Python:获取列表的可能性并更改循环数 [英] Python :get possibilities of lists and change the number of loops

问题描述

我希望能够更改循环数并将所有可能性存储在(列表列表)中.让我通过一个例子来解释:

I'd like to be able to change the number of loops and store all possibilities in a (list of lists). Let me explain by an example:

您有以下列表:

 initial_list=[1,2]

然后您有一个参数n，即循环数

Then you have a parameter n, which is the number of loops

例如，如果n = 2，我希望我的程序返回此列表:

For example if n=2, I'd like my programm to return this list:

 final_list=[[x,x,1,2],[x,1,x,2],[x,1,2,x],[1,x,x,2],[1,x,2,x][1,2,x,x]]

所以当 n=3 时，程序会对 3 'x' 做同样的事情

So for n=3, the program would do the same with 3 'x', etc

此外，我希望我的initial_list可以具有不同的长度，例如[1,2,3]或[1,2,3,4]等

Also, i'd like my initial_list could have different length, like [1,2,3] or [1,2,3,4], etc

如果参数n被硬编码(通过n循环)，我知道该怎么做，但是如果n改变，是否可以这样做?

I know how to do that if the parameter n is hardcoded ( by n loop) but is it possible to do that if n change?

感谢和抱歉，我的解释不正确^^

Thanks and sorry for my bad explaination ^^

推荐答案

您可以使用递归函数:

def get_lists(lst, n, sol):
    if len(lst) == 0 and n == 0:
        return [sol]
    solutions = []
    if len(lst) > 0:
        solutions += get_lists(lst[1:], n, sol + [lst[0]])
    if n > 0:
        solutions += get_lists(lst, n - 1, sol + ['x'])
    return solutions

n = 2
initial_list = [1, 2]
print(get_lists(initial_list, n, []))
>>> [[1, 2, 'x', 'x'], [1, 'x', 2, 'x'], [1, 'x', 'x', 2], ['x', 1, 2, 'x'], ['x', 1, 'x', 2], ['x', 'x', 1, 2]]

工作方式:

1. 您输入了原始列表和x的 n 的数量，以及一个空列表作为原始解决方案"
2. 它检查列表的长度或n是否大于0
3. 如果不是，则返回当前解决方案
4. 否则，它将添加到解决方案中，并返回到步骤2

1. You input your original list and number of x's n, as well as an empty list as the original 'solution'
2. It checks if either the length of your list, or n is larger than 0
3. If not, it returns the current solution
4. Otherwise, it adds onto the solution, and goes back to step 2

这比排列的想法稍微复杂一些，但是，当您的列表或 n 变大时，这会快得多，因为您不必担心删除双打.

This is slightly more complicated than the permutations idea, however, when your list or n becomes large, this will be much faster because you won't have to worry about removing doubles.

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