我的 Tetration（复数）函数必须更好地矢量化（Python）

Question

受 3blue1brown 的启发，我正在尝试用 Python绘制 Tetration函数的逃逸（散度）图——类似于维基百科上这张漂亮的图形。def tetration_com(base, tol=10**-15, max_step=10**6):  # returns t, the infinite tetration of base.  # if t does not converge, the function returns an escape value, aka how fast it diverges..  t = 1.0  step = 0  escape = None  ln_base = cmath.log(base)  t_last = 0  try:    while(abs(t - t_last) > tol):      if(step > max_step):        raise OverflowError      t_last = t      t = cmath.exp(ln_base*t)   # [ base^t == e^(ln(base)*t) ]      step += 1  except(OverflowError):    t = None    escape = 1000/step    # the escape value is is inversely related to the number of steps it took    # us to diverge to infinity  return t, escape我试图让它与 meshgrid 一起工作，以便绘制 xy 平面上的逃逸图。Python 不喜欢输出是 2 个解压缩的变量（限制或转义）——我绝对可以通过拆分成两个函数来解决这个问题。但另一个问题是复杂的数学运算（cmath.log，cmath.exp）只适用于标量......我试图向量化函数：nx, ny = 700, 500x, y = np.linspace(-3.5, 3.5, nx), np.linspace(-2.5, 2.5, ny)xv, yv = np.meshgrid(x, y)tetration_vec = np.vectorize(tetration_com)t, escape = tetration_vec(xv + yv*1j, max_step=500)但它永远在运行。关于如何处理复杂数学运算和矢量化的任何建议？

Answer 1

这是我最后如何绘制逃生图：

def tetration_com(base, tol=10**-15, max_step=10**6, max_val=10**2):

  # returns t, the infinite tetration of base.

  # if t does not converge, the function returns an escape value

  # aka how fast it diverges..

  t = 1.0

  step = 0

  escape = None

  t_last = 0

  try:

    while(abs(t - t_last) > tol):

      if(step > max_step or abs(t) > max_val):

        raise OverflowError

      t_last = t

      t = pow(base, t)

      step += 1

  except(OverflowError):

    t = None

    escape = 1000/step

    # the escape value is is inversely related to the number of steps it took

    # us to diverge to infinity

  return t, escape

向量化辅助函数：

def tetra_graph_escape(real, imag, tol=10**-15, max_step=10**3, max_val=10**2):

  return np.array([np.array([tetration_com(r + im*1j, tol=tol, max_step=max_step, max_val=max_val)[1]

                             for im in imag]) for r in real])

绘图：

# graph our escape:

nx, ny = 700, 500

x, y = np.linspace(-3.5, 3.5, nx), np.linspace(-2.5, 2.5, ny)

val, escape = tetra_graph_conv(x, y), tetra_graph_escape(x, y)

import matplotlib.pyplot as plt

for r in range(len(escape)):

  for c in range(len(escape[0])):

    if escape[r][c] is None:

      escape[r][c] = -100

escape[460][250]

plt.contour(escape)