C++ 决策树实现问题：在代码中思考
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【中文标题】C++ 决策树实现问题：在代码中思考【英文标题】：C++ Decision Tree Implementation Question: Think In Code 【发布时间】：2011-08-04 12:07:00 【问题描述】：我已经编码了几年，但我仍然没有掌握伪编码的窍门，也没有真正用代码思考问题。由于这个问题，我很难弄清楚在创建学习决策树时要做什么。
这里有几个我看过的网站相信我还有很多
Decision Tree Tutorials
DMS Tutorials
还有几本书，例如 Ian Millington 的《AI for Games》，其中包括对决策树中使用的不同学习算法的详细介绍，以及《游戏编程的行为数学》（基本上都是关于决策树和理论的）。我了解决策树的概念以及熵、ID3 以及如何将遗传算法交织在一起并让决策树决定 GA 的节点。 他们提供了很好的洞察力，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逻辑，但如果我对程序没有目的或没有熵或学习算法，那它就没有用了涉及。
我要问的是，有人可以帮我弄清楚我需要做什么来创建这个学习决策树。我的节点在自己的类中流过函数来创建树，但是我如何将熵放入其中，如果它有一个类，一个结构，我不知道如何将它放在一起。伪代码和我对所有这些理论和数字的去向的想法。只要我知道我需要编写什么代码，我就可以将代码放在一起。任何指导将不胜感激。
基本上，我该怎么做。
添加学习算法，例如 ID3 和熵。应该如何设置？
一旦我弄清楚了如何处理这一切，我计划将其实现到一个状态机中，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，我只是想这可以是独立的，一旦我弄清楚了，我就可以将它移到另一个项目中。
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;
决策树.h：
//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 ExistingNodeID, int NewNodeID);
TreeNodes* m_RootNode;
DecisionTree();
virtual ~DecisionTree();
;
Decisions.cpp：
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 ExistingNodeID, 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, ExistingNodeID, NewNodeID))
return true;
else
//try second branch if it exists
if(CurrentNode->NewBranch2 != NULL)
return(SearchAddNode2(CurrentNode->NewBranch2, ExistingNodeID, 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, ExistingNodeID, 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 ExistingNodeID, 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, ExistingNodeID, NewNodeID))
return true;
else
//try second branch if it exists
if(CurrentNode->NewBranch2 != NULL)
return(SearchAddNode2(CurrentNode->NewBranch2, ExistingNodeID, 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;
TreeNodes.h：
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;
;
TreeNodes.cpp：
//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;
【问题讨论】：
好问题，所以我投了赞成票。根据您尝试实现的库（例如 castor），它允许您在 C++ 中本地实现一些逻辑编程范式，这可能会引起您的兴趣。 mpprogramming.com/cpp 天哪，有很多代码需要随机志愿者阅读...... 是的，但这确实表明我在某种意义上知道自己在做什么。我看到的许多问题都不足以解决，或者它们没有表明它们确实有效。我确实工作了！哈哈。我不希望很多人经历每一件小事，只是为了了解发生了什么。 而且在节目上看起来没那么大哈！searchAddNode 中没有您的“创建节点”代码。不管if..
【参考方案1】：
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 的图像来看，实际的决策逻辑应该在节点中，而不是在树中。您可以通过具有多态节点来建模，每个决策都有一个。通过对树结构进行一些更改并对决策委托进行少量修改，您的代码应该没问题。
【讨论】：
【参考方案2】：基本上，您需要将所有内容分解为多个阶段，然后将您尝试实现的算法的每个部分模块化。
您可以使用函数/类和存根在伪代码或代码本身中执行此操作。
然后，您可以在某个函数中编写算法的每个部分，甚至在将它们全部加在一起之前测试该函数。您基本上会得到在算法实现中用于特定目的的各种函数或类。因此，在您构建树的情况下，您将拥有一个计算节点熵的函数，另一个将数据划分为每个节点的子集的函数，等等。
我在这里谈论的是一般情况，而不是专门针对决策树构造。如果您需要有关决策树算法和相关步骤的具体信息，请查看 Mitchell 的机器学习一书。
【讨论】：
【参考方案3】：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)
您将不得不使用一些代码在每个级别存储节点，例如
决策树[attr][val]=new_node
【讨论】：
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