I am trying to make a 3d surface plot of experimental data using matplotlib. I would like to plot different Z values against the same X and Y axes. When I try the simple code below, I get the error
"plot_surface() missing 1 required positional argument: 'Z' ".
It seems that the Axes3D package only work if Z is given as a function of X and Y, rather than an actual data matrix. Does anybody know a way around this?
Please note that the Zmatrix that I need is actual data, but I just used a random matrix for illustration here.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
X=[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
Y= [0,2500,5000,7500,10000,15000,20000,25000,30000,35000,40000,45000,50000,55000,60000,65000,70000]
Zmatrix=np.random.rand(len(X),len(Y))
Axes3D.plot_surface(X,Y,Zmatrix)
There were sone issues with your code:
First you have to get a meshgrid of X and Y (all combinations as matrices). Next swap len(X) and len(Y) inside the Zmatrix. And first define ax = Axes3D(plt.gcf()) and plot_surface afterwards on ax.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
X=[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
Y= [0,2500,5000,7500,10000,15000,20000,25000,30000,35000,40000,45000,50000,55000,60000,65000,70000]
Xm, Ym = np.meshgrid(X, Y)
Zmatrix=np.random.rand(len(Y),len(X))
ax = Axes3D(plt.gcf())
ax.plot_surface(Xm, Ym, Zmatrix)
Here is an example of surface plot.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import random
def fun(x, y):
return x**2 + y
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(-3.0, 3.0, 0.05)
X, Y = np.meshgrid(x, y)
zs = np.array([fun(x,y) for x,y in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)
ax.plot_surface(X, Y, Z)
ax.set_xlabel('X Label')
ax.set_ylabel('Y Label')
ax.set_zlabel('Z Label')
plt.show()
Related
I'm very new in Python and trying to plot a single curve on a surface.
Here is where I came so far and plotted a surface in s domain:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import cmath
x = np.linspace(-400, 0, 100)
y = np.linspace(-100, 100, 100)
X, Y = np.meshgrid(x,y)
fc=50
wc=2*np.pi*fc
s = X + Y*1j
Z= abs(1/(1+s/wc))
fig = plt.figure()
ax = fig.gca(projection='3d')
surf = ax.plot_surface(X, Y, Z)
ax.plot(X, Y, Z)
plt.ylabel('Im')
plt.show()
I now need to plot the curve for X = 0 in different color which means the curve on the same surface along the imaginary axis. surf = ax.plot_surface(0, Y, Z) did not work. Does anybody have experience with such plot?
I'm assuming you meant you wanted to plot y=0 instead of x=0 (since x=0 would be pretty boring).
Since you want to plot a single slice of your data, you can't use the meshgrid format (or if you can, it would require some weird indexing that I don't want to figure out).
Here's how I would plot the y=0 slice:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import cmath
x = np.linspace(-400, 0, 100)
y = np.linspace(-100, 100, 100)
X, Y = np.meshgrid(x,y)
fc=50
wc=2*np.pi*fc
s = X + Y*1j
Z= abs(1/(1+s/wc))
fig = plt.figure()
ax = fig.gca(projection='3d')
surf = ax.plot_surface(X, Y, Z)
# create data for y=0
z = abs(1/(1+x/wc))
ax.plot(x,np.zeros(np.shape(x)),z)
plt.ylabel('Im')
plt.show()
So I have an array of values that I am trying to plot using the plot_surface command. Specifically I have a 30x30 array with one in the middle and zeros elsewhere. When I plot it this is what I obtain:
I would like however for the value to be represented as a cuboid. Is that possible?
Thank you
edit: Code that shows the figure
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
N=30
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(0, N)
z = np.zeros((N,N))
z[15,15] = 1
X, Y = np.meshgrid(x, y)
ax.plot_surface(X, Y, z, cmap='YlOrBr')
plt.show(block=False)
I think a 3D bar plot will give what you're looking for.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
N=30
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = y = np.arange(0, N)
z_bottom = np.zeros((N,N))
z_top = z_bottom.copy()
z_top[15,15] = 1
X, Y = np.meshgrid(x, y)
ax.bar3d(X.ravel(), Y.ravel(), z_bottom.ravel(), 1, 1, z_top.ravel())
plt.show(block=False)
The full documentation is here if you want to play with the colors and so forth.
I am using mplot3d from the mpl_toolkits library. When displaying the 3D surface on the figure I'm realized the axis were not positioned as I wished they would.
Let me show, I have added to the following screenshot the position of each axis:
Is there a way to change the position of the axes in order to get this result:
Here's the working code:
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
ax = Axes3D(plt.figure())
def f(x,y) :
return -x**2 - y**2
X = np.arange(-1, 1, 0.02)
Y = np.arange(-1, 1, 0.02)
X, Y = np.meshgrid(X, Y)
Z = f(X, Y)
ax.plot_surface(X, Y, Z, alpha=0.5)
# Hide axes ticks
ax.set_xticks([-1,1])
ax.set_yticks([-1,1])
ax.set_zticks([-2,0])
ax.set_yticklabels([-1,1],rotation=-15, va='center', ha='right')
plt.show()
I have tried using xaxis.set_ticks_position('left') statement, but it doesn't work.
No documented methods, but with some hacking ideas from https://stackoverflow.com/a/15048653/1149007 you can.
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = ax = fig.add_subplot(111, projection='3d')
ax.view_init(30, 30)
def f(x,y) :
return -x**2 - y**2
X = np.arange(-1, 1, 0.02)
Y = np.arange(-1, 1, 0.02)
X, Y = np.meshgrid(X, Y)
Z = f(X, Y)
ax.plot_surface(X, Y, Z, alpha=0.5)
# Hide axes ticks
ax.set_xticks([-1,1])
ax.set_yticks([-1,1])
ax.set_zticks([-2,0])
ax.xaxis._axinfo['juggled'] = (0,0,0)
ax.yaxis._axinfo['juggled'] = (1,1,1)
ax.zaxis._axinfo['juggled'] = (2,2,2)
plt.show()
I can no idea of the meaning of the third number in triples. If set zeros nothing changes in the figure. So should look in the code for further tuning.
You can also look at related question Changing position of vertical (z) axis of 3D plot (Matplotlib)? with low level hacking of _PLANES property.
Something changed, code blow doesn't work, all axis hide...
ax.xaxis._axinfo['juggled'] = (0,0,0)
ax.yaxis._axinfo['juggled'] = (1,1,1)
ax.zaxis._axinfo['juggled'] = (2,2,2)
I suggest using the plot function to create a graph
I have searched for this in google, but found solutions for 2d points in real time.How can I achieve this for stream of 3d point.
Here I should be able to add new points to plot.
I tried this, its just plots series of data. How to update?
You could just plot in interactive mode, for example the following keeps adding new points,
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
plt.ion()
plt.show()
x = np.linspace(0.,np.pi*4.,100)
ax.set_xlim([0.,13.])
ax.set_ylim([-1.5,1.5])
ax.set_zlim([-1.5,1.5])
for i in x:
ax.scatter(i, np.sin(i), np.cos(i))
print(i)
plt.pause(0.01)
UPDATE: added example of labelling
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
plt.ion()
plt.show()
lsp = np.linspace(0.,np.pi*4.,100)
ax.set_xlim([0.,13.])
ax.set_ylim([-1.5,1.5])
ax.set_zlim([-1.5,1.5])
for i, x in enumerate(lsp):
y = np.sin(x)
z = np.cos(x)
ax.scatter(x, y, z)
if i%10 == 0:
ax.text(x, y, z, str(np.round(x,3))+", "
+str(np.round(y,3))+", "
+str(np.round(z,3)))
plt.pause(0.01)
I am using matplotlib for doing this
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
fig = plt.figure()
ax = Axes3D(fig)
x = [6,3,6,9,12,24]
y = [3,5,78,12,23,56]
ax.plot(x, y, zs=0, zdir='z', label='zs=0, zdir=z')
plt.show()
Now this builds a graph that is horizontal in the 3d space. How do I make the graph vertical so that it faces the user?
What I want to do is build multiple such vertical graphs that are separated by some distance and are facing the user.
bp's answer might work fine, but there's a much simpler way.
Your current graph is 'flat' on the z-axis, which is why it's horizontal. You want it to be vertical, which means that you want it to be 'flat' on the y-axis. This involves the tiniest modification to your code:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
fig = plt.figure()
ax = Axes3D(fig)
x = [6,3,6,9,12,24]
y = [3,5,78,12,23,56]
# put 0s on the y-axis, and put the y axis on the z-axis
ax.plot(xs=x, ys=[0]*len(x), zs=y, zdir='z', label='ys=0, zdir=z')
plt.show()
Then you can easily have multiple such graphs by using different values for the ys parameter (for example, ys=[2]*len(x) instead would put the graph slightly behind).
Mayavi, in particular the mlab module, provides powerful 3D plotting that will work on large and or complex data, and should be easy to use on numpy arrays.
You can set the view angle of the 3d plot with the view_init() function. The example below is for version 1.1 of matplotlib.
from mpl_toolkits.mplot3d import axes3d
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
x = [6,3,6,9,12,24]
y = [3,5,78,12,23,56]
ax.plot(x, y, zs=0, zdir='z', label='zs=0, zdir=z')
ax.view_init(90, -90)
plt.show()
According to the documentation you want to use the ax.plot_surface(x,y,z) method. More information and chart types here.
The following should work:
x = [1,2,3]
y = [4,5,6]
z = [7,8,9]
data = zip(x,y,z)
#map data on the plane
X, Y = numpy.meshgrid(arange(0, max(x), 1), arange(0, max(y), 1))
Z = numpy.zeros((len(Y), len(X)), 'Float32')
for x_,y_,z_ in data:
Z[x_, y_] = z_ #this should work, but only because x and y are integers
#and arange was done with a step of 1, starting from 0
fig = p.figure()
ax = p3.Axes3D(fig)
ax.plot_surface(X, Y, Z)