Python中的递归堆栈

如何解决Python中的递归堆栈

我正在尝试了解以下python递归函数的调用堆栈。该函数提供给定集合的所有组合。

def subsets(A):
    if not A:
        yield []
    else:
        for s in subsets(A[1:]):
            yield s
            yield [A[0]] + s
l1=[1,2,3]
print(list(subsets(l1)))

程序正在根据需要生成输出：

[[],[1],[2],[1,2],[3],3],[2,3]]

但是我无法可视化调用堆栈。如何打印[1,3]？ 据我了解如下

call stack 1 : values are : for s in [2,a[0] = 1,a = [1,3]
call stack 2 : values are : for s in [3],a[0] = 2,a = [2,3]
call stack 3 : values are : for s in [],a[0] = 3,a = [3]

是否可以帮助我理解具有s和a值的调用堆栈流？

解决方法

最初调用susbsets时，参数A为[1,2,3]。需要注意的重要一点是，该函数将在开始产生当前参数[2,3]的值之前，递归地重复使用参数[3]，[]和[1,3]进行调用。然后，所有这些递归调用返回，我们使用值A的{​​{1}}执行以下代码：

[1,3]

因此，我们希望最后两个值在包含列表中为for s in subsets(A[1:]): yield s # produces [2,3] yield [A[0]] + s # produces [1,3] 。

,

您在生成器上调用list()。这将一次又一次调用生成器，直到耗尽为止。让我们跟踪执行流程。这可能是了解生成器的好练习。我将所有内容格式化为代码块，以便可以使用适当的缩进来阐明生成器调用的层次结构。

subsets([1,3]) is called,so A is [1,3]. 
    This list is not empty,so the else block is executed.
    A[1:] is [2,3],so to determine the first s,subsets([2,3]) is called.
        Now A is [2,so A[1:] is [3],so to determine s,subsets([3]) is called.
            Now A is [3],so A[1:] is [],subsets([]) is called.
                Now A is [],so the if block is executed.
                This yields [].
            The loop starts with s = [].
            This yields [] again.
        Now this loop starts,also with s = [],because this is what subsets([3]) has just yielded.
        So this yields [] as well.
    So subsets([2,3]) has yielded [],so this loop also starts with s = [].
    This yields [] yet again.
So subsets([1,and now this generator is called again (because of list()),picking up the action after the previously executed yield statement.
    So we reach the next statement: yield [A[0]] + s.
    This yields [1].
subsets([1,3]) is called again,picking up at the end of the first run through the for loop,so to determine the next s,picking up at yield [A[0]] + s.
        This yields [2].
    So the loop starts again,with s = [2].
    This yields [2].
subsets([1,picking up at yield [A[0]] + s,with s = [2].
    This yields [1,2].
subsets([1,picking up at the end of the for loop,subsets([3]) is called again,picking up at yield [A[0]] + s.
            This yields [3].
        So the loop starts again,with s = [3].
        This yields [3].
    So the loop starts again,with s = [3].
    This yields [3].
subsets([1,with s = [3].
    This yields [1,3].
subsets([1,with s = [3].
        This yields [2,3].
    So the loop starts again,with s = [2,3].
    This yields [2,3]. 
subsets([1,3].
    This yields [1,subsets([]) is called again,picking up at the end of the if block,so we reach the end of the generator,which means it is exhausted and yields nothing anymore.
            So there is no further iteration of the for loop,hence subsets([3]) is also exhausted.
        So subsets([2,3]) is also exhausted.
    So subsets([1,3]) is also exhausted.

,

这是 powerset

的mathematical definition的直接实现

空的yield []没有子集（[A[0]] + s)）。

对于非空，您有两个选择

1. 保留第一个元素，并连接其余所有可能的子集（产量yield s

2. 不要保留第一个元素，并返回其余的可能子集（A = [1,3]）

因此，使用[1] + subsets([2,3])，您将拥有subsets([2,3])和subsets([2,3])的并集。同样，[2] + subsets([3])是subsets([3])和subsets([3])的并集。最后，[]是[3]或ssh <username>@<IP> 。这给出了3个步骤，每个步骤都有2种可能的结果，给出了8种可能的组合。