我有一个关于Keras中model.evaluate（）和model.predict（）函数的问题。我在Keras中构建了一个简单的LSTM模型，并希望在测试数据集上测试模型性能。我考虑了以下两种方法来计算测试数据集上的度量：

• 使用model.evaluate（）方法
• 使用model.predict（）方法获得拟合值并手动计算度量

然而，我最终得到了不同的结果。此外，model.evaluate（）方法的结果也取决于batch_size参数的值。根据我的理解和这篇文章，它们应该有相同的结果。下面是可以复制结果的代码：

import tensorflow as tf
from keras.models import Model
from keras.layers import Dense, LSTM, Activation, Input
import numpy as np
from tqdm.notebook import tqdm
import keras.backend as K
from keras.callbacks import ModelCheckpoint, EarlyStopping

class VLSTM:
    def __init__(self, input_shape=(6, 1), nb_output_units=1, nb_hidden_units=128, dropout=0.0, 
                 recurrent_dropout=0.0, nb_layers=1):
        self.input_shape = input_shape
        self.nb_output_units = nb_output_units
        self.nb_hidden_units = nb_hidden_units
        self.nb_layers = nb_layers
        self.dropout = dropout
        self.recurrent_dropout = recurrent_dropout

    def build(self):
        inputs = Input(shape=self.input_shape)
        outputs = LSTM(self.nb_hidden_units)(inputs)
        outputs = Dense(1, activation=None)(outputs)
        return Model(inputs=[inputs], outputs=[outputs])
    
def RMSE(output, target):
    return K.sqrt(K.mean((output - target) ** 2))

n_train = 500
n_val = 100
n_test = 250 

X_train = np.random.rand(n_train, 6, 1)
Y_train = np.random.rand(n_train, 1)
X_val = np.random.rand(n_val, 6, 1)
Y_val = np.random.rand(n_val, 1)
X_test = np.random.rand(n_test, 6, 1)
Y_test = np.random.rand(n_test, 1)

input_shape = (X_train.shape[1], X_train.shape[2])
model = VLSTM(input_shape=input_shape)
m = model.build()
m.compile(loss=RMSE,
              optimizer='adam',
              metrics=[RMSE])

callbacks = []
callbacks.append(EarlyStopping(patience=30))

# train model
hist = m.fit(X_train, Y_train, \
             batch_size=32, epochs=10, shuffle=True, \
             validation_data=(X_val, Y_val), callbacks=callbacks)

# Use evaluate method with default batch size
test_mse = m.evaluate(X_test, Y_test)[1]
print("Mse is {} using evaluate method with default batch size".format(test_mse))

# Use evaluate method with batch size 1
test_mse = m.evaluate(X_test, Y_test, batch_size=1)[1]
print("Mse is {} using evaluate method with batch size = 1".format(test_mse))

# Use evaluate method with batch size = n_test
test_mse = m.evaluate(X_test, Y_test, batch_size=n_test)[1]
print("Mse is {} using evaluate method with batch size = n_test".format(test_mse))

# Use pred method and compute RMSE mannually
Y_test_pred = m.predict(X_test)
test_mse = np.sqrt( ((Y_test_pred - Y_test) ** 2).mean())
print("Mse is {} using evaluate method with batch size = 1".format(test_mse))

运行代码后，结果如下：
使用默认批量评估方法，MSE为0.3068242073059082
使用批量= 1的评估方法，MSE为0.26647186279296875
使用批量= n_test的评估方法，Mse为0.30763307213783264
用预测方法计算均方误差为0.3076330596820157
看起来使用mode.predict（）和model.evaluate（）与batch size = n_test给出了相同的结果。有人能解释一下吗？提前感谢！