如何在Python中实现通用树?这有内置的数据结构吗?
当前回答
另一个基于Bruno回答的树的实现:
class Node:
def __init__(self):
self.name: str = ''
self.children: List[Node] = []
self.parent: Node = self
def __getitem__(self, i: int) -> 'Node':
return self.children[i]
def add_child(self):
child = Node()
self.children.append(child)
child.parent = self
return child
def __str__(self) -> str:
def _get_character(x, left, right) -> str:
if x < left:
return '/'
elif x >= right:
return '\\'
else:
return '|'
if len(self.children):
children_lines: Sequence[List[str]] = list(map(lambda child: str(child).split('\n'), self.children))
widths: Sequence[int] = list(map(lambda child_lines: len(child_lines[0]), children_lines))
max_height: int = max(map(len, children_lines))
total_width: int = sum(widths) + len(widths) - 1
left: int = (total_width - len(self.name) + 1) // 2
right: int = left + len(self.name)
return '\n'.join((
self.name.center(total_width),
' '.join(map(lambda width, position: _get_character(position - width // 2, left, right).center(width),
widths, accumulate(widths, add))),
*map(
lambda row: ' '.join(map(
lambda child_lines: child_lines[row] if row < len(child_lines) else ' ' * len(child_lines[0]),
children_lines)),
range(max_height))))
else:
return self.name
还有一个如何使用它的例子:
tree = Node()
tree.name = 'Root node'
tree.add_child()
tree[0].name = 'Child node 0'
tree.add_child()
tree[1].name = 'Child node 1'
tree.add_child()
tree[2].name = 'Child node 2'
tree[1].add_child()
tree[1][0].name = 'Grandchild 1.0'
tree[2].add_child()
tree[2][0].name = 'Grandchild 2.0'
tree[2].add_child()
tree[2][1].name = 'Grandchild 2.1'
print(tree)
它应该输出:
Root node
/ / \
Child node 0 Child node 1 Child node 2
| / \
Grandchild 1.0 Grandchild 2.0 Grandchild 2.1
其他回答
Treelib也很方便完成这项任务。文档可以在treelib找到。
from treelib import Node, Tree
tree = Tree() # creating an object
tree.create_node("Harry", "harry") # root node
tree.create_node("Jane", "jane", parent="harry") #adding nodes
tree.create_node("Bill", "bill", parent="harry")
tree.create_node("Diane", "diane", parent="jane")
tree.create_node("Mary", "mary", parent="diane")
tree.create_node("Mark", "mark", parent="jane")
tree.show()
Harry
├── Bill
└── Jane
├── Diane
│ └── Mary
└── Mark
另一个基于Bruno回答的树的实现:
class Node:
def __init__(self):
self.name: str = ''
self.children: List[Node] = []
self.parent: Node = self
def __getitem__(self, i: int) -> 'Node':
return self.children[i]
def add_child(self):
child = Node()
self.children.append(child)
child.parent = self
return child
def __str__(self) -> str:
def _get_character(x, left, right) -> str:
if x < left:
return '/'
elif x >= right:
return '\\'
else:
return '|'
if len(self.children):
children_lines: Sequence[List[str]] = list(map(lambda child: str(child).split('\n'), self.children))
widths: Sequence[int] = list(map(lambda child_lines: len(child_lines[0]), children_lines))
max_height: int = max(map(len, children_lines))
total_width: int = sum(widths) + len(widths) - 1
left: int = (total_width - len(self.name) + 1) // 2
right: int = left + len(self.name)
return '\n'.join((
self.name.center(total_width),
' '.join(map(lambda width, position: _get_character(position - width // 2, left, right).center(width),
widths, accumulate(widths, add))),
*map(
lambda row: ' '.join(map(
lambda child_lines: child_lines[row] if row < len(child_lines) else ' ' * len(child_lines[0]),
children_lines)),
range(max_height))))
else:
return self.name
还有一个如何使用它的例子:
tree = Node()
tree.name = 'Root node'
tree.add_child()
tree[0].name = 'Child node 0'
tree.add_child()
tree[1].name = 'Child node 1'
tree.add_child()
tree[2].name = 'Child node 2'
tree[1].add_child()
tree[1][0].name = 'Grandchild 1.0'
tree[2].add_child()
tree[2][0].name = 'Grandchild 2.0'
tree[2].add_child()
tree[2][1].name = 'Grandchild 2.1'
print(tree)
它应该输出:
Root node
/ / \
Child node 0 Child node 1 Child node 2
| / \
Grandchild 1.0 Grandchild 2.0 Grandchild 2.1
我将根树实现为字典{child:parent}。比如根节点为0，树可能是这样的:
tree={1:0, 2:0, 3:1, 4:2, 5:3}
这种结构使得沿着一条路径从任意节点向上到根结点非常容易，这与我正在处理的问题有关。
嗨，你可以试试itertree(我是作者)。
该包与任何树包的方向相同，但关注点略有不同。在巨大的树(>100000个项目)上的性能要好得多，它处理迭代器具有有效的过滤机制。
>>>from itertree import *
>>>root=iTree('root')
>>># add some children:
>>>root.append(iTree('Africa',data={'surface':30200000,'inhabitants':1257000000}))
>>>root.append(iTree('Asia', data={'surface': 44600000, 'inhabitants': 4000000000}))
>>>root.append(iTree('America', data={'surface': 42549000, 'inhabitants': 1009000000}))
>>>root.append(iTree('Australia&Oceania', data={'surface': 8600000, 'inhabitants': 36000000}))
>>>root.append(iTree('Europe', data={'surface': 10523000 , 'inhabitants': 746000000}))
>>># you might use __iadd__ operator for adding too:
>>>root+=iTree('Antarktika', data={'surface': 14000000, 'inhabitants': 1100})
>>># for building next level we select per index:
>>>root[0]+=iTree('Ghana',data={'surface':238537,'inhabitants':30950000})
>>>root[0]+=iTree('Niger', data={'surface': 1267000, 'inhabitants': 23300000})
>>>root[1]+=iTree('China', data={'surface': 9596961, 'inhabitants': 1411780000})
>>>root[1]+=iTree('India', data={'surface': 3287263, 'inhabitants': 1380004000})
>>>root[2]+=iTree('Canada', data={'type': 'country', 'surface': 9984670, 'inhabitants': 38008005})
>>>root[2]+=iTree('Mexico', data={'surface': 1972550, 'inhabitants': 127600000 })
>>># extend multiple items:
>>>root[3].extend([iTree('Australia', data={'surface': 7688287, 'inhabitants': 25700000 }), iTree('New Zealand', data={'surface': 269652, 'inhabitants': 4900000 })])
>>>root[4]+=iTree('France', data={'surface': 632733, 'inhabitants': 67400000 }))
>>># select parent per TagIdx - remember in itertree you might put items with same tag multiple times:
>>>root[TagIdx('Europe'0)]+=iTree('Finland', data={'surface': 338465, 'inhabitants': 5536146 })
创建的树可以被渲染:
>>>root.render()
iTree('root')
└──iTree('Africa', data=iTData({'surface': 30200000, 'inhabitants': 1257000000}))
└──iTree('Ghana', data=iTData({'surface': 238537, 'inhabitants': 30950000}))
└──iTree('Niger', data=iTData({'surface': 1267000, 'inhabitants': 23300000}))
└──iTree('Asia', data=iTData({'surface': 44600000, 'inhabitants': 4000000000}))
└──iTree('China', data=iTData({'surface': 9596961, 'inhabitants': 1411780000}))
└──iTree('India', data=iTData({'surface': 3287263, 'inhabitants': 1380004000}))
└──iTree('America', data=iTData({'surface': 42549000, 'inhabitants': 1009000000}))
└──iTree('Canada', data=iTData({'surface': 9984670, 'inhabitants': 38008005}))
└──iTree('Mexico', data=iTData({'surface': 1972550, 'inhabitants': 127600000}))
└──iTree('Australia&Oceania', data=iTData({'surface': 8600000, 'inhabitants': 36000000}))
└──iTree('Australia', data=iTData({'surface': 7688287, 'inhabitants': 25700000}))
└──iTree('New Zealand', data=iTData({'surface': 269652, 'inhabitants': 4900000}))
└──iTree('Europe', data=iTData({'surface': 10523000, 'inhabitants': 746000000}))
└──iTree('France', data=iTData({'surface': 632733, 'inhabitants': 67400000}))
└──iTree('Finland', data=iTData({'surface': 338465, 'inhabitants': 5536146}))
└──iTree('Antarktika', data=iTData({'surface': 14000000, 'inhabitants': 1100}))
过滤可以这样做:
>>>item_filter = Filter.iTFilterData(data_key='inhabitants', data_value=iTInterval(0, 20000000))
>>>iterator=root.iter_all(item_filter=item_filter)
>>>for i in iterator:
>>> print(i)
iTree("'New Zealand'", data=iTData({'surface': 269652, 'inhabitants': 4900000}), subtree=[])
iTree("'Finland'", data=iTData({'surface': 338465, 'inhabitants': 5536146}), subtree=[])
iTree("'Antarktika'", data=iTData({'surface': 14000000, 'inhabitants': 1100}), subtree=[])
Python不像Java那样具有相当广泛的“内置”数据结构。但是，因为Python是动态的，所以很容易创建通用树。例如，二叉树可能是:
class Tree:
def __init__(self):
self.left = None
self.right = None
self.data = None
你可以这样使用它:
root = Tree()
root.data = "root"
root.left = Tree()
root.left.data = "left"
root.right = Tree()
root.right.data = "right"
如果每个节点需要任意数量的子节点，则使用子节点列表:
class Tree:
def __init__(self, data):
self.children = []
self.data = data
left = Tree("left")
middle = Tree("middle")
right = Tree("right")
root = Tree("root")
root.children = [left, middle, right]
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