在 Python 中使用 Monte Carlo 方法估算球体的体积时无法计算小于 1 的数字

Question

For an assignment, I need to write a program that can estimate the volume of an n-dimensional sphere using the Monte-Carlo method. I think I'm very close to having it done, but there is one problem.

I feel like everything works fine, except the part where I need to count the number of sum_x s that are <1. It simply doesn't seem to add all of them together. How can I make it count all the sum_x s that are <1?

What can I do to fix this problem and make it count up all sum_x s < 1?

import random

def vol_sphere(n,s): #n is amount of dimensions, and s is the amount of points taken

    #grabbing random points
    for j in range(s):
        sum_x = 0
        count = 0
        for i in range(n):
            i = random.randrange(0,101)/100 #this might seem weird but I'm doing it so I can get decimals   
            x = i**2
            sum_x += x
        print("The sum of the x^2's is:", sum_x)
        if sum_x <= 1:
            count += 1
    v = count/s
    volume = (2**n)*v
    print(volume)
    
vol_sphere(3,1000000)

Answer 1

You are resetting count to zero every iteration through the loop. Move the line count = 0 before the loop.

Also, why not use random.uniform(-1,1) to generate coordinates uniformly between -1 and 1? Your current method is only generating positive coordinates, but the 2**n in your volume calculation implies you are thinking of the cube each of whose sides goes from -1 to 1.

Answer 2

In addition to Nick's answer (you reset count in the loop, when you should reset it just once; and you should use random.ramdom() to get a float uniform value from [0,1), instead of that random.randrange(0,101)/100 which introduce a rough discretization, and cost CPU — Nick's said random.uniform(-1,1), which is exactly the same here, since we don't really care for the sign, but cost 60% to 100% more CPU than random.random), you should also consider relying on numpy batch computation, to avoid using for loop yourself.

import random
import sys
import numpy as np

def vol_sphere(n,s): #n is amount of dimensions and s is amount of points taken

    #grabbing random points
    count = 0
    for j in range(s):
        sum_x = 0
        for i in range(n):
            # I renamed your random variable ii, because I'm not very
            # comfortable with reusing the iteration variable. Even tho 
            # you can do it, as long as you don't really need i's value
            ii = random.random() #this might seem weird but I'm doing it so I can get decimals   
            x = ii**2
            sum_x += x
# Commented print (those printing cost all the CPU)
#        print("The sum of the x^2's is:", sum_x)
        if sum_x <= 1:
            count += 1
    v = count/s
    volume = (2**n)*v
    print(volume)

def vol_sphere2(n,s):
    # Same, but does computation by batches of 10000 (acceptable tradeoff
    # between CPU and memory)
    count=0
    reals=0
    for j in range(0, s, 10000):
        X2=np.random.random((10000, n))**2 # Generate 10000×n random number, squared at once
        # X2.sum(axis=1) is the column of 10000 sums of n columns of X2
        # X2.sum(axis=1)<=1 is a column of 10000 True/False saying which are<=1
        # (X2.sum(axis=1)<=1).sum() is the number of True
        count += (X2.sum(axis=1)<=1).sum()
        reals+=10000 # if s is not a factor of 10000, then we will do slightly more than s points.
    print(count/reals*2**n)

vol_sphere2(n,s)

On my computer version 2 is 20 times faster than version 1.

Edit: just for fun, I add a one-liner answer

((np.random.random((s,n))**2).sum(axis=1)<=1).sum()/s*2**n

Which use some memory, but not much code lines.

For example, an estimation of pi is

((np.random.random((100000,2))**2).sum(axis=1)<=1).sum()/100000*2**2