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我有这段代码，我想对所有折叠（测试和训练）应用 10 折交叉验证，下面的代码仅适用于一折。如何为所有折叠编写一个循环（9 折用于训练，1 折用于测试，循环用于所有折叠）？

dataset=pd.read_csv("C:/Users/User/Desktop/minority class equal.csv",encoding='cp1252')

dataset.shape 
#output(53480, 37) 

array = dataset.values
X = np.expand_dims(array[:,0:36], axis=-1)
Y = np.expand_dims(array[:,36], axis=-1)
kf = KFold(n_splits=10)
kf.get_n_splits(X)

ACC_array = np.array([])
sensitivity_array = np.array([])
specificity_array = np.array([])

for trainindex, testindex in kf.split(X):
Xtrain, Xtest = X[trainindex], X[testindex]
Ytrain, Ytest = Y[trainindex], Y[testindex]

Ytrain = np.expand_dims(Ytrain, axis=-1)
Ytest = np.expand_dims(Ytest, axis=-1)
print(Xtrain.shape, Ytrain.shape)

n_timesteps, n_features, n_outputs =Xtrain.shape[0], Xtrain.shape[1], Ytrain.shape[1]

model = Sequential()
model.add(Conv1D(filters=64, kernel_size=1, activation='relu',input_shape=(n_features,1)))

model.add(Conv1D(filters=64, kernel_size=1, activation='relu'))
model.add(Dropout(0.5))
model.add(MaxPooling1D(pool_size=2))
model.add(Flatten())

model.add(Dense(100, activation='relu'))

model.add(Dense(n_outputs, activation='sigmoid'))

model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])

model.fit(Xtrain, Ytrain, epochs=20, batch_size=100)
Xtest = Xtest.astype(np.float32)

Predictions = model.predict(Xtest,batch_size =1024)
rounded = [round(x[0]) for x in Predictions]
Y_predection = pd.DataFrame(rounded)
Y_predection = Y_predection.iloc[:, 0]

confusion = metrics.confusion_matrix(Ytest,Y_predection)
TP= confusion[1, 1]
TN= confusion[0, 0]
FP= confusion[0, 1]
FN= confusion[1, 0]

Accuracy=(TP + TN) / float(TP + TN + FP + FN)
sensitivity = TP / float(FN + TP)
specificity= TN / float(TN + FP)

ACC_array = np.append(ACC_array,Accuracy)
sensitivity_array = np.append(sensitivity_array,sensitivity)
specificity_array = np.append(specificity_array,specificity)

ACC_mean=np.mean(ACC_array, axis=0 )
print ('mean Accuracy',ACC_mean*100)
sensitivity_mean=np.mean(sensitivity_array, axis=0 )
print ('mean sensitivity',sensitivity_mean*100)
specificity_mean=np.mean(specificity_array, axis=0 )
print ('mean specificity',specificity_mean*100)

此代码没有任何错误，但仅针对（测试为 1 倍，训练为 9 倍）应用了 10 倍交叉验证。

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