Python - How do I write a more efficient, Pythonic reduce?

I'm trying to build a very lightweight Node class to serve as a Python-based hierarchy search tool. See the definition below.

from functools import reduce
from operator import or_


class Node:

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

   开发者_如何学Go 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 seems to be a textbook case of functional programming (pulled directly from Why Functional Programming Matters).

Question: is there a more efficient or Pythonic way of writing contains? I know that map is usually replaced by list comprehension, but I hadn't seen a better way of doing reduce-based recursion.

Thanks,

Mike

===EDITED ... HERE'S THE REDONE CLASS TAKING INTO ACCOUNT THE ANSWER AND COMMENTS===

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

I think (not tested) that you instead of reduce should use any like this, which will stop on the first hit:

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

By the way, did you mean for a.contains(b) to return False when a == b and len(a.children) > 0?

Edit: If your tree contains a loop, like this:

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

then

a.contains(b)

will crash the program. You may want to check for this either in contains or in add_child, depending on which you use the most.