输入:树形结构是按父/子帐户的分层顺序分隔的财务帐户列表。任何给定帐户可以有任意数量的 parent / child 。在 Python 结构中,每个子项都是一个列表,可以包含任意数量的字典和/或文本值。字典表示指向其他帐户的子项,而文本值表示没有进一步后代的子项。以下是一些 JSON 格式的示例输入(要测试它,请将其转换回 Python):
[
{
"Assets":[
{
"Bank":[
"Car",
"House"
]
},
{
"Savings":[
"Emergency",
{
"Goals":[
"Roof"
]
}
]
},
"Reserved"
]
}
]
在幕后有一个输入文件,其中包含如下所示的帐户定义:
Assets:Bank:House
Assets:Savings:Emergency
Assets:Savigs:Goals:Roof
我有现有的代码可以解析并创建上面看到的树结构。
目标:最终目标是通过搜索树来利用给定的字符串输入提供自动完成功能。使用上面的示例输入,以下输入将产生各自的输出:
"Assets" => ["Bank, "Savings", "Reserved"]
"Assets:Bank" => ["Car", "House"]
"Assets:Savings:Goals" => ["Roof"]
部分解决方案:递归是我遇到问题的地方。我能够创建可以处理为“根”帐户提供结果的代码,但我不确定如何使其递归地为子帐户提供结果。代码如下:
def search_tree(account, tree):
# Check to see if we're looking for a root level account
if isinstance(account, str) and ":" not in account:
# Collect all keys in the child dictionaries
keys = {}
for item in tree:
if isinstance(item, dict):
keys[item.keys()[0]] = item
# Check to see if the input matches any children
if account in keys:
# Collect all children of this account
children = []
for child in keys[account][account]:
if isinstance(child, str):
children.append(child)
else:
children.append(child.keys()[0])
return children
# tree = .....
account = "Assets"
print search_tree(account, tree) # Would produce ["Bank", "Savings", "Reserved"]
# In the future I would provide "Assets:Bank" as the account string and get back the following: ["Car", "House"]
我如何进行递归搜索以搜索到n个 child ?
最佳答案
我不会真正回答你的问题(关于你的特定标准输出输出要求),但我会帮助你展示如何搜索树结构
首先描述你的树结构
- 树 = 节点列表
- nodeType1 = 由nodeName=>children组成的字典
- nodeType2 = 没有子节点(叶节点)的简单基字符串(节点名称)
现在我们可以开始编写递归解决方案
def search(key,tree):
if isinstance(tree,(list,tuple)): # this is a tree
for subItem in tree: # search each "node" for our item
result = search(key,subItem)
if result:
return result
elif isinstance(tree,dict): # this is really a node (nodeType1)
nodeName,subTree = next(tree.iteritems())
if nodeName == key: # match ... in your case the key has many parts .. .you just need the "first part"
print "Found:",key
return subTree
else: # did not find our key so search our subtree
return search(key,subTree)
elif isinstance(tree,basestring): #leaf node
if tree == key: # found our key leaf node
print "Found",key
return tree
这实际上只是一个非常通用的解决方案,它可用于搜索单个条目(即“房屋”或“帐户”......它不记录用于到达解决方案的路径)
现在让我们回到检查你的问题陈述
key 是多部分 key Part1:part2:part3
那么让我们开始解决这个问题
def search_multipartkey(key,T,separator=":"):
result = T
for part in key.split(separator):
result = search(part,result)
if not result:
print "Unable to find part:",part
return False
else:
print "Found part %s => %s"%(part,result)
return result
你几乎肯定可以对此进行改进,但这提供了一个很好的起点(尽管它不是像某人希望的那样递归)
关于python - 搜索具有嵌套值的树结构?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38724662/