Python版B树

话说以前的树都用java写的,最近发现python有点生疏了,于是用python写了个B树实现,B树在索引领域用得还是蛮多了,如果没记错mysql的默认索引好像就是B树…

首先是数据实体对象,很简单,只存放key,value


class Entity(object):
'''数据实体'''

def __init__(self,key,value):
self.key = key
self.value = value

然后节点对象


class Node(object):
'''B树的节点'''

def __init__(self):
self.parent = None
self.entitys = []
self.childs = []

def find(self,key):
'''通过key查找并返回一个数据实体'''

for e in self.entitys:
if key == e.key:
return e


def delete(self,key):
'''通过key删除一个数据实体,并返回它和它的下标(下标,实体)'''
for i,e in enumerate(self.entitys):
if e.key == key:
del self.entitys[i]
return (i,e)


def isLeaf(self):
'''判断该节点是否是一个叶子节点'''

return len(self.childs) == 0


def addEntity(self,entity):
'''添加一个数据实体'''

self.entitys.append(entity)
self.entitys.sort(key=lambda x:x.key)


def addChild(self,node):
'''添加一个子节点'''

self.childs.append(node)
node.parent = self
self.childs.sort(key=lambda x:x.entitys[0].key)

最后是Tree类


class Tree(object):
'''B树'''

def __init__(self,size=6):
self.size = size
self.root = None
self.length = 0


def add(self,key,value=None):
'''插入一条数据到B树'''

self.length += 1

if self.root:
current = self.root

while not current.isLeaf():
for i,e in enumerate(current.entitys):
if e.key > key:
current = current.childs[i]
break
elif e.key == key:
e.value = value
self.length -= 1
return
else:
current = current.childs[-1]

current.addEntity(Entity(key,value))

if len(current.entitys) > self.size:
self.__spilt(current)
else:
self.root = Node()
self.root.addEntity(Entity(key,value))


def get(self,key):
'''通过key查询一个数据'''

node = self.__findNode(key)

if node:
return node.find(key).value


def delete(self,key):
'''通过key删除一个数据项并返回它'''

node = self.__findNode(key)

if node:
i,e = node.delete(key)

#在节点不是叶子节点时需要做修复(取对应下标的子节点的最大的一个数据项来补)
if not node.isLeaf():
child = node.childs[i]
j,entity = child.delete(child.entitys[-1].key)
node.addEntity(entity)

while not child.isLeaf():
node = child
child = child.childs[j]
j,entity = child.delete(child.entitys[-1].key)
node.addEntity(entity)

self.length -= 1
return e.value


def isEmpty(self):
return self.length == 0


def __findNode(self, key):
'''通过key值查询一个数据在哪个节点,找到就返回该节点'''

if self.root:
current = self.root

while not current.isLeaf():
for i, e in enumerate(current.entitys):
if e.key > key:
current = current.childs[i]
break
elif e.key == key:
return current
else:
current = current.childs[-1]

if current.find(key):
return current


def __spilt(self,node):
'''
分裂一个节点,规则为:
1、中间的数据项移到父节点
2、新建一个右兄弟节点,将中间节点右边的数据项移到新节点
'''

middle = len(node.entitys) / 2

top = node.entitys[middle]

right = Node()

for e in node.entitys[middle + 1:]:
right.addEntity(e)

for n in node.childs[middle + 1:]:
right.addChild(n)

node.entitys = node.entitys[:middle]
node.childs = node.childs[:middle + 1]

parent = node.parent

if parent:
parent.addEntity(top)
parent.addChild(right)

if len(parent.entitys) > self.size:
self.__spilt(parent)
else:
self.root = Node()
self.root.addEntity(top)
self.root.addChild(node)
self.root.addChild(right)

测试代码


if __name__ == '__main__':
t = Tree(4)
t.add(20)
t.add(40)
t.add(60)
t.add(70,'c')
t.add(80)
t.add(10)
t.add(30)
t.add(15,'python')
t.add(75,'java')
t.add(85)
t.add(90)
t.add(25)
t.add(35,'c#')
t.add(50)
t.add(22,'c++')
t.add(27)
t.add(32)

print t.get(15)
print t.get(75)
print t.delete(35)
print t.delete(22)
t.add(22,'lua')
print t.get(22)
print t.length
    原文作者:B树
    原文地址: https://blog.csdn.net/dieslrae/article/details/84168693
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞

发表评论

电子邮件地址不会被公开。 必填项已用*标注