受 3blue1brown 的啟發(fā)，我正在嘗試用 Python繪制 Tetration函數(shù)的逃逸（散度）圖——類(lèi)似于維基百科上這張漂亮的圖形。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我試圖讓它與 meshgrid 一起工作，以便繪制 xy 平面上的逃逸圖。Python 不喜歡輸出是 2 個(gè)解壓縮的變量（限制或轉(zhuǎn)義）——我絕對(duì)可以通過(guò)拆分成兩個(gè)函數(shù)來(lái)解決這個(gè)問(wèn)題。但另一個(gè)問(wèn)題是復(fù)雜的數(shù)學(xué)運(yùn)算（cmath.log，cmath.exp）只適用于標(biāo)量......我試圖向量化函數(shù)：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)但它永遠(yuǎn)在運(yùn)行。關(guān)于如何處理復(fù)雜數(shù)學(xué)運(yùn)算和矢量化的任何建議？