Matplotlib立方体面上的轮廓

pwuypxnk  于 2023-05-01  发布在  其他
关注(0)|答案(2)|浏览(165)

我正在尝试使用Python Matplotlib使用contourf函数“绘制”立方体的面。这可能吗?
这与here的想法类似,但显然我不能使用补丁。同样,我不认为我可以使用add_collection3d like this,因为它只支持PolyCollection,LineColleciton和PatchCollection。
我一直在尝试在fig.gca(projection='3d')上使用contourf。下面的玩具例子。

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

plt.close('all')
fig = plt.figure()
ax = fig.gca(projection='3d')

############################################
#  plotting the 'top' layer works okay...  #
############################################

X = np.linspace(-5, 5, 43)
Y = np.linspace(-5, 5, 28)
X, Y = np.meshgrid(X, Y)

varone=np.random.rand(75,28,43)
Z=varone[0,:,:]
cset = ax.contourf(X, Y, Z, zdir='z', offset=1,
                levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')
#see [1]
plt.show()

#################################################
#  but now trying to plot a vertical slice....  #
#################################################

plt.close('all')
fig = plt.figure()
ax = fig.gca(projection='3d')

Z=varone[::-1,:,-1]
X = np.linspace(-5, 5, 28)
Y = np.linspace(-5, 5, 75)
X, Y = np.meshgrid(X, Y)

#this 'projection' doesn't result in what I want, I really just want to rotate it
cset = ax.contourf(X, Y, Z, offset=5,zdir='x',
                levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')

#here's what it should look like....
ax=fig.add_subplot(1, 2,1)
cs1=ax.contourf(X,Y,Z,levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')
#see [2]    
plt.show()

1从示例中可以轻松获得顶面:

但是我不知道怎么做两边。此图的左侧是截面的外观(但已旋转)。..

对其他Python方法开放。我实际绘制的数据是地球物理netcdf文件。

ss2ws0br

ss2ws0br1#

您必须将数据分配到正确的轴。锯齿形的结果是,现在你在x = const,并且在z方向上振荡(来自01之间生成的随机数据)。
如果在示例中以不同方式分配矩阵,则最终会得到所需的结果:

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

plt.close('all')
fig = plt.figure()
ax = fig.gca(projection='3d')

X = np.linspace(-5, 5, 43)
Y = np.linspace(-5, 5, 28)
X, Y = np.meshgrid(X, Y)

varone=np.random.rand(75,28,43) * 5.0 - 10.0
Z=varone[0,:,:]

cset = [[],[],[]]

# this is the example that worked for you:
cset[0] = ax.contourf(X, Y, Z, zdir='z', offset=5,
                      levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')

# now, for the x-constant face, assign the contour to the x-plot-variable:
cset[1] = ax.contourf(Z, Y, X, zdir='x', offset=5,
                      levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')

# likewise, for the y-constant face, assign the contour to the y-plot-variable:
cset[2] = ax.contourf(X, Z, Y, zdir='y', offset=-5,
                      levels=np.linspace(np.min(Z),np.max(Z),30),cmap='jet')

# setting 3D-axis-limits:    
ax.set_xlim3d(-5,5)
ax.set_ylim3d(-5,5)
ax.set_zlim3d(-5,5)

plt.show()

结果如下所示:

kgqe7b3p

kgqe7b3p2#

下面给出的答案并不完全令人满意。实际上,x、y和z方向上的平面再现相同的场。
下文中,允许表示每个平面中的正确场的函数。

import numpy as np
import matplotlib.pyplot as plt

def plot_cube_faces(arr, ax):
    """
    External faces representation of a 3D array with matplotlib

    Parameters
    ----------
    arr: numpy.ndarray()
        3D array to handle
    ax: Axes3D object
        Axis to work with
    """
    x0 = np.arange(arr.shape[0])
    y0 = np.arange(arr.shape[1])
    z0 = np.arange(arr.shape[2])
    x, y, z = np.meshgrid(x0, y0, z0)
    
    xmax, ymax, zmax = max(x0), max(y0), max(z0)
    vmin, vmax = np.min(arr), np.max(arr)

    ax.contourf(x[:, :, 0], y[:, :, 0], arr[:, :, -1].T,
                zdir='z', offset=zmax, vmin=vmin, vmax=vmax)
    ax.contourf(x[0, :, :].T, arr[:, 0, :].T, z[0, :, :].T,
                zdir='y', offset=0, vmin=vmin, vmax=vmax)
    ax.contourf(arr[-1, :, :].T, y[:, 0, :].T, z[:, 0, :].T,
                zdir='x', offset=xmax, vmin=vmin, vmax=vmax)

x0 = np.arange(30)
y0 = np.arange(20)
z0 = np.arange(10)
x, y, z = np.meshgrid(x0, y0, z0)
arr = (x + y + z) // 10

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
plot_cube_faces(arr, ax)
plt.show()

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