How to fix the TypeError: predict_proba() missing 1 required positional argument: 'X'

Question

对于这个问题，我不知道为什么它说 predict_proba() missing 1 required positional argument: 'X'

谁能帮我解决这个问题？

这是我的代码：

df1 = pd.read_csv('new_customer_info.csv')
df1 = df1.drop(columns = ['person_home_ownership', 'loan_intent', 'loan_grade'])
df1.dropna(inplace=True)
df1['cb_person_default_on_file'] = df1['cb_person_default_on_file'].map({'Y':1 ,'N':0})

X = df1.to_numpy()
from sklearn import linear_model
regression = linear_model.LinearRegression()
regression.fit(X_train,y_train)

y_pred = LogisticRegression.predict_proba(X)

Answer 1

您正在直接使用 class，请使用 class LogisticRegression 的对象，该对象在您的代码中定义为 regression = linear_model.LogisticRegression()

解决方案：

y_pred = regression.predict_proba(X)

注：你也在混合线性回归和逻辑回归。请记住 predict_proba 不适用于回归算法 (LinearRegression)

How to fix the TypeError: predict_proba() missing 1 required positional argument: 'X'

How to fix the TypeError: predict_proba() missing 1 required positional argument: 'X'

python

linear-regression

scikit-learn

logistic-regression

data-science