Inspired by 3blue1brown我正在尝试绘制 Tetration 的逃逸(发散)图使用 Python 运行函数 –– 类似于这个漂亮的图形 on Wikipedia .
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
我正在尝试使其与网格一起使用，以便在 x-y 平面上绘制逃逸图。 Python 不喜欢输出是 2 个未打包的变量(限制或转义)——我绝对可以通过拆分为两个函数来解决这个问题。
但另一个问题是复杂数学运算(cmath.log、cmath.exp)仅适用于标量......
我尝试向量化该函数:
nx, ny = 700, 500
x, 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)
但它会永远运行。
关于如何处理复杂数学运算和向量化有什么建议吗？
最佳答案
以下是我最终如何绘制我的逃生图:
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
矢量化辅助函数:
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)
四次逃逸的等高线图:
关于python - 我的 Tetration(复数)函数必须更好地矢量化(Python)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/61921444/