如何在不使用shell的情况下打印出生成器对象?

时间:2018-04-18 14:52:42

标签: python graph depth-first-search brute-force function-calls

在线看到这个例子,它是一个dfs函数,可以找到图形中的所有周期。

我试图将循环部分添加到函数中,因此我不必使用shell来获得结果。 我是发电机对象的新手,所以我不确定如何显示周期。

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]]

这是我的尝试:

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]))
    cycles = (list([node]+path for node in g for path in dfs(g, node, node)))
    print("cycles",cycles)
return path

dfs(graph, 1, 1)

尝试了几个不同的开始和结束节点。

我的图表与上面相同,

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

输出=      生成器对象dfs位于0x000001D9CB846EB8

有什么想法吗?

2 个答案:

答案 0 :(得分:2)

这是你在找什么?

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)

你不需要在生成器中返回,所以你可以使用列表推导或常规循环来循环它。

答案 1 :(得分:0)

首先,请修改你的缩进。

我认为你的问题是缺少[]列表理解工作的结果。

尝试更改线路 cycles = (list([node]+path for node in g for path in dfs(g, node, node)))cycles = [[node]+path for node in g for path in dfs(g, node, node)]