找出while循环的一般情况

Question

我正在使用 python 列表作为以下代码：

x=[0.1,0.1,0.1]
dx=0.1
R=1
while x[0] < R:
  while x[1] < R:
    if np.sqrt(x[0] ** 2 + x[1] ** 2 + x[2] ** 2) < R:
      x[2] = x[2] + dx           
     counter = counter + 1
    else:

       x[1] = x[1] + dx
       x[0]= dx
 print(counter)
  x[0] = x[0] + dx
  x[1] = dx

但对于更大的列表，例如：

x=[0.1,0.1,0.1,0.1]
dx=0.1
#we have to add another while loop 
while x[0]<R:
     while x[1]<R:
         while x[2]<R:
           if np.sqrt(x[0] ** 2 + x[1] ** 2 + x[2] ** 2) < R:
                  x[3] = x[3] + dx           
                  counter = counter + 1
            else:

                   x[2] = x[2] + dx
                   x[1]= dx
         x[1] = x[1] + dx
         x[2] = dx
    x[0]=x[0]+dx
    x[1]=dx

等等 我想要做的是找到一种方法来为任意数量的元素的任何列表实现此代码（一般情况） 但我不知道如何将这些 while 循环转换为任意数量的维度（数组中的元素数量）

以防万一这段代码会执行以下操作：

example: for dx=0.1 and R=1 and we start with 0.1

start with x=[0.1, 0.1, 0.1] (after the first loop) x=[0.9, 0.1, 0.1] And then [0.1, 0.2, 0.1] And so on until [0.9, 0.9, 0.1] After we will get [0.1,0.1,0.2] And we will start again with [0.2, 0.1, 0.2] and so on

任何帮助将不胜感激

Answer 1

我同意 cmets 的观点，几乎可以肯定有更好的方法来做到这一点。但是，这个（快速组合在一起）递归函数将完成并编辑您的数组 x

def f(x, dx, n=0):
    if n == len(x) - 2:
        while x[n] < R:
           if np.sqrt(sum(i**2 for i in x[:-1])) < R:  # I think you mean this
              x[n+1] += dx           
              counter += 1
           else:
              x[n] += dx
              x[n-1] = dx
    else:
        while x[n] < R:
            f(x, dx, n+1)  # recursion
            x[n] += dx
            x[n+1] = dx