6

背景：

我正在做一个项目，我需要为文本处理编写一些规则。在为这个项目工作了几天并实施了一些规则之后，我意识到我需要确定规则的顺序。没问题，我们有拓扑排序来帮忙。但后来我意识到我不能指望图表总是满的。所以我想出了这个想法，给定具有一组依赖项（或单个依赖项）的单个规则，我需要检查依赖项的依赖关系。听起来很熟悉？是的。这个主题与图的深度优先搜索非常相似。
我不是数学家，也没有学过 CS 因此，图论对我来说是一个新领域。尽管如此，我实现了一些有效的东西（见下文）（我怀疑效率低下）。

编码：

这是我的搜索和收益算法。如果你在下面的例子中运行它，你会看到它不止一次地访问了一些节点。因此，推测效率低下。
关于输入的一句话。我写的规则基本上都是python类，有一个类属性depends。我因不使用而受到批评inspect.getmro-但这会使事情变得非常复杂，因为该类需要相互继承（请参见此处的示例）

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

好的，现在您已经查看了代码，下面是一些您可以测试的输入：

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：

  • 我尝试对我的工作进行分类，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，并且仍然使用生成器？

更新：

只是为了省去你运行代码的麻烦，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')]

与此同时，虽然我想出了一个更好的解决方案，但上面的问题仍然存在。所以请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

上面的输出是：

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

所以你现在可以看到节点 E 只被访问一次。

4

2 回答 2

2

使用 Gareth 和其他类型的 Stackoverflow 用户的反馈，这就是我想出的。它更清晰，也更笼统：

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
于 2013-10-16T07:31:42.147 回答
1

这是在不复制访问节点的情况下进行广度优先搜索的另一种方法。

import pylab
import networkx as nx

G = nx.DiGraph()
G.add_nodes_from([x for x in 'ABCDEF'])
G.nodes()

返回 ['A', 'C', 'B', 'E', 'D', 'F']

G.add_edge('A','B')
G.add_edge('A','C')
G.add_edge('B','D')
G.add_edge('B','E')
G.add_edge('C','E')
G.add_edge('C','F')

以下是如何在不复制节点的情况下遍历树。

nx.traversal.dfs_successors(G)

返回 {'A': ['C', 'B'], 'B': ['D'], 'C': ['E', 'F']} 即可绘制图形。

nx.draw(G,node_size=1000)
于 2013-10-27T06:55:55.850 回答