Python - 如何编写更高效的 Pythonic 化简？

Question

我正在尝试构建一个非常轻量级的 Node 类来充当基于 Python 的层次结构搜索工具。请参阅下面的定义。

from functools import reduce
from operator import or_


class Node:

    def __init__(self, name):
        self.name = name
        self.children = []

    def add_child(self, child_node):
        self.children.append(child_node)

    def contains(self, other_node):
        if self == other_node:
            return True
        elif other_node in self.children:
            return True
        else:
            return reduce(or_, [child.contains(other_node)
                                for child in self.children], False)

    def is_contained_by(self, other_node):
        return other_node.contains(self)

    def __eq__(self, other_node):
        return self.name == other_node.name

    def __de__(self, other_node):
        return self.name != other_node.name

contains 似乎是函数式编程的教科书案例（直接从 为什么函数式编程很重要）。

问题：有没有更有效或Pythonic的方式来编写contains？我知道 map 通常被列表理解所取代，但我还没有看到更好的方法来进行基于 reduce 的递归。

谢谢，

迈克

===已编辑...这是考虑答案和评论的重做课程===

class Node:

    def __init__(self, name):
        self.name = name
        self.children = []

    def add_child(self, child_node):
        # Hattip to lazyr for catching this.
        if self.contains(child_node) or child_node.contains(self):
            raise TreeError('A relationship is already defined.')    
        else:
            self.children.append(child_node)                

    def contains(self, other_node):
        # Hattip to lazyr for pointing out any() and to Jochen Ritzel for
        # eliminating the silly child check.
        return (self == other_node or
                any(child.contains(other_node) for child in self.children))

    def is_contained_by(self, other_node):
        return other_node.contains(self)

    def __eq__(self, other_node):
        return self.name == other_node.name

    def __de__(self, other_node):
        return self.name != other_node.name

    def __repr__(self):
        return self.name

Answer 1

我认为（未经测试）您应该使用像这样的 any 而不是 reduce ，它将在第一次点击时停止：

return any(child.contains(other_node) for child in self.children)

顺便说一句，您的意思是 a.contains(b) 当 a == b 且 len(a.children) > 时返回 False 0？

编辑：如果您的树包含循环，如下所示：

a = Node("a")
b = Node("b")
a.add_child(a)
a.add_child(b)

那么

a.contains(b)

将使程序崩溃。您可能需要在 contains 或 add_child 中检查这一点，具体取决于您最常使用的内容。