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testing_project/plot_logistic.py
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testing_project/plot_logistic.py
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"""Usage:
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```
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plot_trajectory(100, 3.6, 0.1)
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plot_bifurcation(2.5, 4.2, 0.001)
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```
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"""
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import numpy as np
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from matplotlib import pyplot as plt
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from logistic import iterate_f
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def plot_trajectory(n, r, x0, fname="single_trajectory.png"):
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"""
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Saves a plot of a single trajectory of the logistic function
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inputs
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n: int (number of iterations)
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r: float (r value for the logistic function)
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x0: float (between 0 and 1, starting point for the iteration)
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fname: str (filename to which to save the image)
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returns
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fig, ax (matplotlib objects)
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"""
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xs = iterate_f(n, x0, r)
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fig, ax = plt.subplots(figsize=(10, 5))
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ax.plot(list(range(n)), xs)
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fig.suptitle('Logistic Function')
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fig.savefig(fname)
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return fig, ax
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def plot_bifurcation(start, end, step, fname="bifurcation.png", it=100000,
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last=300):
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"""
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Saves a plot of the bifurcation diagram of the logistic function. The
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`start`, `end`, and `step` parameters define for which r values to
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calculate the logistic function. If you space them too closely, it might
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take a very long time, if you dont plot enough, your bifurcation diagram
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won't be informative. Choose wisely!
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inputs
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start, end, step: float (which r values to calculate the logistic
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function for)
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fname: str (filename to which to save the image)
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it: int (how many iterations to run for each r value)
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last: int (how many of the last iterates to plot)
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returns
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fig, ax (matplotlib objects)
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"""
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r_range = np.arange(start, end, step)
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x = []
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y = []
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for r in r_range:
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xs = iterate_f(it, 0.1, r)
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all_xs = xs[len(xs) - last::].copy()
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unique_xs = np.unique(all_xs)
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y.extend(unique_xs)
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x.extend(np.ones(len(unique_xs)) * r)
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fig, ax = plt.subplots(figsize=(20, 10))
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ax.scatter(x, y, s=0.1, color='k')
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ax.set_xlabel("r")
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fig.savefig(fname)
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return fig, ax
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