[英]Strange behavior in Python code
I am trying to write a simple method to link the nodes of a tree together in this way: 我试图写一个简单的方法以这种方式将树的节点链接在一起:
For example, if we have this tree: 例如,如果我们有这棵树:
A
/ | \
B C D
/ \ / \
E F G H
|
I
This should be the result of the method: 这应该是该方法的结果:
Here is the method code: 这是方法代码:
prevToken = None
def depthFirstTraverseTokenLinking(tree):
global prevToken
if len(tree.children) == 0:
tree.prevToken = prevToken
if prevToken != None :
prevToken.nextToken = tree # Is something wrong with this line?
prevToken = tree
return
for c in tree.children:
depthFirstTraverseTokenLinking(c)
tree.prevToken = tree.children[0].prevToken
tree.nextToken = tree.children[-1].nextToken
For some strange reason, the non-leaves aren't linked to the next leaves, for example: 由于一些奇怪的原因,非叶子没有链接到下一个叶子,例如:
Although 虽然
I wonder why is that happening? 我想知道为什么会这样? The last lines at the end of the recursive function should grantee that a parent will have the same next as its last child!
递归函数末尾的最后几行应该授予父级将与其最后一个子级相同的下一行!
The problem is, when you visit C, you traverse only it's children E & F. 问题是,当你访问C时,你只会遍历它的孩子E&F。
"I" hasn't been visited yet, so C.children[-1].nextToken == None
because only visiting "I" will set F.nextToken
“我”尚未访问过,所以
C.children[-1].nextToken == None
因为只访问“我”会设置F.nextToken
Solution: you'll have to do a run on all leaves first, then a second run on the internal nodes. 解决方案:您必须首先在所有叶子上运行,然后在内部节点上运行第二次运行。
For example: 例如:
prevToken = None
def depthFirstTraverseTokenLinking(tree):
depthFirstTraverseTokenLinkingPhase1(tree)
depthFirstTraverseTokenLinkingPhase2(tree)
def depthFirstTraverseTokenLinkingPhase1(tree):
global prevToken
if len(tree.children) == 0:
tree.prevToken = prevToken
if prevToken != None :
prevToken.nextToken = tree # Is something wrong with this line?
prevToken = tree
return
for c in tree.children:
depthFirstTraverseTokenLinkingPhase1(c)
def depthFirstTraverseTokenLinkingPhase2(tree):
if len(tree.children) == 0:
return
for c in tree.children:
depthFirstTraverseTokenLinkingPhase2(c)
if tree.children[0].prevToken is not None:
tree.prevToken = tree.children[0].prevToken
else:
tree.prevToken = tree.children[0]
if tree.children[-1].nextToken is not None:
tree.nextToken = tree.children[-1].nextToken
else:
tree.nextToken = tree.children[-1]
Also note the change for the prevToken
/ nextToken
of internal nodes. 另请注意内部节点的
prevToken
/ nextToken
的更改。 This is needed if you want them to link to the actual first/last leaf. 如果您希望它们链接到实际的第一个/最后一个叶子,则需要这样做。
Alternatively, use generators an an instance-checking loop 或者,使用生成器和实例检查循环
The generator yields the node as the base case if the node has no children, else another generator to travel down the tree. 如果节点没有子节点,则生成器将节点作为基本情况,否则另一个生成器沿树向下传播。 Caveat here is that node.children is ordered from left to right.
这里要注意的是node.children是从左到右排序的。
def leafs(node):
if len(node.children) == 0:
yield node
else:
for child in node.children:
yield leafs(child)
...and a loop with stack of generators... This got uglier as I wrote it - I think you could clean it up a bit and get rid of the while True... ......还有一堆带有发电机的循环...当我写这篇文章时它变得更加丑陋 - 我想你可以把它清理一下然后摆脱它真的......
current_node = leafs(a)
stack = []
last_node = None
while True:
if isinstance(current_node, types.GeneratorType):
stack.append(current_node)
current_node = current_node.next()
else:
if last_node and last_node != current_node:
last_node.nextToken = current_node
current_node.prevToken = last_node
last_node = current_node
try:
current_node = stack[-1].next()
except StopIteration:
stack.pop()
except IndexError:
break
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