决策树（离散数据）的python实现_决策树可以输入离散特征吗-优快云博客

决策树（离散数据）的python实现

1.计算熵和信息增益率等，来确定按特征进行分类的顺序
2.利用多重字典一层一层建立决策树

熵：Ent(D)=−∑pi*log(pi）
条件熵：条件熵𝐻(X|Y)相当于联合熵𝐻(𝑋,𝑌)减去单独的熵𝐻(Y)，即：𝐻(X|Y)=𝐻(𝑋,𝑌)−𝐻(Y)
信息增益率：信息增益/分裂信息度量

1.创建特征和分类数据

## 矩阵、列表皆可
dataSet = [[ 1. ,  2.1],
        [ 2. ,  1.1],
        [ 1.3,  1. ],
        [ 1. ,  1. ],
        [ 2. ,  1. ]]
classSet = ["a", "a", "b", "b", "a"]
# dataSet = mat([[ 1. ,  2.1],[ 2. ,  1.1],[ 1.3,  1. ],[ 1. ,  1. ],[ 2. ,  1. ]])
# classSet = mat(["a", "a", "b", "b", "a"])

2.计算信息熵

from numpy import *
def entropy(dataSet,classSet):
    ## 信息熵
    ## 输入特征和分类 并一律转化为矩阵
    dataSet=mat(dataSet)
    classSet=mat(classSet)
    classSet=asarray(classSet).tolist() ## 分类矩阵转化为列表
    numfeature= len(dataSet)  ## 特征数：5
    classCounts = {}
    for class0 in classSet[0]:
        if  class0  not in classCounts.keys():
            classCounts[class0] = 0
        classCounts[class0] += 1      # 每类个数， {'Y': 4, 'N': 3}
    entropy = 0.0
    for key in classCounts:
        prob = float(classCounts[key])/numfeature
        entropy -= prob * log2(prob)
    return entropy

entropy(dataSet,classSet)
##  -(2/5)*log2(2/5)-(3/5)*log2(3/5)=0.9709505944546686

熵为：0.9709505944546686

3.计算条件熵

def splitDataSet(dataSet,classSet, axis, value):
	## 每类特征分别按特征值进行聚类
    a=asarray(mat(dataSet).T).tolist()
    classSet=asarray(mat(classSet)).tolist()[0]
    a.append(list(classSet))
    a=mat(a).T
    dataSet=asarray(a).tolist()
    retDataSet = []
    for featVec in dataSet:
        ## 按各特征进行聚类，
        ## 如[[1,2,a],[1,3,a],[2,3,c]] 前两列特征，后一列分类
        ## 聚类后[[2,a],[3,a]]和[[3,c]] 
        if str(featVec[axis]) == str(value):
            reduceFeatVec = featVec[:axis]              
            reduceFeatVec.extend(featVec[axis+1:])
            retDataSet.append(reduceFeatVec)   
#         print(retDataSet)
    return retDataSet

def conditional_entropy(dataSet,classSet):
	## 条件熵
    dataSet=asarray(dataSet).tolist()
    dataSet=mat(dataSet);
    numFeatures = len(dataSet[0])                 #特征个数
    conditional_entlist=[]
    entropylist=[]
    for i,features in enumerate(np.asarray(dataSet.T).tolist()):   
        uniqueVals = set(features)                      # 每组特征值去重     
        entropy1 = 0.0
        conditional_entropy = 0.0
        for value in uniqueVals:
            subDataSet = splitDataSet(dataSet,classSet, i, value)  #每个唯一值对应的剩余feature的组成子集
            prob = len(subDataSet)/float(len(dataSet))
            conditional_entropy += prob * entropy(mat(subDataSet)[:,:-1],mat(subDataSet)[:,-1].T)
            entropy1 += -prob * log2(prob)
        conditional_entlist.append(conditional_entropy)
        entropylist.append(entropy1)
    return conditional_entlist,entropylist
    
conditional_entropy(dataSet,classSet)
## 条件熵：2/5*(-(1/2)*log2(1/2)-(1/2)*log2(1/2))+2/5*(-(1)*log2(1)-(0)*log2(0))+1/5*(-(1)*log2(1)-(0)*log2(0))=0.4
## 分裂信息度量：-(2/5)*log2(2/5)-(2/5)*log2(2/5)-(1/5)*log2(1/5)=1.5219280948873621

条件熵、分裂信息度量为：([0.4, 0.5509775004326937], [1.5219280948873621, 1.3709505944546687])

4.计算信息增益和信息增益率

def information_gain(dataSet,classSet):
    ### 信息增益和信息增益率 返回信息增益率最大的特征
#     entropy=entropy(dataSet,classSet)
    maxInf_gainRatio=0
    entropy0=entropy(dataSet,classSet)
    conditional_entlist,entropylist=conditional_entropy(dataSet,classSet)
    information_gain=entropy0-conditional_entlist
    print("信息增益：",information_gain)
    print("信息增益率：")
    for i in range(len(entropylist)):
        
        if ( entropylist[i]== 0): # fix the overflow bug
            continue
        Inf_gainRatio = information_gain[i] / entropylist[i]
        print(Inf_gainRatio)
        #这个feature的infoGainRatio      
        if (Inf_gainRatio > maxInf_gainRatio):          #选择最大的gain ratio
            maxInf_gainRatio = Inf_gainRatio
            bestFeature = i
    return(bestFeature)
    
information_gain(dataSet,classSet)

信息增益：[0.57095059 0.41997309]
信息增益率：0.37514952012034747 、 0.30633714717416943
返回的信息增益率最好的特征是第0个

5.创建树（多重字典）

def createTree(dataSet, feat_name):
    dataSet = asarray(mat(dataSet)).tolist() 
    classList = [example[-1] for example in dataSet]    
    if classList.count(classList[0]) == len(classList):       # 类别完全相同，停止划分，返回其类别
        return classList[0]                             
    if len(dataSet[0]) == 0:                                # 遍历完所有特征时返回"End of the classification
        return "End of the classification"
    bestfeat = information_gain(mat(dataSet)[:,:-1],mat(dataSet)[:,-1].T)             # 信息增益率最大的特征位置 0-->1 
  
    bestfeat_name = feat_name[bestfeat]                         # 信息增益率最大的特征名
    dm_tree = {bestfeat_name:{}}                              # 多重字典构建树    {'t1': {'1.0': {'t2': {'1.0': 'b', '2.1': 'a'}}, '1.3': 'b', '2.0': 'a'}}
    del(feat_name[bestfeat])                                      
    feat_data = [featdata[bestfeat] for featdata in dataSet]      ## ['1.0', '2.0', '1.3', '1.0', '2.0']
    feat_set = set(feat_data)                                      ## {'1.0', '1.3', '2.0'}   
    for value in feat_set:
        featname = feat_name[:]                                
        dm_tree[bestfeat_name][value] = createTree(splitDataSet(mat(dataSet)[:,:-1],mat(dataSet)[:,-1].T, bestfeat, value), featname)
    return dm_tree

c = hstack((mat(dataSet),mat(classSet).T))  ## 合并特征和分类矩阵
o=createTree(c,["t1","t2"])                 ## 决策树字典
drawtree.createPlot(o)                      ## 绘制决策树，drawtree：绘制树的自定义模块

决策树如下：

6.利用树进行分类

def treeclass(dm_tree,t_data,feat_name):
    for i in range(len(t_data)):
        try:
            key1=[key1 for key1 in dm_tree[feat_name[i]].keys()]
            print(key1)
        except:
            pass
        else:   
            if str(t_data[i]) in key1:
                tree2=dm_tree[feat_name[i]]
                print(tree2)
                if str(type(tree2[str(t_data[i])])) == "<class 'dict'>":
                    a=t_data.pop(i)
                    feat_name.pop(i)
                    print(t_data,tree2[str(a)],feat_name)
                    treeclass(tree2[str(a)],t_data,feat_name)
                else:
                    print(tree2[str(t_data[i])])
                    return tree2[str(t_data[i])]