Scikit learn 利用sklearn的id3算法训练决策树

我正在尝试使用id3算法训练决策树。 其目的是获得所选特征的索引，估计发生率，并建立总混淆矩阵

该算法应将数据集拆分为训练集和测试集，并使用4倍交叉验证

我是新来的，我读过关于sklearn的教程和关于学习过程的理论，但我还是很困惑

我试着做的是：

from sklearn.model_selection import cross_val_predict,KFold,cross_val_score, 
train_test_split, learning_curve
from sklearn.tree import DecisionTreeClassifier
from sklearn.metrics import confusion_matrix


X_train, X_test, y_train, y_test = train_test_split(x, y, test_size=0.2, random_state=1)
clf = DecisionTreeClassifier(criterion='entropy', random_state=0)
clf.fit(X_train,y_train)
results = cross_val_score(estimator=clf, X=X_train, y=y_train, cv=4)
print("Accuracy: %0.2f (+/- %0.2f)" % (results.mean(), results.std()))
y_pred = cross_val_predict(estimator=clf, X=x, y=y, cv=4)
conf_mat = confusion_matrix(y,y_pred)
print(conf_mat)
dot_data = tree.export_graphviz(clf, out_file='tree.dot') 

我有一些问题：

如何获得培训中使用的功能索引列表？我必须穿过clf的树吗？找不到任何api方法来检索它们

我必须使用“适合”、“交叉评分”和“交叉预测”吗？似乎他们都做了一些学习过程，但我无法从他们中的一个获得clf拟合、精度和混淆矩阵

我是否必须使用测试集来估计数据集的折叠或划分数据集的折叠

要检索培训过程中使用的功能列表，您可以通过以下方式从x中获取列：

feature\u list=x.columns

正如你所知道的，并不是每个特征在预测中都有用。您可以看到，在训练模型之后，使用

clf.功能\u重要性\u

要素列表中的要素索引与要素重要性列表中的要素索引相同

如果使用交叉验证，则无法立即检索分数。
cross_val_分数完成了交易，但获得分数的更好方法是使用cross_Valid。它的工作方式与cross_val_score相同，但您可以检索更多的分数值，只需使用make_score创建所需的每个分数并通过，下面是一个示例：

from sklearn.model_selection import train_test_split,  cross_validate
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score 
import pandas as pd, numpy as np       

x_train, x_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
dtc = DecisionTreeClassifier()
dtc_fit = dtc.fit(x_train, y_train)

def tn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 0]
def fp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 1]
def fn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 0]
def tp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 1]

scoring = {
    'tp' : make_scorer(tp), 
    'tn' : make_scorer(tn), 
    'fp' : make_scorer(fp), 
    'fn' : make_scorer(fn), 
    'accuracy' : make_scorer(accuracy_score),
    'precision': make_scorer(precision_score),
    'f1_score' : make_scorer(f1_score),
    'recall'   : make_scorer(recall_score)
}

sc = cross_validate(dtc_fit, x_train, y_train, cv=5, scoring=scoring)

print("Accuracy: %0.2f (+/- %0.2f)" % (sc['test_accuracy'].mean(), sc['test_accuracy'].std() * 2))
print("Precision: %0.2f (+/- %0.2f)" % (sc['test_precision'].mean(), sc['test_precision'].std() * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (sc['test_f1_score'].mean(), sc['test_f1_score'].std() * 2))
print("Recall: %0.2f (+/- %0.2f)" % (sc['test_recall'].mean(), sc['test_recall'].std() * 2), "\n")

stp = math.ceil(sc['test_tp'].mean())
stn = math.ceil(sc['test_tn'].mean())
sfp = math.ceil(sc['test_fp'].mean())
sfn = math.ceil(sc['test_fn'].mean())

confusion_matrix = pd.DataFrame(
    [[stn, sfp], [sfn, stp]],
    columns=['Predicted 0', 'Predicted 1'],
    index=['True 0', 'True 1']
)
print(conf_m)

from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import StratifiedShuffleSplit
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score
import pandas as pd, numpy as np

precision = []; recall = []; f1score = []; accuracy = []

sss = StratifiedShuffleSplit(n_splits=10, test_size=0.2)    
dtc = DecisionTreeClassifier()

for train_index, test_index in sss.split(X, y):
    X_train, X_test = X.iloc[train_index], X.iloc[test_index]
    y_train, y_test = y.iloc[train_index], y.iloc[test_index]

    dtc.fit(X_train, y_train)
    pred = dtc.predict(X_test)

    precision.append(precision_score(y_test, pred))
    recall.append(recall_score(y_test, pred))
    f1score.append(f1_score(y_test, pred))
    accuracy.append(accuracy_score(y_test, pred))   

print("Accuracy: %0.2f (+/- %0.2f)" % (np.mean(accuracy),np.std(accuracy) * 2))
print("Precision: %0.2f (+/- %0.2f)" % (np.mean(precision),np.std(precision) * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (np.mean(f1score),np.std(f1score) * 2))
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(recall),np.std(recall) * 2))

使用交叉函数时，函数本身会为测试和训练创建折叠。如果您想管理列车折叠和测试折叠，您可以使用K_折叠类自己完成。
如果你需要保持班级平衡，总是需要一个好的得分决策树分类，你必须使用分层折叠。如果要随机洗牌折叠中包含的值，可以使用StratifiedShuffleSplit。这里有一个例子：

from sklearn.model_selection import train_test_split,  cross_validate
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score 
import pandas as pd, numpy as np       

x_train, x_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
dtc = DecisionTreeClassifier()
dtc_fit = dtc.fit(x_train, y_train)

def tn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 0]
def fp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 1]
def fn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 0]
def tp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 1]

scoring = {
    'tp' : make_scorer(tp), 
    'tn' : make_scorer(tn), 
    'fp' : make_scorer(fp), 
    'fn' : make_scorer(fn), 
    'accuracy' : make_scorer(accuracy_score),
    'precision': make_scorer(precision_score),
    'f1_score' : make_scorer(f1_score),
    'recall'   : make_scorer(recall_score)
}

sc = cross_validate(dtc_fit, x_train, y_train, cv=5, scoring=scoring)

print("Accuracy: %0.2f (+/- %0.2f)" % (sc['test_accuracy'].mean(), sc['test_accuracy'].std() * 2))
print("Precision: %0.2f (+/- %0.2f)" % (sc['test_precision'].mean(), sc['test_precision'].std() * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (sc['test_f1_score'].mean(), sc['test_f1_score'].std() * 2))
print("Recall: %0.2f (+/- %0.2f)" % (sc['test_recall'].mean(), sc['test_recall'].std() * 2), "\n")

stp = math.ceil(sc['test_tp'].mean())
stn = math.ceil(sc['test_tn'].mean())
sfp = math.ceil(sc['test_fp'].mean())
sfn = math.ceil(sc['test_fn'].mean())

confusion_matrix = pd.DataFrame(
    [[stn, sfp], [sfn, stp]],
    columns=['Predicted 0', 'Predicted 1'],
    index=['True 0', 'True 1']
)
print(conf_m)

from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import StratifiedShuffleSplit
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score
import pandas as pd, numpy as np

precision = []; recall = []; f1score = []; accuracy = []

sss = StratifiedShuffleSplit(n_splits=10, test_size=0.2)    
dtc = DecisionTreeClassifier()

for train_index, test_index in sss.split(X, y):
    X_train, X_test = X.iloc[train_index], X.iloc[test_index]
    y_train, y_test = y.iloc[train_index], y.iloc[test_index]

    dtc.fit(X_train, y_train)
    pred = dtc.predict(X_test)

    precision.append(precision_score(y_test, pred))
    recall.append(recall_score(y_test, pred))
    f1score.append(f1_score(y_test, pred))
    accuracy.append(accuracy_score(y_test, pred))   

print("Accuracy: %0.2f (+/- %0.2f)" % (np.mean(accuracy),np.std(accuracy) * 2))
print("Precision: %0.2f (+/- %0.2f)" % (np.mean(precision),np.std(precision) * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (np.mean(f1score),np.std(f1score) * 2))
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(recall),np.std(recall) * 2))

我希望我已经回答了你需要的一切

要检索培训过程中使用的功能列表，您可以通过以下方式从x中获取列：

feature\u list=x.columns

正如你所知道的，并不是每个特征在预测中都有用。您可以看到，在训练模型之后，使用

clf.功能\u重要性\u

要素列表中的要素索引与要素重要性列表中的要素索引相同

如果使用交叉验证，则无法立即检索分数。
cross_val_分数完成了交易，但获得分数的更好方法是使用cross_Valid。它的工作方式与cross_val_score相同，但您可以检索更多的分数值，只需使用make_score创建所需的每个分数并通过，下面是一个示例：

from sklearn.model_selection import train_test_split,  cross_validate
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score 
import pandas as pd, numpy as np       

x_train, x_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
dtc = DecisionTreeClassifier()
dtc_fit = dtc.fit(x_train, y_train)

def tn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 0]
def fp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 1]
def fn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 0]
def tp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 1]

scoring = {
    'tp' : make_scorer(tp), 
    'tn' : make_scorer(tn), 
    'fp' : make_scorer(fp), 
    'fn' : make_scorer(fn), 
    'accuracy' : make_scorer(accuracy_score),
    'precision': make_scorer(precision_score),
    'f1_score' : make_scorer(f1_score),
    'recall'   : make_scorer(recall_score)
}

sc = cross_validate(dtc_fit, x_train, y_train, cv=5, scoring=scoring)

print("Accuracy: %0.2f (+/- %0.2f)" % (sc['test_accuracy'].mean(), sc['test_accuracy'].std() * 2))
print("Precision: %0.2f (+/- %0.2f)" % (sc['test_precision'].mean(), sc['test_precision'].std() * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (sc['test_f1_score'].mean(), sc['test_f1_score'].std() * 2))
print("Recall: %0.2f (+/- %0.2f)" % (sc['test_recall'].mean(), sc['test_recall'].std() * 2), "\n")

stp = math.ceil(sc['test_tp'].mean())
stn = math.ceil(sc['test_tn'].mean())
sfp = math.ceil(sc['test_fp'].mean())
sfn = math.ceil(sc['test_fn'].mean())

confusion_matrix = pd.DataFrame(
    [[stn, sfp], [sfn, stp]],
    columns=['Predicted 0', 'Predicted 1'],
    index=['True 0', 'True 1']
)
print(conf_m)

from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import StratifiedShuffleSplit
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score
import pandas as pd, numpy as np

precision = []; recall = []; f1score = []; accuracy = []

sss = StratifiedShuffleSplit(n_splits=10, test_size=0.2)    
dtc = DecisionTreeClassifier()

for train_index, test_index in sss.split(X, y):
    X_train, X_test = X.iloc[train_index], X.iloc[test_index]
    y_train, y_test = y.iloc[train_index], y.iloc[test_index]

    dtc.fit(X_train, y_train)
    pred = dtc.predict(X_test)

    precision.append(precision_score(y_test, pred))
    recall.append(recall_score(y_test, pred))
    f1score.append(f1_score(y_test, pred))
    accuracy.append(accuracy_score(y_test, pred))   

print("Accuracy: %0.2f (+/- %0.2f)" % (np.mean(accuracy),np.std(accuracy) * 2))
print("Precision: %0.2f (+/- %0.2f)" % (np.mean(precision),np.std(precision) * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (np.mean(f1score),np.std(f1score) * 2))
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(recall),np.std(recall) * 2))

使用交叉函数时，函数本身会为测试和训练创建折叠。如果您想管理列车折叠和测试折叠，您可以使用K_折叠类自己完成。
如果你需要保持班级平衡，总是需要一个好的得分决策树分类，你必须使用分层折叠。如果要随机洗牌折叠中包含的值，可以使用StratifiedShuffleSplit。这里有一个例子：

from sklearn.model_selection import train_test_split,  cross_validate
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score 
import pandas as pd, numpy as np       

x_train, x_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
dtc = DecisionTreeClassifier()
dtc_fit = dtc.fit(x_train, y_train)

def tn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 0]
def fp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[0, 1]
def fn(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 0]
def tp(y_true, y_pred): return confusion_matrix(y_true, y_pred)[1, 1]

scoring = {
    'tp' : make_scorer(tp), 
    'tn' : make_scorer(tn), 
    'fp' : make_scorer(fp), 
    'fn' : make_scorer(fn), 
    'accuracy' : make_scorer(accuracy_score),
    'precision': make_scorer(precision_score),
    'f1_score' : make_scorer(f1_score),
    'recall'   : make_scorer(recall_score)
}

sc = cross_validate(dtc_fit, x_train, y_train, cv=5, scoring=scoring)

print("Accuracy: %0.2f (+/- %0.2f)" % (sc['test_accuracy'].mean(), sc['test_accuracy'].std() * 2))
print("Precision: %0.2f (+/- %0.2f)" % (sc['test_precision'].mean(), sc['test_precision'].std() * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (sc['test_f1_score'].mean(), sc['test_f1_score'].std() * 2))
print("Recall: %0.2f (+/- %0.2f)" % (sc['test_recall'].mean(), sc['test_recall'].std() * 2), "\n")

stp = math.ceil(sc['test_tp'].mean())
stn = math.ceil(sc['test_tn'].mean())
sfp = math.ceil(sc['test_fp'].mean())
sfn = math.ceil(sc['test_fn'].mean())

confusion_matrix = pd.DataFrame(
    [[stn, sfp], [sfn, stp]],
    columns=['Predicted 0', 'Predicted 1'],
    index=['True 0', 'True 1']
)
print(conf_m)

from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import StratifiedShuffleSplit
from sklearn.metrics import classification_report, confusion_matrix, accuracy_score, f1_score, precision_score, make_scorer, recall_score
import pandas as pd, numpy as np

precision = []; recall = []; f1score = []; accuracy = []

sss = StratifiedShuffleSplit(n_splits=10, test_size=0.2)    
dtc = DecisionTreeClassifier()

for train_index, test_index in sss.split(X, y):
    X_train, X_test = X.iloc[train_index], X.iloc[test_index]
    y_train, y_test = y.iloc[train_index], y.iloc[test_index]

    dtc.fit(X_train, y_train)
    pred = dtc.predict(X_test)

    precision.append(precision_score(y_test, pred))
    recall.append(recall_score(y_test, pred))
    f1score.append(f1_score(y_test, pred))
    accuracy.append(accuracy_score(y_test, pred))   

print("Accuracy: %0.2f (+/- %0.2f)" % (np.mean(accuracy),np.std(accuracy) * 2))
print("Precision: %0.2f (+/- %0.2f)" % (np.mean(precision),np.std(precision) * 2))
print("f1_score: %0.2f (+/- %0.2f)" % (np.mean(f1score),np.std(f1score) * 2))
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(recall),np.std(recall) * 2))

---

我希望我已经回答了你需要的一切

非常感谢你！1.我想在结果中得到所选的特征，所以我假设我可以使用特征的重要性？2.如果我使用“accurancy”参数进行评分，它会如何改变算法？难道我不需要提供一个评分函数来实现使用熵的增益信息，使其成为“id3”吗？3.交叉验证是否保持了类别平衡？1。是的，你说得对。只需观察特征u重要性u您可以选择预测中最重要的特征，以降低预测模型的复杂性（此步骤也称为特征选择）2。当你评估一个模型时，你不仅要使用准确度，还要使用精确度、回忆和其他分数。此分数可以更改您的功能选择过程3。交叉验证支持类平衡。使用我发布的解决方案，您也可以检索自己的混淆矩阵。简而言之，我已经向您展示了交叉验证在这一点上是如何工作的。您是否使用pandas作为pd？我发现一个错误“AttributeError:module'pandas'没有属性'DataFrame'”，这帮了大忙谢谢，还有一个问题-你说过精度、回忆和其他分数与学习过程相关，那么为什么在拟合后选择它们呢？它们还与id3特别相关吗？是的，我使用熊猫作为pd，对不起，我忘了包括它。无论如何，精确性、召回率和其他分数都是在验证步骤之后测量的。这些在每个预测模型中都是相关的，比如ID3。有些模型，如DecisionTree，可能会出现过度拟合（）。精度为0.99的模型可能拟合过度。你只要看分数就可以看到这些东西非常感谢你！1.我想在结果中得到所选的特征，所以我假设我可以使用特征的重要性？2.如果我使用“accurancy”参数进行评分，它会如何改变算法？难道我不需要提供一个评分函数来实现使用熵的增益信息，使其成为“id3”吗？3.交叉验证是否保持了类别平衡？1。是的，你说得对。只需观察特征的重要性，就可以选择预测中最重要的特征，以降低预测的复杂性