我有这个代码:import matplotlib.pyplot as plt import mpl_toolkits.mplot3d.axes3d as plt3dimport mpl_toolkits.mplot3d.art3d as artplt3dimport matplotlib.animation as animationimport numpy as npimport math def animateBeta(): def translate(segment_, xTr_, yTr_, zTr_): translationMatrix = np.array([ [1, 0, xTr_], [0, 1, yTr_], [0, 0, zTr_] ]) return np.matmul(translationMatrix, segment_) def rotate(vec_, xRotate_, yRotate_, zRotate_): xMat = np.array([ [1,0,0], [0,math.cos(xRotate_), -math.sin(xRotate_)], [0, math.sin(xRotate_), math.cos(xRotate_)] ]) yMat = np.array([ [math.cos(yRotate_), 0, math.sin(yRotate_)], [0, 1, 0], [-math.sin(yRotate_), 0, math.cos(yRotate_)] ]) zMat = np.array([ [math.cos(zRotate_), -math.sin(zRotate_), 0], [math.sin(zRotate_), math.cos(zRotate_), 0], [0, 0, 1] ]) rotationMatrix = np.matmul(zMat, np.matmul(yMat, xMat)) return np.matmul(rotationMatrix, vec_) segment = [ [0,0], [0,0], [1, -1] ] fig = plt.figure() ax = plt3d.Axes3D(fig) line, = ax.plot(segment[0], segment[1], zs=segment[2], color = 'b') artplt3d.line_2d_to_3d(line) print(line.__class__) def animate(i): s0 = [0,0,1] s1 = [0,0,-1] s1 = translate(rotate(translate(s1, -s0[0], -s0[1], -s0[2]), -i*(math.pi/180),0,-i*(math.pi/180)), s0[0], s0[1], s0[2]) segment = np.concatenate((np.reshape(s0, (3,-1)),np.reshape(s1, (3,-1))), axis=1)当使用注释data = ax.plot(segment[0], segment[1], segment[2], color = 'b')行而不是以下两行时,它可以工作(但我试图这样做,以便在顶部绘制新行时不会绘制先前的行)。如果您按原样使用代码,则动画看起来很奇怪。我有一个理论并line_2d_to_3d没有按预期发挥作用,但我不确定。
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