Notebook

这段程序帮我们理解什么是期望更新。

In [1]:

#######################################################################
# Copyright (C)                                                       #
# 2018 Shangtong Zhang(zhangshangtong.cpp@gmail.com)                  #
# Permission given to modify the code as long as you keep this        #
# declaration at the top                                              #
#######################################################################

import numpy as np
import matplotlib
%matplotlib inline
import matplotlib.pyplot as plt
from tqdm import tqdm

# for figure 8.7, run a simulation of 2 * @b steps
def b_steps(b):
    # set the value of the next b states
    # it is not clear how to set this
    distribution = np.random.randn(b)

    # true value of the current state
    true_v = np.mean(distribution)

    samples = []
    errors = []

    # sample 2b steps
    for t in range(2 * b):
        v = np.random.choice(distribution)
        samples.append(v)
        errors.append(np.abs(np.mean(samples) - true_v))

    return errors

In [2]:

runs = 100
branch = [2, 10, 100, 1000]
for b in branch:
    errors = np.zeros((runs, 2 * b))
    for r in tqdm(np.arange(runs)):
        errors[r] = b_steps(b)
    errors = errors.mean(axis=0)
    x_axis = (np.arange(len(errors)) + 1) / float(b)
    """
    为什么要 np.zeros((runs, 2 * b)) ?
    为了将不同的 b 的数画在 0, 1b, 2b 的坐标上
    x_axis 将其横坐标都归一化到了 (0,2b) 的范围
    """
    plt.plot(x_axis, errors, label='b = %d' % (b))

plt.xlabel('number of computations')
plt.xticks([0, 1.0, 2.0], ['0', 'b', '2b'])
plt.ylabel('RMS error')
plt.legend()

plt.show()

100%|████████████████| 100/100 [00:00<00:00, 4359.48it/s]
100%|████████████████| 100/100 [00:00<00:00, 3458.39it/s]
100%|█████████████████| 100/100 [00:00<00:00, 257.87it/s]
100%|██████████████████| 100/100 [00:14<00:00,  6.98it/s]

对于期望更新来讲，在进行了 b 次运算可以将误差降为 0 ；
但是如果 b 很大，使用采样更新，无需 b 次运算，就可让误差贴近 0 。

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