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今天小編給大家分享一下python如何實現(xiàn)決策樹分類算法的相關(guān)知識點,內(nèi)容詳細(xì),邏輯清晰,相信大部分人都還太了解這方面的知識,所以分享這篇文章給大家參考一下,希望大家閱讀完這篇文章后有所收獲,下面我們一起來了解一下吧。
決策樹是一種十分常用的分類算法,屬于監(jiān)督學(xué)習(xí);也就是給出一批樣本,每個樣本都有一組屬性和一個分類結(jié)果。算法通過學(xué)習(xí)這些樣本,得到一個決策樹,這個決策樹能夠?qū)π碌臄?shù)據(jù)給出合適的分類
假設(shè)現(xiàn)有用戶14名,其個人屬性及是否購買某一產(chǎn)品的數(shù)據(jù)如下:
編號 | 年齡 | 收入范圍 | 工作性質(zhì) | 信用評級 | 購買決策 |
---|---|---|---|---|---|
01 | <30 | 高 | 不穩(wěn)定 | 較差 | 否 |
02 | <30 | 高 | 不穩(wěn)定 | 好 | 否 |
03 | 30-40 | 高 | 不穩(wěn)定 | 較差 | 是 |
04 | >40 | 中等 | 不穩(wěn)定 | 較差 | 是 |
05 | >40 | 低 | 穩(wěn)定 | 較差 | 是 |
06 | >40 | 低 | 穩(wěn)定 | 好 | 否 |
07 | 30-40 | 低 | 穩(wěn)定 | 好 | 是 |
08 | <30 | 中等 | 不穩(wěn)定 | 較差 | 否 |
09 | <30 | 低 | 穩(wěn)定 | 較差 | 是 |
10 | >40 | 中等 | 穩(wěn)定 | 較差 | 是 |
11 | <30 | 中等 | 穩(wěn)定 | 好 | 是 |
12 | 30-40 | 中等 | 不穩(wěn)定 | 好 | 是 |
13 | 30-40 | 高 | 穩(wěn)定 | 較差 | 是 |
14 | >40 | 中等 | 不穩(wěn)定 | 好 | 否 |
為了方便處理,對模擬數(shù)據(jù)按以下規(guī)則轉(zhuǎn)換為數(shù)值型列表數(shù)據(jù):
年齡:<30賦值為0;30-40賦值為1;>40賦值為2
收入:低為0;中為1;高為2
工作性質(zhì):不穩(wěn)定為0;穩(wěn)定為1
信用評級:差為0;好為1
#創(chuàng)建數(shù)據(jù)集 def createdataset(): dataSet=[[0,2,0,0,'N'], [0,2,0,1,'N'], [1,2,0,0,'Y'], [2,1,0,0,'Y'], [2,0,1,0,'Y'], [2,0,1,1,'N'], [1,0,1,1,'Y'], [0,1,0,0,'N'], [0,0,1,0,'Y'], [2,1,1,0,'Y'], [0,1,1,1,'Y'], [1,1,0,1,'Y'], [1,2,1,0,'Y'], [2,1,0,1,'N'],] labels=['age','income','job','credit'] return dataSet,labels
調(diào)用函數(shù),可獲得數(shù)據(jù):
ds1,lab = createdataset() print(ds1) print(lab)
[[0, 2, 0, 0, ‘N’], [0, 2, 0, 1, ‘N’], [1, 2, 0, 0, ‘Y’], [2, 1, 0, 0, ‘Y’], [2, 0, 1, 0, ‘Y’], [2, 0, 1, 1, ‘N’], [1, 0, 1, 1, ‘Y’], [0, 1, 0, 0, ‘N’], [0, 0, 1, 0, ‘Y’], [2, 1, 1, 0, ‘Y’], [0, 1, 1, 1, ‘Y’], [1, 1, 0, 1, ‘Y’], [1, 2, 1, 0, ‘Y’], [2, 1, 0, 1, ‘N’]]
[‘age’, ‘income’, ‘job’, ‘credit’]
信息熵也稱為香農(nóng)熵,是隨機(jī)變量的期望。度量信息的不確定程度。信息的熵越大,信息就越不容易搞清楚。處理信息就是為了把信息搞清楚,就是熵減少的過程。
def calcShannonEnt(dataSet): numEntries = len(dataSet) labelCounts = {} for featVec in dataSet: currentLabel = featVec[-1] if currentLabel not in labelCounts.keys(): labelCounts[currentLabel] = 0 labelCounts[currentLabel] += 1 shannonEnt = 0.0 for key in labelCounts: prob = float(labelCounts[key])/numEntries shannonEnt -= prob*log(prob,2) return shannonEnt
樣本數(shù)據(jù)信息熵:
shan = calcShannonEnt(ds1) print(shan)
0.9402859586706309
信息增益:用于度量屬性A降低樣本集合X熵的貢獻(xiàn)大小。信息增益越大,越適于對X分類。
def chooseBestFeatureToSplit(dataSet): numFeatures = len(dataSet[0])-1 baseEntropy = calcShannonEnt(dataSet) bestInfoGain = 0.0;bestFeature = -1 for i in range(numFeatures): featList = [example[i] for example in dataSet] uniqueVals = set(featList) newEntroy = 0.0 for value in uniqueVals: subDataSet = splitDataSet(dataSet, i, value) prop = len(subDataSet)/float(len(dataSet)) newEntroy += prop * calcShannonEnt(subDataSet) infoGain = baseEntropy - newEntroy if(infoGain > bestInfoGain): bestInfoGain = infoGain bestFeature = i return bestFeature
以上代碼實現(xiàn)了基于信息熵增益的ID3決策樹學(xué)習(xí)算法。其核心邏輯原理是:依次選取屬性集中的每一個屬性,將樣本集按照此屬性的取值分割為若干個子集;對這些子集計算信息熵,其與樣本的信息熵的差,即為按照此屬性分割的信息熵增益;找出所有增益中最大的那一個對應(yīng)的屬性,就是用于分割樣本集的屬性。
計算樣本最佳的分割樣本屬性,結(jié)果顯示為第0列,即age屬性:
col = chooseBestFeatureToSplit(ds1) col
0
def majorityCnt(classList): classCount = {} for vote in classList: if vote not in classCount.keys():classCount[vote] = 0 classCount[vote] += 1 sortedClassCount = sorted(classList.iteritems(),key=operator.itemgetter(1),reverse=True)#利用operator操作鍵值排序字典 return sortedClassCount[0][0] #創(chuàng)建樹的函數(shù) def createTree(dataSet,labels): classList = [example[-1] for example in dataSet] if classList.count(classList[0]) == len(classList): return classList[0] if len(dataSet[0]) == 1: return majorityCnt(classList) bestFeat = chooseBestFeatureToSplit(dataSet) bestFeatLabel = labels[bestFeat] myTree = {bestFeatLabel:{}} del(labels[bestFeat]) featValues = [example[bestFeat] for example in dataSet] uniqueVals = set(featValues) for value in uniqueVals: subLabels = labels[:] myTree[bestFeatLabel][value] = createTree(splitDataSet(dataSet, bestFeat, value), subLabels) return myTree
majorityCnt
函數(shù)用于處理一下情況:最終的理想決策樹應(yīng)該沿著決策分支到達(dá)最底端時,所有的樣本應(yīng)該都是相同的分類結(jié)果。但是真實樣本中難免會出現(xiàn)所有屬性一致但分類結(jié)果不一樣的情況,此時majorityCnt
將這類樣本的分類標(biāo)簽都調(diào)整為出現(xiàn)次數(shù)最多的那一個分類結(jié)果。
createTree
是核心任務(wù)函數(shù),它對所有的屬性依次調(diào)用ID3信息熵增益算法進(jìn)行計算處理,最終生成決策樹。
利用樣本數(shù)據(jù)構(gòu)造決策樹:
Tree = createTree(ds1, lab) print("樣本數(shù)據(jù)決策樹:") print(Tree)
樣本數(shù)據(jù)決策樹:
{‘age’: {0: {‘job’: {0: ‘N’, 1: ‘Y’}},
1: ‘Y’,
2: {‘credit’: {0: ‘Y’, 1: ‘N’}}}}
給出一個新的用戶信息,判斷ta是否購買某一產(chǎn)品:
年齡 | 收入范圍 | 工作性質(zhì) | 信用評級 |
---|---|---|---|
<30 | 低 | 穩(wěn)定 | 好 |
<30 | 高 | 不穩(wěn)定 | 好 |
def classify(inputtree,featlabels,testvec): firststr = list(inputtree.keys())[0] seconddict = inputtree[firststr] featindex = featlabels.index(firststr) for key in seconddict.keys(): if testvec[featindex]==key: if type(seconddict[key]).__name__=='dict': classlabel=classify(seconddict[key],featlabels,testvec) else: classlabel=seconddict[key] return classlabel
labels=['age','income','job','credit'] tsvec=[0,0,1,1] print('result:',classify(Tree,labels,tsvec)) tsvec1=[0,2,0,1] print('result1:',classify(Tree,labels,tsvec1))
result: Y
result1: N
以下代碼用于繪制決策樹圖形,非決策樹算法重點,有興趣可參考學(xué)習(xí)
import matplotlib.pyplot as plt decisionNode = dict(box, fc="0.8") leafNode = dict(box, fc="0.8") arrow_args = dict(arrow) #獲取葉節(jié)點的數(shù)目 def getNumLeafs(myTree): numLeafs = 0 firstStr = list(myTree.keys())[0] secondDict = myTree[firstStr] for key in secondDict.keys(): if type(secondDict[key]).__name__=='dict':#測試節(jié)點的數(shù)據(jù)是否為字典,以此判斷是否為葉節(jié)點 numLeafs += getNumLeafs(secondDict[key]) else: numLeafs +=1 return numLeafs #獲取樹的層數(shù) def getTreeDepth(myTree): maxDepth = 0 firstStr = list(myTree.keys())[0] secondDict = myTree[firstStr] for key in secondDict.keys(): if type(secondDict[key]).__name__=='dict':#測試節(jié)點的數(shù)據(jù)是否為字典,以此判斷是否為葉節(jié)點 thisDepth = 1 + getTreeDepth(secondDict[key]) else: thisDepth = 1 if thisDepth > maxDepth: maxDepth = thisDepth return maxDepth #繪制節(jié)點 def plotNode(nodeTxt, centerPt, parentPt, nodeType): createPlot.ax1.annotate(nodeTxt, xy=parentPt, xycoords='axes fraction', xytext=centerPt, textcoords='axes fraction', va="center", ha="center", bbox=nodeType, arrowprops=arrow_args ) #繪制連接線 def plotMidText(cntrPt, parentPt, txtString): xMid = (parentPt[0]-cntrPt[0])/2.0 + cntrPt[0] yMid = (parentPt[1]-cntrPt[1])/2.0 + cntrPt[1] createPlot.ax1.text(xMid, yMid, txtString, va="center", ha="center", rotation=30) #繪制樹結(jié)構(gòu) def plotTree(myTree, parentPt, nodeTxt):#if the first key tells you what feat was split on numLeafs = getNumLeafs(myTree) #this determines the x width of this tree depth = getTreeDepth(myTree) firstStr = list(myTree.keys())[0] #the text label for this node should be this cntrPt = (plotTree.xOff + (1.0 + float(numLeafs))/2.0/plotTree.totalW, plotTree.yOff) plotMidText(cntrPt, parentPt, nodeTxt) plotNode(firstStr, cntrPt, parentPt, decisionNode) secondDict = myTree[firstStr] plotTree.yOff = plotTree.yOff - 1.0/plotTree.totalD for key in secondDict.keys(): if type(secondDict[key]).__name__=='dict':#test to see if the nodes are dictonaires, if not they are leaf nodes plotTree(secondDict[key],cntrPt,str(key)) #recursion else: #it's a leaf node print the leaf node plotTree.xOff = plotTree.xOff + 1.0/plotTree.totalW plotNode(secondDict[key], (plotTree.xOff, plotTree.yOff), cntrPt, leafNode) plotMidText((plotTree.xOff, plotTree.yOff), cntrPt, str(key)) plotTree.yOff = plotTree.yOff + 1.0/plotTree.totalD #創(chuàng)建決策樹圖形 def createPlot(inTree): fig = plt.figure(1, facecolor='white') fig.clf() axprops = dict(xticks=[], yticks=[]) createPlot.ax1 = plt.subplot(111, frameon=False, **axprops) #no ticks #createPlot.ax1 = plt.subplot(111, frameon=False) #ticks for demo puropses plotTree.totalW = float(getNumLeafs(inTree)) plotTree.totalD = float(getTreeDepth(inTree)) plotTree.xOff = -0.5/plotTree.totalW; plotTree.yOff = 1.0; plotTree(inTree, (0.5,1.0), '') plt.savefig('決策樹.png',dpi=300,bbox_inches='tight') plt.show()
以上就是“python如何實現(xiàn)決策樹分類算法”這篇文章的所有內(nèi)容,感謝各位的閱讀!相信大家閱讀完這篇文章都有很大的收獲,小編每天都會為大家更新不同的知識,如果還想學(xué)習(xí)更多的知識,請關(guān)注億速云行業(yè)資訊頻道。
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