python - Matplotlib绘图脉冲在3D中的传播

我想在每一步都以这种方式绘制脉冲传播图，它绘制出脉冲形状。换句话说，对于y的每个值，我想要一个x-z图的序列。像这样的东西(无颜色):


如何使用matplotlib(或Mayavi)做到这一点？这是我到目前为止所做的:

def drawPropagation(beta2, C, z):
    """ beta2 in ps / km
        C is chirp
        z is an array of z positions """
    T = numpy.linspace(-10, 10, 100)
    sx = T.size
    sy = z.size

    T = numpy.tile(T, (sy, 1))
    z = numpy.tile(z, (sx, 1)).T

    U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))

    fig = pyplot.figure()
    ax = fig.add_subplot(1,1,1, projection='3d')
    surf = ax.plot_wireframe(T, z, abs(U))

最佳答案

改成:

ax.plot_wireframe(T, z, abs(U), cstride=1000)

并致电:

drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))

将创建以下图形:

如果您需要用白色填充曲线，请执行以下操作:

import numpy
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import pyplot
from matplotlib.collections import PolyCollection

def drawPropagation(beta2, C, z):
    """ beta2 in ps / km
        C is chirp
        z is an array of z positions """
    T = numpy.linspace(-10, 10, 100)
    sx = T.size
    sy = z.size

    T = numpy.tile(T, (sy, 1))
    z = numpy.tile(z, (sx, 1)).T

    U = 1 / numpy.sqrt(1 - 1j*beta2*z * (1 + 1j * C)) * numpy.exp(- 0.5 * (1 + 1j * C) * T * T / (1 - 1j*beta2*z*(1 + 1j*C)))

    fig = pyplot.figure()
    ax = fig.add_subplot(1,1,1, projection='3d')
    U = numpy.abs(U)

    verts = []
    for i in xrange(T.shape[0]):
        verts.append(zip(T[i, :], U[i, :]))

    poly = PolyCollection(verts, facecolors=(1,1,1,1), edgecolors=(0,0,1,1))
    ax.add_collection3d(poly, zs=z[:, 0], zdir='y')
    ax.set_xlim3d(numpy.min(T), numpy.max(T))
    ax.set_ylim3d(numpy.min(z), numpy.max(z))
    ax.set_zlim3d(numpy.min(U), numpy.max(U))

drawPropagation(1.0, 1.0, numpy.linspace(-2, 2, 10))
pyplot.show()

关于python - Matplotlib绘图脉冲在3D中的传播，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/13240633/