如何在不使用shell的情况下打印生成器对象？

Question

在网上看到这个例子，它是一个dfs函数，它可以在一个图中找到所有的循环。在

我正在尝试将循环部分添加到函数中，这样就不必使用shell来获得结果。 我不熟悉生成器对象，所以不知道如何显示循环。在

使用shell的版本：

def dfs(graph, start, end):
    fringe = [(start, [])]
    while fringe:
        state, path = fringe.pop()
        if path and state == end:
            yield path
            continue
    for next_state in graph[state]:
        if next_state in path:
            continue
        fringe.append((next_state, path+[next_state]))

>>> graph = { 1: [2, 3, 5], 2: [1], 3: [1], 4: [2], 5: [2] }
>>> cycles = [[node]+path  for node in graph for path in dfs(graph, node, node)]
>>> len(cycles)
7
>>> cycles
[[1, 5, 2, 1], [1, 3, 1], [1, 2, 1], [2, 1, 5, 2], [2, 1, 2], [3, 1, 3], [5, 2, 1, 5]]

以下是我的尝试：

^{pr2}$

尝试了几个不同的开始和结束节点，结果都一样。在

我的图表和上面一样

 graph = { 1: [2, 3, 5], 2: [1], 3: [1], 4: [2], 5: [2] }

输出= 位于0x000001D9CB846EB8的生成器对象dfs

有什么想法吗？在

Answer 1

首先，请修正你的压痕。在

我认为你的问题是由于缺少[]来理解列表。在

试着换线 cycles = (list([node]+path for node in g for path in dfs(g, node, node)))到cycles = [[node]+path for node in g for path in dfs(g, node, node)]

Answer 2

这就是你要找的吗？在

def dfs(g, start, end):
    fringe = [(start, [])]
    while fringe:
        state, path = fringe.pop()
        if path and state == end:

            yield path
            continue
        for next_state in g[state]:
            if next_state in path:
                continue
            fringe.append((next_state, path+[next_state]))


graph = {1: [2, 3, 5], 2: [1], 3: [1], 4: [2], 5: [2]}
cycles = [[node]+path for node in graph for path in dfs(graph, node, node)]
print(cycles)

您不需要在生成器中返回，所以可以使用列表理解或常规循环遍历它。在