Python中的DFS算法与生成器

Question

背景:

我正在研究一个项目,我需要为文本处理编写一些规则.在完成这个项目几天并实施一些规则后,我意识到我需要确定规则的顺序.没问题,我们有拓扑排序帮忙.但后来我意识到我不能指望图表总是满满的.所以我想出了这个想法,给定一个带有一组依赖关系(或单个依赖关系)的规则,我需要检查依赖关系的依赖关系.听起来很熟悉？是.该主题与图的深度优先搜索非常相似.
我不是数学家,也不研究CS因此,图论对我来说是一个新领域.尽管如此,我实施了一些工作(见下文)(我怀疑效率低下).

代码:

这是我的搜索和产量算法.如果您在下面的示例中运行它,您将看到它多次访问某些节点.因此,推测效率低下.
关于输入的一个词.我写的规则基本上是python类,它们有一个类属性depends.我被批评没有使用inspect.getmro- 但这会使事情变得非常复杂,因为这个类需要相互继承(参见这里的例子)

def _yield_name_dep(rules_deps):
    global recursion_counter
    recursion_counter = recursion_counter +1 
    # yield all rules by their named and dependencies
    for rule, dep in rules_deps.items():
        if not dep:
            yield rule, dep
            continue
        else:
            yield rule, dep
            for ii in dep:
                i = getattr(rules, ii)
                instance = i()
                if instance.depends:
                    new_dep={str(instance): instance.depends}
                    for dep in _yield_name_dep(new_dep):
                        yield dep    
                else:
                    yield str(instance), instance.depends

好了,既然你盯着代码,这里有一些你可以测试的输入:

demo_class_content ="""
class A(object):
    depends = ('B')
    def __str__(self):
        return self.__class__.__name__

class B(object):
    depends = ('C','F')
    def __str__(self):
        return self.__class__.__name__

class C(object):
    depends = ('D', 'E')
    def __str__(self):
        return self.__class__.__name__

class D(object):
    depends = None
    def __str__(self):
        return self.__class__.__name__   

class F(object):
    depends = ('E')
    def __str__(self):
        return self.__class__.__name__

class E(object):
    depends = None  
    def __str__(self):
        return self.__class__.__name__
"""       

with open('demo_classes.py', 'w') as clsdemo:
    clsdemo.write(demo_class_content)

import demo_classes as rules

rule_start={'A': ('B')}

def _yield_name_dep(rules_deps):
    # yield all rules by their named and dependencies
    for rule, dep in rules_deps.items():
        if not dep:
            yield rule, dep
            continue
        else:
            yield rule, dep
            for ii in dep:
                i = getattr(rules, ii)
                instance = i()
                if instance.depends:
                    new_dep={str(instance): instance.depends}
                    for dep in _yield_name_dep(new_dep):
                        yield dep    
                else:
                    yield str(instance), instance.depends

if __name__ == '__main__':
    # this is yielding nodes visited multiple times, 
    # list(_yield_name_dep(rule_start))
    # hence, my work around was to use set() ...
    rule_dependencies = list(set(_yield_name_dep(rule_start)))
    print rule_dependencies

问题:

• 我尝试对我的作品进行分类,我认为我所做的与DFS类似.你真的可以这样分类吗？
• 如何改进此功能以跳过访问过的节点,仍然使用生成器？

更新:

为了省去运行代码的麻烦,上面函数的输出是:

>>> print list(_yield_name_dep(rule_wd))
[('A', 'B'), ('B', ('C', 'F')), ('C', ('D', 'E')), ('D', None), ('E', None), ('F', 'E'), ('E', None)]
>>> print list(set(_yield_name_dep(rule_wd)))
[('B', ('C', 'F')), ('E', None), ('D', None), ('F', 'E'), ('C', ('D', 'E')), ('A', 'B')]

在我提出更好的解决方案的同时,上述问题仍然存在.所以随意批评我的解决方案:

visited = []
def _yield_name_dep_wvisited(rules_deps, visited):
    # yield all rules by their name and dependencies
    for rule, dep in rules_deps.items():
        if not dep and rule not in visited:
            yield rule, dep
            visited.append(rule)
            continue
        elif rule not in visited:
            yield rule, dep
            visited.append(rule)
            for ii in dep:
                i = getattr(grules, ii)
                instance = i()
                if instance.depends:
                    new_dep={str(instance): instance.depends}
                    for dep in _yield_name_dep_wvisited(new_dep, visited):
                        if dep not in visited:
                            yield dep    

                elif str(instance) not in visited:
                    visited.append(str(instance))
                    yield str(instance), instance.depends

以上的输出是:

>>>list(_yield_name_dep_wvisited(rule_wd, visited))
[('A', 'B'), ('B', ('C', 'F')), ('C', ('D', 'E')), ('D', None), ('E', None), ('F', 'E')]

因此,您现在可以看到节点E只被访问过一次.

Answer 1

根据 Gareth 和 Stackoverflow 其他用户的反馈，我得出了以下结论。它更清晰，也更普遍：

def _dfs(start_nodes, rules, visited):
    """
    Depth First Search
    start_nodes - Dictionary of Rule with dependencies (as Tuples):    

        start_nodes = {'A': ('B','C')}

    rules - Dictionary of Rules with dependencies (as Tuples):
    e.g.
    rules = {'A':('B','C'), 'B':('D','E'), 'C':('E','F'), 
             'D':(), 'E':(), 'F':()}
    The above rules describe the following DAG:

                    A
                   / \
                  B   C
                 / \ / \
                D   E   F
    usage:
    >>> rules = {'A':('B','C'), 'B':('D','E'), 'C':('E','F'), 
                 'D':(), 'E':(), 'F':()}
    >>> visited = []
    >>> list(_dfs({'A': ('B','C')}, rules, visited))
    [('A', ('B', 'C')), ('B', ('D', 'E')), ('D', ()), ('E', ()), 
    ('C', ('E', 'F')), ('F', ())]
    """

    for rule, dep in start_nodes.items():
        if rule not in visited:
            yield rule, dep
            visited.append(rule)
            for ii in dep:
                new_dep={ ii : rules[ii]}
                for dep in _dfs(new_dep, rules, visited):
                    if dep not in visited:
                        yield dep