Giz*_*lly 4 python machine-learning scikit-learn
我需要将一个值设置为特定的阈值并生成一个混淆矩阵。数据位于 csv 文件 (11,1 MB) 中,此下载链接为:https : //drive.google.com/file/d/1cQFp7HteaaL37CefsbMNuHqPzkINCVzs/view ? usp=sharing ?
首先,我收到一条错误消息:““AttributeError: predict_proba is not available whenprobability=False”“所以我用它来纠正:
svc = SVC(C=1e9,gamma= 1e-07)
scv_calibrated = CalibratedClassifierCV(svc)
svc_model = scv_calibrated.fit(X_train, y_train)
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我在互联网上看到了很多,但我不太明白具体的阈值是如何被个人化的。听起来挺难的。现在,我看到错误的输出:
array([[ 0, 0],
[5359, 65]])
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我不知道出了什么问题。
我需要帮助,我是新手。谢谢
from sklearn.model_selection import train_test_split
df = pd.read_csv('fraud_data.csv')
X = df.iloc[:,:-1]
y = df.iloc[:,-1]
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)
def answer_four():
from sklearn.metrics import confusion_matrix
from sklearn.svm import SVC
from sklearn.calibration import CalibratedClassifierCV
from sklearn.model_selection import train_test_split
svc = SVC(C=1e9,gamma= 1e-07)
scv_calibrated = CalibratedClassifierCV(svc)
svc_model = scv_calibrated.fit(X_train, y_train)
# set threshold as -220
y_pred = (svc_model.predict_proba(X_test)[:,1] >= -220)
conf_matrix = confusion_matrix(y_pred, svc_model.predict(X_test))
return conf_matrix
answer_four()
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这个函数应该返回一个混淆矩阵,一个有 4 个整数的 2x2 numpy 数组。
这段代码产生了预期的输出,除了在前面的代码中我错误地使用了混淆矩阵的事实之外,我还应该使用decision_function并使输出过滤220阈值。
def answer_four():
from sklearn.metrics import confusion_matrix
from sklearn.svm import SVC
from sklearn.calibration import CalibratedClassifierCV
from sklearn.model_selection import train_test_split
#SVC without mencions of kernel, the default is rbf
svc = SVC(C=1e9, gamma=1e-07).fit(X_train, y_train)
#decision_function scores: Predict confidence scores for samples
y_score = svc.decision_function(X_test)
#Set a threshold -220
y_score = np.where(y_score > -220, 1, 0)
conf_matrix = confusion_matrix(y_test, y_score)
####threshold###
#input threshold in the model after trained this model
#threshold is a limiar of separation of class
return conf_matrix
answer_four()
#output:
array([[5320, 24],
[ 14, 66]])
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