Generating a 3D sine curve using python - python
I want to generate a 3D sine curve in python. Does lumpy support it or is there another library I can use for it?
Generating a 2D curve is straightforward something like this --
x = numpy.linspace(0, 20, 0.1)
y = numpy.sin(x)
I now have x and y I can save to disk.
Now I want to do something similar but for a 3D sine curve over x, y and z axes. I am hoping someone might have done it before and help me.
EDIT: I have realized that for my use case I want a 2D curve in 3D space. So the other axes can be constant. So I am simply generating a 2D curve and adding a constant third parameter value to get the x,y,z values.
You can try something like:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
x = np.arange(0, 20, 0.1)
y = np.sin(x)
z = y*np.sin(x)
fig = plt.figure()
ax = plt.axes(projection='3d')
c = x + y
ax.scatter(x, y, z, c=c)
or maybe you want z = x*np.sin(x) or even z = np.sin(y)
Edit: maybe this is the best solution z = np.sin(np.sqrt(x**2+y**2)) from here
Have a play and to find what you want. Pretty funky stuff and depends on exactly what output you are looking for.
Related
3D surface/volume plot of list
I would like to represent a set of 3D points as a surface. The points are in an array format for x, y, z. I managed to plot the points in 3D in a sub-optimal solution by adding an array c which just contains 1's for each (x,y,z). There's surely a better solution for that already. What I want to do now is connect the points somehow and fill out a volume which contains these points in 3D. import numpy as np import matplotlib.pyplot as plt x = np.array([3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8]) y = np.array([15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,21,21,21,15,15,15,15,15,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,26,26,17,17,18,18,18,18,18,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,20,20,20,21,21,21,21,21,21,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,31,31,31,31,31,31,31,31,31,31,32,32,32,32,32,32,32,32,32,32,33,33,33,33,33,33,33,33,33,33,22,23,23,23,23,24,24,24,24,24,24,24]) z = np.array([5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,8,9,10,1,2,3,4,5,1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,4,5,6,7,8,9,10,9,10,1,2,1,2,3,4,5,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,6,7,8,9,10,1,2,3,1,2,3,4,5,6,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,1,2,3,4,1,2,3,4,5,6,7]) c = np.array([1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]) fig = plt.figure() ax = fig.add_subplot(111, projection='3d') img = ax.scatter(x, y, z, c=c) plt.show() This is what I get: What I approximately try to get: So basically I'm looking for something that encloses all my points. I hope this is not a duplicate, at least I did not find any answer that worked out for me. Edit: Just added Mayavi to the tags, since I'm trying to find a workaround with the package.
Adjusting Plotted Values of Contour Plots
I'm making contour plots which are basically analytical or numerical solutions to a fluid dynamic system. I don't think the technical stuff really matters too much, but here's my plots. The first plot is the numerical (Matrix system) solution, and the second plot is the nice closed form (single forumla) solution.
As can be seen, my second plot has the bubbles on the right hand side. Looking at the legend/scale, I have negative values. I'd like to not have negative values, or not plot them, although I'm not sure how to adjust this within my code. I've spent some time looking into how to adjust the z values to being positive only, but I can't seem to get it. I'll drop my plot code, and then my nice closed form function that is used in the plot.
import numpy as np
import matplotlib.pyplot as plt
import scipy as sp
import scipy.special as sp1
from mpl_toolkits.mplot3d import Axes3D
def v(r,z,gamma):
a=r*(1-z/gamma)
sums = 0
for n in range(1,26):
sums += ((sp1.iv(1(n*np.pi*r)/gamma))/(n*sp1.iv(1(n*np.pi)/gamma)))*np.sin(n*np.pi*z/gamma)
return a-(2/np.pi)*sums
def plot_contour(a, filename=None, zlabel='v(r,z)',cmap=plt.cm.gnuplot):
fig = plt.figure(figsize=(5,4))
ax = fig.add_subplot(111)
x = np.arange(a.shape[0])
y = np.arange(a.shape[1])
X, Y = np.meshgrid(x, y)
Z = a[X, Y]
cset = ax.contourf(X, Y, Z, 20, cmap=cmap)
ax.set_xlabel('r')
ax.set_ylabel('z')
ax.set_title('\u0393=2.5')
ax.axis('off')
ax.set_aspect(1)
cb = fig.colorbar(cset, shrink=0.5, aspect=5)
cb.set_label(zlabel)
if filename:
fig.savefig(filename,dpi=1600)
plt.close(fig)
return filename
else:
return ax
...
plot_contour(v1, 'gamma25e+1')
This is all the necessary code. The rest of it is the matrix solution stuff, which is just a bunch of linear algebra. Any help on what I need to add or adjust to prevent negative values from showing up on the second plot. It should look exactly like the first.
I've spent some time looking into how to adjust the z values to being positive only what you can do depends greatly on what you want to do with the results below zero, if your sole purpose is to make the points below zero show as zero, you can simply make them zero, however that would be showing a wrong result. x = np.arange(a.shape[0]) y = np.arange(a.shape[1]) X, Y = np.meshgrid(x, y) Z = a[X, Y] Z[Z < 0] = 0 another solution is to subtract the minimum value of you data so that the minimum value of the result is 0. x = np.arange(a.shape[0]) y = np.arange(a.shape[1]) X, Y = np.meshgrid(x, y) Z = a[X, Y] Z -= np.amin(Z)
colormap from a matrix in python
I have a 2D output matrix (say, Z) which was calculated as a function of two variables x,y. x varies in a non-uniform manner like [1e-5,5e-5,1e-4,5e-4,1e-3,5e-3,1e-2] y varies in a uniform manner like [300,400,500,600,700,800] [ say, Z = np.random.rand(7,6) ] I was trying to plot a colormap of the matrix Z by first creating a meshgrid for x,y and then using the pcolormesh. Since, my x values are non-uniform, I do not kn ow how to proceed. Any inputs would be greatly appreciated.
No need for meshgrids; regarding the non-uniform axes: In your case a log-scale works fine:
import numpy as np
from matplotlib import pyplot as plt
x = [1e-5,5e-5,1e-4,5e-4,1e-3,5e-3,1e-2]
y = [300,400,500,600,700,800]
# either enlarge x and y by one number (right-most
# endpoint for those bins), or make Z smaller as I did
Z = np.random.rand(6,5)
fig = plt.figure()
ax = fig.gca()
ax.pcolormesh(x,y,Z.T)
ax.set_xscale("log")
fig.show()
Plot and function with three variables in python
An equation which is represent as below sin(x)*sin(y)*sin(z)+cos(x)*sin(y)*cos(z)=0 I know the code to plot function for z=f(x,y) using matplotlib but to plot above function I don’t know the code, but I tried MATLAB MuPad code which is as follows Plot(sin(x)*sin(y)*sin(z)+cos(x)*sin(y)*cos(z),#3d)
This will be much easier if you can isolate z. Your equation is the same as sin(z)/cos(z) = -cos(x)*sin(y)/(sin(x)*sin(y)) so z = atan(-cos(x)*sin(y)/(sin(x)*sin(y))).
Please don't mistake me, but I think your given equation to plot can be reduced to a simple 2D plot. sin(x)*sin(y)*sin(z)+cos(x)*sin(y)*cos(z) = 0 sin(y)[sin(x)*sin(z)+cos(x)*cos(z)] = 0 sin(y)*cos(x-z) = 0 Hence sin(y) = 0 or cos(x-z)=0 Hence y = n*pi (1) or x-z=(2*n + 1)pi/2 Implies, x = z + (2*n + 1)pi/2 (2) For (1), it will be a straight line (the plot of y vs n) and in second case, you will get parallel lines which cuts x-axis at (2*n + 1)pi/2 and distance between two parallel lines would be pi. (Assuming you keep n constant). Assuming, your y can't be zero, you could simplify the plot to a 2D plot with just x and z. And answering your original question, you need to use mplot3d to plot 3D plots. But as with any graphing tool, you need values or points of x, y, z. (You can compute the possible points by programming). Then you feed those points to the plot, like below. from mpl_toolkits import mplot3d import numpy as np import matplotlib.pyplot as plt fig = plt.figure() ax = plt.axes(projection="3d") xs = [] # X values ys = [] # Y values zs = [] # Z values ax.plot3D(xs, ys, zs) plt.show()
How to do 4D plot in matplotlib without np.meshgrid?
I know that I can do a 4D plot in matplotlib with the following code, with the fourth dimension shown as a colormap: import numpy as np from mpl_toolkits.mplot3d import Axes3D import matplotlib.pyplot as plt from matplotlib import cm fig = plt.figure() ax = fig.add_subplot(111,projection= '3d' ) x = np.arange(100)/ 101 y = np.sin(x) + np.cos(x) X,Y = np.meshgrid(x,y) Z = (X**2) / (Y**2) A = np.sin(Z) ax.plot_surface(X,Y, Z, facecolors=cm.Oranges(A)) plt.show() But what if my data is not a function of the other data? How do I do this without np.meshgrid? (In other words, my Z series cannot be a function of the output of the X,Y which is the output of np.meshgrid(x,y), because Z is not a function of X and Y.)
A surface plot is a mapping of 2D points to a 1D value, i.e. for each pair of (x,y) coordinates you need exactly one z coordinate. So while it isn't strictly necessary to have Z being a function of X and Y, those arrays to plot need to have the same number of elements. For a plot_surface the restriction is to have X and Y as gridded 2D data. Z does not have to be 2D but needs to have the same number of elements. This requirement can be weakened using a plot_trisurf where the only requirement is that there is a strict mapping of x,y,z, i.e. the ith value in X and Y corresponds to the ith value in Z. In any case, even if there is no analytic function to map X and Y to Z, Z still needs to be some kind of mapping. Otherwise it is even questionable what information the resulting plot would convey.