拿哪一个取决于你。从情节本身很难说，但我的猜测是它是线性的(最简单的选择)。在

使用您的代码并添加z轴，您可以这样做import matplotlib as mpl

from mpl_toolkits.mplot3d import Axes3D

import numpy as np

import matplotlib.pyplot as plt

from math import exp,sin,cos

from pylab import *

mpl.rcParams['legend.fontsize'] = 10

fig = plt.figure()

ax = fig.gca(projection='3d')

a=0.05

b=0.10

# took the liberty of reducing the max value for th

# as it was giving you values of the order of e42

th=np.linspace(0, 50, 10000)

x=a*exp(b*th)*cos(th)

y=a*exp(b*th)*sin(th)

z=np.linspace(0,2, 10000) # creating the z array with the same length as th

ax.plot(x, y, z) # adding z as an argument for the plot

ax.legend()

plt.show()

您可以使用a和b参数来获得所需的椭圆形。您还可以使用z的定义，使其增长为指数或对数。。或者完全是别的什么。在

顺便说一句，您的导入有点多余，可能一个包中的某些函数被另一个包跟踪了。在