Question

我正在模拟将999个硬币翻转1000次，并绘制样本均值的分布，这可能需要很长时间（大约21秒）。有一个更好的方法吗？例如，一种更快的循环运行方式。向量化会有用吗？

import datetime
import numpy as np

sample_mean_dis = []
start_time = datetime.datetime.now()
# to draw a distribution of sample mean
for i in range(1000):
    if not (i%100):
        print('iterate: ', i)
    sums_1000coins = []
    # simulate 1k repetition of experiment_1
    # and consider this opertation as a sample
    # and compute the sample mean
    for i in range(1000):
        # this is simulating experiment_1 which flip 999 coins
        # and sum heads
        coins = np.random.randint(2, size=999)
        sums_1000coins.append(np.sum(1 == coins))
    sample_mean_dis.append(np.mean(sums_1000coins))
end_time = datetime.datetime.now()
elapsedTime = end_time - start_time
print("Elapsed time: %d seconds" % (elapsedTime.total_seconds()))

Answer 1

要掷出999个硬币并查看哪个硬币正面朝上，read 999 bits of random data（有点像硬币可能是Sub Combine_workbooks() Path = "C:\Users\PChen\Desktop\Test\" FileName = Dir(Path & "*.xlsx") Workbooks.Add Do While FileName <> "" Workbooks.Open FileName:=Path & FileName, ReadOnly:=True Call AuthOpenAll Worksheets("Data").Activate Worksheets("Data").Copy After:=Workbooks("book1").Worksheets("sheet1") Workbooks(FileName).Close savechanges:=False FileName = Dir() Loop End Sub或0，概率为50/50），然后{ {3}}。

以上内容可能会返回接近499.5的数字

要翻转999个硬币1000次，请在import random bin(random.getrandbits(999)).count("1")循环中执行上述操作：

for

num_heads = [bin(random.getrandbits(999)).count("1") for _ in range(1000)]将是1000个整数count how many bits are set to 1的列表，大约为499.5（999/2）。

如何使样本分布的模拟速度更快

1 个答案: