具有不同数量的For循环的函数(python)

Question

我的问题很难解释.

我想创建一个包含嵌套for循环的函数,
其数量与传递给函数的参数成比例.

这是一个假设的例子:

Function(2)

......会涉及......

for x in range (y):
    for x in range (y):
        do_whatever()

另一个例子...

  Function(6)

......会涉及......

for x in range (y):
    for x in range (y):
        for x in range (y):
            for x in range (y):
                for x in range (y):
                    for x in range (y):
                        whatever()

for循环(y)的变量实际上并未在嵌套代码中使用.

你的第一个想法可能是创建一个for循环,其范围是数字参数的功能......
这不能正常工作,因为产品将是巨大的.我需要实例,其中有8个嵌套for循环.
该产品对于for循环中的范围而言太大.

还有其他参数需要传递给函数,但我可以自己处理.

这是代码(它创建了Snowflake Fractal)

from turtle import *
length = 800
speed(0)

def Mini(length):
    for x in range (3):
        forward(length)
        right(60)

penup()
setpos(-500, 0)
pendown()   

choice = input("Enter Complexity:")

if choice == 1:
    for x in range (3):
        forward(length)
        left(120)

elif choice == 2:
    for x in range (3):
        Mini(length/3)
        left(120)

if choice == 3:
    for x in range (6):
        Mini(length/9)
        right(60)
        Mini(length/9)
        left(120)

if choice == 4:
    for y in range (6):
        for x in range (2):
            Mini(length/27)
            right(60)
            Mini(length/27)
            left(120)
        right(180)
        for x in range (2):
            Mini(length/27)
            right(60)
            Mini(length/27)
            left(120)

if choice == 5:
    for a in range (6):
        for z in range (2):
            for y in range (2):
                for x in range (2):
                    Mini(length/81)
                    right(60)
                    Mini(length/81)
                    left(120)
                right(180)
                for x in range (2):
                    Mini(length/81)
                    right(60)
                    Mini(length/81)
                    left(120)
            right(180)
        right(180)

if choice == 6:
    for c in range (6):
        for b in range (2):
            for a in range (2):
                for z in range (2):
                    for y in range (2):
                        for x in range (2):
                            Mini(length/243)
                            right(60)
                            Mini(length/243)
                            left(120)
                        right(180)
                        for x in range (2):
                            Mini(length/243)
                            right(60)
                            Mini(length/243)
                            left(120)
                    right(180)
                right(180)
            right(180)
        right(180)

if choice == 7:
    for a in range (6):
        for b in range(2):
            for c in range (2):
                for d in range (2):
                    for e in range (2):
                        for f in range (2):
                            for y in range (2):
                                for x in range (2):
                                    Mini(length/729)
                                    right(60)
                                    Mini(length/729)
                                    left(120)
                                right(180)
                                for x in range (2):
                                    Mini(length/729)
                                    right(60)
                                    Mini(length/729)
                                    left(120)
                            right(180)
                        right(180)
                    right(180)
                right(180)
            right(180)
        right(180)

我很感激你可以给我任何帮助,
但如果你建议一个不同的方法(如递归),
请不要只是粘贴代码; 相反,建议一些可以让我朝着正确方向前进的想法.

(该算法适用于专家数学作业)

---

规格:
Python 2.7.1
Turtle
IDLE
Windows7

Answer 1

我不清楚为什么你不能使用边界的产品和做

for x in range(y exp n)

其中n是循环数....你说y exp n会很大,但我确定python可以处理它.

但是,那就是说,某种递归算法怎么样？

def loop_rec(y, n):
    if n >= 1:
        for x in range(y):
            loop_rec(y, n - 1)
    else:
       whatever()

Answer 2

这个问题可以通过递归来解决.我只是在这里写一个算法,因为我相信这可能是一个普遍的问题.

function Recurse (y, number) 
   if (number > 1)
      Recurse ( y, number - 1 )
   else
      for x in range (y)
      whatever()

Answer 3

这可以在没有使用递归的情况下完成 itertools.product

import itertools
def function(n):
    for x in itertools.product(range(n),repeat=n):
        whatever()

Answer 4

干得好。让范围成为您的范围，在需要时对结果进行操作。

ranges=((1,4),(0,3),(3,6))
from operator import mul
operations=reduce(mul,(p[1]-p[0] for p in ranges))-1
result=[i[0] for i in ranges]
pos=len(ranges)-1
increments=0
print result
while increments < operations:
    if result[pos]==ranges[pos][1]-1:
        result[pos]=ranges[pos][0]
        pos-=1
    else:
        result[pos]+=1
        increments+=1
        pos=len(ranges)-1 #increment the innermost loop
        print result

[1, 0, 3]
[1, 0, 4]
[1, 0, 5]
[1, 1, 3]
[1, 1, 4]
[1, 1, 5]
[1, 2, 3]
[1, 2, 4]
[1, 2, 5]
[2, 0, 3]
[2, 0, 4]
[2, 0, 5]
[2, 1, 3]
[2, 1, 4]
[2, 1, 5]
[2, 2, 3]
[2, 2, 4]
[2, 2, 5]
[3, 0, 3]
[3, 0, 4]
[3, 0, 5]
[3, 1, 3]
[3, 1, 4]
[3, 1, 5]
[3, 2, 3]
[3, 2, 4]
[3, 2, 5]
[1, 0, 4]

使用以下内容进行测试会得到相同的结果：

for x in range(*ranges[0]):
    for y in range(*ranges[1]):
        for z in range(*ranges[2]):
            print [x,y,z]

Answer 5

递归将是你最好的选择.考虑它应该在基本情况和递归情况下做什么.

根据要求,代码被遗漏了.