我正在尝试建立一个逻辑回归模型，它可以预测新实例的类
下面是我所做的：

path = 'diabetes.csv'
df = pd.read_csv(path, header = None)
print "Classifying with Logistic Regression"
values = df.values
X = values[1:,0:8]
y = values[1:,8]
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.10, random_state=42)
model=LogisticRegression()
model.fit(X_train,y_train)

X_test = []
X_test.append(int(pregnancies_info))
X_test.append(int(glucose_info))
X_test.append(int(blood_press_info))
X_test.append(int(skin_thickness_info))
X_test.append(int(insulin_info))
X_test.append(float(BMI_info))
X_test.append(float(dpf_info))
X_test.append(int(age_info))
#X_test = np.array(X_test).reshape(-1, 1)
print X_test
y_pred=model.predict(X_test)
if y_pred == 0:
    Label(login_screen, text="Healthy").pack()
if y_pred == 1:
    Label(login_screen, text="Diabetes Metillus").pack()

pregnancies_entry.delete(0, END)
glucose_entry.delete(0, END)
blood_press_entry.delete(0, END)
skin_thickness_entry.delete(0, END)
insulin_entry.delete(0, END)
BMI_entry.delete(0, END)
dpf_entry.delete(0, END)
age_entry.delete(0, END)

但我有一个错误：

如果数据具有单个特征，则使用array.reshape（-1,1）重塑数据；如果数据包含单个样本，则使用array.reshape（1，-1）重塑数据。

如果我取消注释这行X\u test=np。数组（X_测试）。重塑（-1，1）出现以下错误：

文件“/anaconda2/lib/python2.7/site packages/sklearn/linear_model/base.py”，第305行，在decision_函数（X.shape[1]，n_features））值中错误：X每个样本有1个特征；8岁