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Question

在python中为树类编写迭代器时，我偶然发现了这个问题，我显然无法访问母类的字段和方法而没有通过引用已经存在的Tree实例来实现迭代器实例，所以我总是需要调用像it.iterate（tree）这样的迭代器非常难看。我想知道是否有某种方法来设计这些东西，因此不需要迭代器及其方法的实例引用。那么，我可以以某种方式从迭代器的实例访问Tree实例的字段而不将Tree实例的引用传递给迭代器吗？

########################################################################
# basic tree class, can be traversed bottom up from right hand corner
########################################################################

import sys

########################################################################
########################################################################
class Node:
    """!
    @brief node class
    """
    def __init__(self, pred=-1, ID=0, label=None):

        self.sucs = list()
        self.pred = pred
        self.ID = ID
        self.suc_count = 0
        self.label = label
########################################################################
    def make_suc(self, ID, label=None):
        """!
        @brief generates new successor node, assigning it a unique ID
        """
        self.suc_count += 1
        sucID = ID+1
        suc = Node(self, sucID, label)
        self.sucs.append(suc)
        return suc
########################################################################
########################################################################
class Tree:
    """!
    @brief tree class
    """
    def __init__(self):
        self.root = Node()
        self.allNodes = dict() # mapping from IDs (strings) to nodes
        self.init()
        self.leaves = list()
########################################################################
    # initializes node dict
    def init(self):
        self.allNodes[0] = self.root
########################################################################
    def find_node(self, ID):
        """!
        @brief looks up a node's ID and returns the node itself
        """
        return self.allNodes[ID]
########################################################################
    def add(self, parent, label=None):
        """!
        @brief adds a new node under parent with label label
        """
        if parent != Node:
            parent = self.find_node(parent)
        suc = parent.make_suc(len(self.allNodes)-1, label)
        self.allNodes[suc.ID] = suc
########################################################################
    def traverse(self, node):
        """!
        @brief traverses tree
        """
        for suc in node.sucs:
            self.traverse(suc)
            print suc.label
########################################################################
    def get_leaves(self, node):
        """!
        @brief when called resets leveas field and build it up anew by
               traversing tree and adding all leaves to it
        """
        self.leaves = list()
        self._find_leaves(node)
        return self.leaves
########################################################################
    def get_dominated(self, node, dom_nodes=[]):
        """!
        @brief finds all dominated nodes
        """
        for suc in node.sucs:
            self.get_dominated(suc, dom_nodes)
            dom_nodes.append(suc)
########################################################################
    def _find_leaves(self, node):
        """!
        @brief traverses tree in in order and adds all leaves to leaves field
               last leaf in list will be right hand corner of tree, due to in
               order travsersal
        """
        if node.suc_count == 0:
            self.leaves.append(node)
        for suc in node.sucs:
            self._find_leaves(suc)
########################################################################
    class TreeRHCIterator:
        """!
        @brief Right hand corner initialised iterator, traverses tree bottom
               up, right to left
        """
        def __init__(self, tree):
            self.current = tree.get_leaves(tree.root)[-1] # last leaf is right corner
            self.csi = len(self.current.sucs)-1 # right most index of sucs
            self.visited = list() # visisted nodes
########################################################################
        def begin(self, tree):
            return tree.get_leaves(tree.root)[-1]
########################################################################
        def end(self, tree):
            return tree.root
########################################################################
        def find_unvisited(self, node, tree):
            """!
            @brief finds rightmost unvisited node transitively dominated by node
            """
            leaves = tree.get_leaves(tree.root)
            # loop through leaves from right to left, as leaves are listed
            # in order, thus rightmost list elememt is rightmost leaf
            for i in range(len(leaves)-1, -1, -1):
                # return the first leaf, that has not been visited yet
                if leaves[i] not in self.visited:
                    return leaves[i]
            # return None if all leaves have been visited
            return None
########################################################################
        def go_up(self, node, tree):
            """!
            @brief sets self.current to pred of self.current,
                   appends current node to visited nodes, reassignes csi
            """
            self.visited.append(self.current)
            self.current = self.current.pred
            if self.current.sucs[0] not in self.visited:
                self.current = self.find_unvisited(self.current, tree)
            self.csi = len(self.current.sucs)-1
            self.visited.append(self.current)
########################################################################
        def iterate(self, tree):
            """!
            @brief advances iterator
            """
            # if current node is a leaf, go to its predecessor
            if self.current.suc_count == 0 or self.current in self.visited:
                self.go_up(self.current, tree)
            # if current node is not a leaf, find the next unvisited
            else:
                self.current = self.find_unvisited(self.current, tree)
########################################################################
########################################################################

这样叫：

tree = Tree()

it = tree.TreeRHCIterator(tree)
end = it.end(tree)

while (it.current != end):
    print it.current.label
    it.iterate(tree)

修改

在实现标准迭代器协议之后，我对它的工作方式有点困惑。在循环遍历树时，以某种方式跳过了起始节点。所以我做了一个测试类来研究行为，并且没有跳过任何元素，即使迭代方法基本上以相同的方式工作。有人可以为我阐明这个吗？

重新设计的迭代器：

########################################################################
# RIGHT-HAND-CORNER-BOTTOM-UP-POST-ORDER-TRAVERSAL-ITERATOR
########################################################################
    class RBPIter:
        """!
        @brief Right hand corner initialised iterator, traverses tree bottom
                     up, right to left
        """
        def __init__(self, tree):
            self.current = tree.get_leaves(tree.root)[-1] # last leaf is right corner
            self.csi = len(self.current.sucs)-1 # right most index of sucs
            self.visited = list() # visisted nodes
            self.tree = tree
            self.label = self.current.label
########################################################################
        def __iter__(self):
            print "iter: ", self.label
            return self
########################################################################
        def begin(self):
            return self.tree.get_leaves(self.tree.root)[-1]
########################################################################
        def end(self):
            return self.tree.root
########################################################################
        def find_unvisited(self, node):
            """!
            @brief finds rightmost unvisited node transitively dominated by node
            """
            leaves = self.tree.get_leaves(self.tree.root)
            # loop through leaves from right to left, as leaves are listed
            # in order, thus rightmost list elememt is rightmost leaf
            for i in range(len(leaves)-1, -1, -1):
                # return the first leaf, that has not been visited yet
                if leaves[i] not in self.visited:
                    self.label = leaves[i].label
                    return leaves[i]
            # return None if all leaves have been visited
            return None
########################################################################
        def go_up(self, node):
            """!
            @brief sets self.current to pred of self.current,
                         appends current node to visited nodes, reassignes csi
            """
            self.visited.append(self.current)
            self.current = self.current.pred
            if self.current.sucs[0] not in self.visited:
                self.current = self.find_unvisited(self.current)
            self.label = self.current.label
            self.csi = len(self.current.sucs)-1
            self.visited.append(self.current)
########################################################################
        def next(self):
            """!
            @brief advances iterator
            """
            print "next: ", self.label
            # if current node is a leaf, go to its predecessor
            if self.current.suc_count == 0 or self.current in self.visited:
                self.go_up(self.current)
            # if current node is not a leaf, find the next unvisited
            else:
                self.current = self.find_unvisited(self.current)
            if self.current == self.end():
                raise StopIteration
            return self
########################################################################
########################################################################

对于以下测试文件，我得到以下输出：

tree1 = Tree()

tree1.add(0, "t")
tree1.add(1, "e")
tree1.add(2, "s")
tree1.add(3, "t")

tree1.add(2, "t")
tree1.add(5, "r")
tree1.add(6, "i")
tree1.add(7, "s")

tree1.add(6, "a")

for node in tree1.RBPIter(tree1):
  print node.label

输出：

iter:  a
next:  a
s
next:  s
i
next:  i
r
next:  r
t
next:  t
t
next:  t
s
next:  s
e
next:  e
t
next:  t

这个树看起来像这样：

！ 1 [a trie]

因此，正如您所见，“a” - 意味着缺少右侧角点节点，我不明白为什么，因为迭代器方法正确返回第一个元素，正如您在调试输出中看到的那样。

Answer 1

你可以使用这样的东西使迭代器看起来很好：

class TreeIter:
    def __init__(self, parametersIfAny):
        code godes here
    def __iter__(self):
        return self
    def __next__(self):
       code that makes the iteration

class Tree:
    def __iter__(self):
        return TreeIter(parametersIfAny)

然后你可以像这样调用它：

tree = Tree()

for node in tree:
    print node.label

如果你需要有许多不同的迭代器，即有序，后序等，我去年必须做这样的事情（虽然有图表）。我当时所做的就是：

class PostOrderIter:
    def __init__(self, tree):
        self.tree = tree #and some more stuff
    def __iter__(self):
        return self
class PostOrder:
    def __init__(self, tree):
        self.tree = tree #and some more stuff
    def __iter__(self):
        return PostOrderIter(self.tree)

用以下方式调用它：

for node in PostOrder(tree):
    print node.label

Python访问母类字段

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1 个答案: