问题描述
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我已经编码了几年了，但是我仍然没有掌握伪编码的知识，也没有真正考虑过代码中的问题。由于这个问题，我很难弄清楚在创建学习型决策树时该怎么做。
这是我浏览过的一些网站，相信我还有很多其他网站
决策树教程
DMS教程
包括伊恩·米林顿（Ian Millington）的游戏AI（AI for Games）等几本书，其中包括对决策树中使用的不同学习算法的体面总结以及用于游戏编程的行为数学（基本上涉及决策树和理论）。我了解了决策树的概念以及熵，ID3，以及有关如何交织遗传算法并让决策树为GA确定节点的一些知识。
他们提供了很好的见解，但不是我真正想要的。
我确实有一些为决策树创建节点的基本代码，而且我相信我知道如何实现实际的逻辑，但是如果我对程序没有目的或没有熵或学习的知识，那将是没有用的涉及的算法。
我要问的是，有人可以帮我弄清楚我需要做些什么来创建这个学习决策树。我将节点放在一个自己的类中，这些类通过函数来创建树，但是如何将熵放入其中，是否应该有一个类，我不确定如何将其组合在一起。伪代码以及所有这些理论和数字的发展思路。只要知道我需要编写的代码，就可以将代码放在一起。任何指导将不胜感激。
基本上，我将如何处理。
添加学习算法，例如ID3和熵。应该如何设置？
一旦确定了如何执行所有这些操作，我计划将其实现为状态机，该状态机以游戏/模拟格式经历不同的状态。所有这些都已经设置好了，我只是认为这可以是独立的，一旦弄清楚了，就可以将其移至另一个项目。
这是我现在拥有的源代码。
提前致谢！
Main.cpp：
int main()
{
//create the new decision tree object
DecisionTree* NewTree = new DecisionTree();
//add root node the very first \'Question\' or decision to be made
//is monster health greater than player health?
NewTree->CreateRootNode(1);
//add nodes depending on decisions
//2nd decision to be made
//is monster strength greater than player strength?
NewTree->AddNode1(1,2);
//3rd decision
//is the monster closer than home base?
NewTree->AddNode2(1,3);
//depending on the weights of all three decisions,will return certain node result
//results!
//Run,Attack,NewTree->AddNode1(2,4);
NewTree->AddNode2(2,5);
NewTree->AddNode1(3,6);
NewTree->AddNode2(3,7);
//Others: Run to Base ++ Strength,Surrender Monster/Player,//needs to be made recursive,that way when strength++ it affects decisions second time around DT
//display information after creating all the nodes
//display the entire tree,i want to make it look like the actual diagram!
NewTree->Output();
//ask/answer question decision making process
NewTree->Query();
cout << \"Decision Made. Press Any Key To Quit.\" << endl;
//pause quit,oh wait how did you do that again...look it up and put here
//release memory!
delete NewTree;
//return random value
//return 1;
}
//the decision tree class
class DecisionTree
{
public:
//functions
void RemoveNode(TreeNodes* node);
void displayTree(TreeNodes* CurrentNode);
void Output();
void Query();
void QueryTree(TreeNodes* rootNode);
void AddNode1(int ExistingNodeID,int NewNodeID);
void AddNode2(int ExistingNodeID,int NewNodeID);
void CreateRootNode(int NodeID);
void MakeDecision(TreeNodes* node);
bool SearchAddNode1(TreeNodes* CurrentNode,int ExistingNodeID,int NewNodeID);
bool SearchAddNode2(TreeNodes* CurrentNode,int NewNodeID);
TreeNodes* m_RootNode;
DecisionTree();
virtual ~DecisionTree();
};
int random(int upperLimit);
//for random variables that will effect decisions/node values/weights
int random(int upperLimit)
{
int randNum = rand() % upperLimit;
return randNum;
}
//constructor
//Step 1!
DecisionTree::DecisionTree()
{
//set root node to null on tree creation
//beginning of tree creation
m_RootNode = NULL;
}
//destructor
//Final Step in a sense
DecisionTree::~DecisionTree()
{
RemoveNode(m_RootNode);
}
//Step 2!
void DecisionTree::CreateRootNode(int NodeID)
{
//create root node with specific ID
// In MO,you may want to use thestatic creation of IDs like with entities. depends on how many nodes you plan to have
//or have instantaneously created nodes/changing nodes
m_RootNode = new TreeNodes(NodeID);
}
//Step 5.1!~
void DecisionTree::AddNode1(int ExistingNodeID,int NewNodeID)
{
//check to make sure you have a root node. can\'t add another node without a root node
if(m_RootNode == NULL)
{
cout << \"ERROR - No Root Node\";
return;
}
if(SearchAddNode1(m_RootNode,ExistingNodeID,NewNodeID))
{
cout << \"Added Node Type1 With ID \" << NewNodeID << \" onto Branch Level \" << ExistingNodeID << endl;
}
else
{
//check
cout << \"Node: \" << ExistingNodeID << \" Not Found.\";
}
}
//Step 6.1!~ search and add new node to current node
bool DecisionTree::SearchAddNode1(TreeNodes *CurrentNode,int NewNodeID)
{
//if there is a node
if(CurrentNode->m_NodeID == ExistingNodeID)
{
//create the node
if(CurrentNode->NewBranch1 == NULL)
{
CurrentNode->NewBranch1 = new TreeNodes(NewNodeID);
}
else
{
CurrentNode->NewBranch1 = new TreeNodes(NewNodeID);
}
return true;
}
else
{
//try branch if it exists
//for a third,add another one of these too!
if(CurrentNode->NewBranch1 != NULL)
{
if(SearchAddNode1(CurrentNode->NewBranch1,NewNodeID))
{
return true;
}
else
{
//try second branch if it exists
if(CurrentNode->NewBranch2 != NULL)
{
return(SearchAddNode2(CurrentNode->NewBranch2,NewNodeID));
}
else
{
return false;
}
}
}
return false;
}
}
//Step 5.2!~ does same thing as node 1. if you wanted to have more decisions,//create a node 3 which would be the same as this maybe with small differences
void DecisionTree::AddNode2(int ExistingNodeID,int NewNodeID)
{
if(m_RootNode == NULL)
{
cout << \"ERROR - No Root Node\";
}
if(SearchAddNode2(m_RootNode,NewNodeID))
{
cout << \"Added Node Type2 With ID \" << NewNodeID << \" onto Branch Level \" << ExistingNodeID << endl;
}
else
{
cout << \"Node: \" << ExistingNodeID << \" Not Found.\";
}
}
//Step 6.2!~ search and add new node to current node
//as stated earlier,make one for 3rd node if there was meant to be one
bool DecisionTree::SearchAddNode2(TreeNodes *CurrentNode,int NewNodeID)
{
if(CurrentNode->m_NodeID == ExistingNodeID)
{
//create the node
if(CurrentNode->NewBranch2 == NULL)
{
CurrentNode->NewBranch2 = new TreeNodes(NewNodeID);
}
else
{
CurrentNode->NewBranch2 = new TreeNodes(NewNodeID);
}
return true;
}
else
{
//try branch if it exists
if(CurrentNode->NewBranch1 != NULL)
{
if(SearchAddNode2(CurrentNode->NewBranch1,NewNodeID));
}
else
{
return false;
}
}
}
return false;
}
}
//Step 11
void DecisionTree::QueryTree(TreeNodes* CurrentNode)
{
if(CurrentNode->NewBranch1 == NULL)
{
//if both branches are null,tree is at a decision outcome state
if(CurrentNode->NewBranch2 == NULL)
{
//output decision \'question\'
///////////////////////////////////////////////////////////////////////////////////////
}
else
{
cout << \"Missing Branch 1\";
}
return;
}
if(CurrentNode->NewBranch2 == NULL)
{
cout << \"Missing Branch 2\";
return;
}
//otherwise test decisions at current node
MakeDecision(CurrentNode);
}
//Step 10
void DecisionTree::Query()
{
QueryTree(m_RootNode);
}
////////////////////////////////////////////////////////////
//debate decisions create new function for decision logic
// cout << node->stringforquestion;
//Step 12
void DecisionTree::MakeDecision(TreeNodes *node)
{
//should I declare variables here or inside of decisions.h
int PHealth;
int MHealth;
int PStrength;
int MStrength;
int distanceFBase;
int distanceFMonster;
////sets random!
srand(time(NULL));
//randomly create the numbers for health,strength and distance for each variable
PHealth = random(60);
MHealth = random(60);
PStrength = random(50);
MStrength = random(50);
distanceFBase = random(75);
distanceFMonster = random(75);
//the decision to be made string example: Player health: Monster Health: player health is lower/higher
cout << \"Player Health: \" << PHealth << endl;
cout << \"Monster Health: \" << MHealth << endl;
cout << \"Player Strength: \" << PStrength << endl;
cout << \"Monster Strength: \" << MStrength << endl;
cout << \"distance Player is From Base: \" << distanceFBase << endl;
cout << \"disntace Player is From Monster: \" << distanceFMonster << endl;
//MH > PH
//MH < PH
//PS > MS
//PS > MS
//DB > DM
//DB < DM
//good place to break off into different decision nodes,not just \'binary\'
//if statement lower/higher query respective branch
if(PHealth > MHealth)
{
}
else
{
}
//re-do question for next branch. Player strength: Monster strength: Player strength is lower/higher
//if statement lower/higher query respective branch
if(PStrength > MStrength)
{
}
else
{
}
//recursive question for next branch. Player distance from base/monster.
if(distanceFBase > distanceFMonster)
{
}
else
{
}
//DECISION WOULD BE MADE
//if statement?
// inside query output decision?
//cout << <<
//QueryTree(node->NewBranch2);
//MakeDecision(node);
}
//Step.....8ish?
void DecisionTree::Output()
{
//take repsective node
displayTree(m_RootNode);
}
//Step 9
void DecisionTree::displayTree(TreeNodes* CurrentNode)
{
//if it doesn\'t exist,don\'t display of course
if(CurrentNode == NULL)
{
return;
}
//////////////////////////////////////////////////////////////////////////////////////////////////
//need to make a string to display for each branch
cout << \"Node ID \" << CurrentNode->m_NodeID << \"Decision display: \" << endl;
//go down branch 1
displayTree(CurrentNode->NewBranch1);
//go down branch 2
displayTree(CurrentNode->NewBranch2);
}
//Final step at least in this case. A way to Delete node from tree. Can\'t think of a way to use it yet but i kNow it\'s needed
void DecisionTree::RemoveNode(TreeNodes *node)
{
//Could probably even make it to where you delete a specific node by using it\'s ID
if(node != NULL)
{
if(node->NewBranch1 != NULL)
{
RemoveNode(node->NewBranch1);
}
if(node->NewBranch2 != NULL)
{
RemoveNode(node->NewBranch2);
}
cout << \"Deleting Node\" << node->m_NodeID << endl;
//delete node from memory
delete node;
//reset node
node = NULL;
}
}
using namespace std;
//tree node class
class TreeNodes
{
public:
//tree node functions
TreeNodes(int nodeID/*,string QA*/);
TreeNodes();
virtual ~TreeNodes();
int m_NodeID;
TreeNodes* NewBranch1;
TreeNodes* NewBranch2;
};
//contrctor
TreeNodes::TreeNodes()
{
NewBranch1 = NULL;
NewBranch2 = NULL;
m_NodeID = 0;
}
//deconstructor
TreeNodes::~TreeNodes()
{ }
//Step 3! Also step 7 hah!
TreeNodes::TreeNodes(int nodeID/*,string NQA*/)
{
//create tree node with a specific node ID
m_NodeID = nodeID;
//reset nodes/make sure! that they are null. I wont have any funny business #s -_-
NewBranch1 = NULL;
NewBranch2 = NULL;
}
解决方法
如果我错了，请纠正我，但从http://dms.irb.hr/tutorial/tut_dtrees.php和http://www.decisiontrees.net/?q=node/21上的图像判断逻辑应该放在节点中，而不是树中。您可以通过具有多态节点来建模，以针对每个决策制定一个。通过对树的构造进行一些更改，并对决策委派进行一些小的修改，您的代码就可以了。
,基本上，您需要将所有内容分解为几个阶段，然后将要实现的算法的每个部分模块化。
您可以使用函数/类和存根以伪代码或代码本身的形式进行操作。
然后，您可以将算法的每个部分编码为某个函数，甚至可以在将该函数加在一起之前对其进行测试。基本上，您将最终获得用于算法实现中特定目的的各种函数或类。因此，在您进行树结构构建的情况下，您将拥有一个函数来计算节点上的熵，另一个函数将数据划分为每个节点上的子集，等等。
我在这里是在一般情况下讨论，而不是专门针对决策树构建。如果您需要有关决策树算法及其相关步骤的特定信息，请查阅Mitchell撰写的有关机器学习的书。
,实现决策树的伪代码如下
createdecisiontree(data,attributes)
Select the attribute a with the highest information gain
for each value v of the attribute a
Create subset of data where data.a.val==v ( call it data2)
Remove the attribute a from the attribute list resulting in attribute_list2
Call CreateDecisionTree(data2,attribute_list2)