a simple if statement:
import sympy as sp
import sympy.vector as sv
import numpy as np
import matplotlib.pyplot as plt
C = sv.CoordSys3D("")
x, y, z = C.base_scalars()
t = sp.symbols("t")
f = sp.sin(2*sp.pi*t)*sp.exp(-(x-3*sp.sin(sp.pi*t))**2 -(y-3*sp.cos(sp.pi*t))**2)
fig = plt.figure(figsize=(6, 6))
ax = fig.add_subplot(projection='3d')
X = np.linspace(-5,5,100)
Y = np.linspace(-5,5,100)
xvals, yvals = np.meshgrid(X,Y)
zvals = sp.lambdify((x,y),f.subs(t,1),"numpy")(xvals,yvals)
# if zvals is just a number, create a proper matrix
if not isinstance(zvals, np.ndarray):
zvals = zvals * np.ones_like(xvals)
ax.plot_surface(xvals,yvals,zvals)
plt.show()
The fact that this doesn't render is bug in lambdify that it doesn't work well for constant expressions.
Your real problem though is that the expression you are trying to plot is just zero:
In [5]: f
Out[5]:
2 2
- (x_ - 3⋅sin(π⋅t)) - (y_ - 3⋅cos(π⋅t))
ℯ ⋅sin(2⋅π⋅t)
In [6]: f.subs(t, 1)
Out[6]: 0
Related
Let's assume I have 3 arrays defined as:
v1=np.linspace(1,100)
v2=np.linspace(1,100)
v3=np.linspace(1,100)
Then I have a function that takes those 3 values and gives me the desired output, let's assume it is like:
f = (v1 + v2*10)/v3
I want to plot that function on a 3D plot with axis v1,v2,v3 and color it's surface depending on its value.
More than the best way to plot it, I was also interested in how to scroll all the values in the in vectors and build the function point by point.
I have been trying with for loops inside other for loops but I am always getting one error.
MANY THANKS
I tried this but i'm always getting a line instead of a surface
import mpl_toolkits.mplot3d.axes3d as axes3d
import sympy
from sympy import symbols, Function
# Parameters I use in the function
L = 132
alpha = 45*math.pi/180
beta = 0
s,t = symbols('s,t')
z = Function('z')(s,t)
figure = plt.figure(figsize=(8,8))
ax = figure.add_subplot(1, 1, 1, projection='3d')
# experiment with various range of data in x and y
x1 = np.linspace(-40,-40,100)
y1 = np.linspace(-40,40,100)
x,y = np.meshgrid(x1,y1)
# My function Z
c1=math.cos(beta)**2
c2=math.cos(alpha)**2
s1=math.sin(alpha)**2
den = math.sqrt((c1*c2)+s1)
z=L*((math.cos(beta)/den)-1)+(s*(math.sin(alpha)))+(t*(1-math.cos(alpha)))
ax.plot_surface(x,y,z,cmap='rainbow')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.show()
In this example I'm going to show you how to achieve your goal. Specifically, I use Numpy because it supports vectorized operations, hence I avoid for loops.
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import Normalize
import matplotlib.cm as cm
# Parameters I use in the function
L = 132
alpha = 45*np.pi/180
beta = 0
figure = plt.figure()
ax = figure.add_subplot(1, 1, 1, projection='3d')
# experiment with various range of data in x and y
x1 = np.linspace(-40,40,100)
y1 = np.linspace(-40,40,100)
x,y = np.meshgrid(x1,y1)
# My function Z
c1=np.cos(beta)**2
c2=np.cos(alpha)**2
s1=np.sin(alpha)**2
den = np.sqrt((c1*c2)+s1)
z=L*((np.cos(beta)/den)-1)+(x*(np.sin(alpha)))+(y*(1-np.cos(alpha)))
# compute the color values according to some other function
color_values = np.sqrt(x**2 + y**2 + z**2)
# normalize color values between 0 and 1
norm = Normalize(vmin=color_values.min(), vmax=color_values.max())
norm_color_values = norm(color_values)
# chose a colormap and create colors starting from the normalized values
cmap = cm.rainbow
colors = cmap(norm_color_values)
surf = ax.plot_surface(x,y,z,facecolors=colors)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
# add a colorbar
figure.colorbar(cm.ScalarMappable(norm=norm, cmap=cmap), label="radius")
plt.show()
I'm trying to create a plot in Python to illustrate how my sequence changes as n grows. I get a dimension error. How can I fix this?
My code:
import matplotlib.pyplot as plt
import numpy as np
x = np.zeros(101)
x[0] = 0
for n in range(0, 101):
x[n] = x[n-1] - n
if x[n] < 0:
x[n] = x[n-1] + n
y = set(x)
print(y)
i = np.linspace(0, 100)
plt.plot(y, i, 'g')
Error:
ValueError: x and y must have same first dimension, but have shapes (1,) and (50,)
The problem is you are using plt.plot with a set and because of that its saying first dimension has shape (1,). The x-argument in plt.plot should be a array-like or scalar you can read more in the documentation so you can convert it to a list:
y = list(set(x))
Also you need to make sure that the for each value in y there is a corresponding value in i (y and i need to be the same shape). So you need to set np.linspace to return len(y) values:
i = np.linspace(0, 100, len(y))
The code:
import matplotlib.pyplot as plt
import numpy as np
x = np.zeros(101)
x[0] = 0
for n in range(0,101):
x[n] = x[n-1] - n
if x[n]<0:
x[n] = x[n-1] + n
y = list(set(x))
i = np.linspace(0, 100, len(y))
plt.plot(y, i,'g')
plt.show()
Output:
import numpy as np
from matplotlib import pyplot as plt
data = np.random.normal(0,1,[100,3])
x = data[:,0]
y = data[:,1]
z = data[:,2]
plt.contour([x,y],z)
When I run this code with dummy data I get:
ValueError: Contour levels must be increasing
Do you have any idea what would this mean and how I could fix it?
plt.contour is a bit particular about its input, the z values must be on values on a rectangular 2D grid, see for example:
import matplotlib.pyplot as plt
import numpy as np
x = np.expand_dims(np.arange(1,11,1), axis=1)
y = np.expand_dims(np.arange(2,21,2), axis=0)
z = y * x
print(x.shape)
print(y.shape)
print(z.shape)
plt.figure()
plt.contour(z)
plt.show()
You can also provide x and y values for plt.contour by using np.meshgrid :
XX,YY = np.meshgrid(x,y)
plt.figure()
plt.contour(XX, YY, z)
plt.show()
If you have z-values with irregular values for x and y, you might use plt.tricontour, see the following example:
from matplotlib.tri import Triangulation
data = np.random.normal(0,1,[100,3])
x = data[:,0]
y = data[:,1]
#z = data[:,2]
z = x * y
tri = Triangulation(x,y)
plt.figure()
plt.tricontour(tri, z, )
plt.scatter(x,y, c=z)
plt.show()
Edit: from JohanC's comment i learned that this can be simplified without importing matplotlib.tri by:
plt.figure()
plt.tricontour(x,y,z)
plt.scatter(x,y, c=z)
plt.show()
I have a simple question but have not found any answer..
Let's have a look at this code :
from matplotlib import pyplot
import numpy
x=[0,1,2,3,4]
y=[5,3,40,20,1]
pyplot.plot(x,y)
It is plotted and all the points ared linked.
Let's say I want to get the y value of x=1,3.
How can I get the x values matching with y=30 ? (there are two)
Many thanks for your help
You could use shapely to find the intersections:
import matplotlib.pyplot as plt
import numpy as np
import shapely.geometry as SG
x=[0,1,2,3,4]
y=[5,3,40,20,1]
line = SG.LineString(list(zip(x,y)))
y0 = 30
yline = SG.LineString([(min(x), y0), (max(x), y0)])
coords = np.array(line.intersection(yline))
print(coords[:, 0])
fig, ax = plt.subplots()
ax.axhline(y=y0, color='k', linestyle='--')
ax.plot(x, y, 'b-')
ax.scatter(coords[:, 0], coords[:, 1], s=50, c='red')
plt.show()
finds solutions for x at:
[ 1.72972973 2.5 ]
The following code might do what you want. The interpolation of y(x) is straight forward, as the x-values are monotonically increasing. The problem of finding the x-values for a given y is not so easy anymore, once the function is not monotonically increasing as in this case. So you still need to know roughly where to expect the values to be.
import numpy as np
import scipy.interpolate
import scipy.optimize
x=np.array([0,1,2,3,4])
y=np.array([5,3,40,20,1])
#if the independent variable is monotonically increasing
print np.interp(1.3, x, y)
# if not, as in the case of finding x(y) here,
# we need to find the zeros of an interpolating function
y0 = 30.
initial_guess = 1.5 #for the first zero,
#initial_guess = 3.0 # for the secon zero
f = scipy.interpolate.interp1d(x,y,kind="linear")
fmin = lambda x: np.abs(f(x)-y0)
s = scipy.optimize.fmin(fmin, initial_guess, disp=False)
print s
I use python 3.
print(numpy.interp(1.3, x, y))
Y = 30
eps = 1e-6
j = 0
for i, ((x0, x1), (y0, y1)) in enumerate(zip(zip(x[:-1], x[1:]), zip(y[:-1], y[1:]))):
dy = y1 - y0
if abs(dy) < eps:
if y0 == Y:
print('There are infinite number of solutions')
else:
t = (Y - y0)/dy
if 0 < t < 1:
sol = x0 + (x1 - x0)*t
print('solution #{}: {}'.format(j, sol))
j += 1
I'm plotting some functions that have several discontinuities. Each function is given as a list. I want to connect points with lines only where the function is continuous.
Here is a simplified example of what plot is doing.
x=linspace(0,1,100)
y=zeros(100)
y[x<0.5] = x[x<0.5]
y[x>=0.5] = 1 + x[x>=0.5]
plot(x, y, '-o')
There is a discontinuity at x=0.5, but plot connects all points with lines regardless.
My functions are different of course. They typically have several discontinuities in different places. The criterion for the discontinuity is simple. Say, if the function jumps by more than 0.5, I assume it is discontinuous at that point.
Is there an option in plot to tell it to drop the connecting lines between the points where the function is discontinuous? I recall being able to do that easily with gnuplot.
use nan to break the line into multiple segments:
import numpy as np
from pylab import *
x=linspace(0,1,100)
y=zeros(100)
y[x<0.5] = x[x<0.5]
y[x>=0.5] = 1 + x[x>=0.5]
pos = np.where(np.abs(np.diff(y)) >= 0.5)[0]
x[pos] = np.nan
y[pos] = np.nan
plot(x, y, '-o')
Edit:
to insert nan at discontinuities:
pos = np.where(np.abs(np.diff(y)) >= 0.5)[0]+1
x = np.insert(x, pos, np.nan)
y = np.insert(y, pos, np.nan)
Here is my suggestion for plotting tan(x):
import matplotlib.pyplot as plt
from math import *
x_lim = 3*pi/2
y_lim = 5
n = 1000
X = []
Y = []
Z = []
for i in range(0,2*n):
x = -x_lim + i*x_lim/n
y = tan(x)
if y<y_lim and y>-y_lim:
X.append(x)
Y.append(y)
else:
if len(X)>0 and len(Y)>0:
Z.append([X,Y])
del X,Y
X = []
Y = []
for i in range(0, len(Z)):
plt.plot(Z[i][0],Z[i][1])
plt.grid(True)
plt.show()