如何在Sequential keras模型中定义输入形状

问题描述 投票:0回答:1

请帮助在keras模型中定义合适的Dense输入形状。也许我必须首先重塑我的数据。我的数据集尺寸如下所示:

Data shapes are X_train: (2858, 2037) y_train: (2858, 1) X_test: (715, 2037) y_test: (715, 1) Number of features (input shape) is 2037

我想像那样定义Sequential keras模型

``

    batch_size = 128
    num_classes = 2
    epochs = 20

    model = Sequential()
    model.add(Dense(512, activation='relu', input_shape=(X_input_shape,)))
    model.add(Dropout(0.2))
    model.add(Dense(512, activation='relu'))

    model.summary()

    model.compile(loss='binary_crossentropy',
                optimizer=RMSprop(),
                from_logits=True,
                metrics=['accuracy'])

``

型号摘要:

``

    Layer (type)                 Output Shape              Param #   
    =================================================================
    dense_20 (Dense)             (None, 512)               1043456   
    _________________________________________________________________
    dropout_12 (Dropout)         (None, 512)               0         
    _________________________________________________________________
    dense_21 (Dense)             (None, 512)               262656    
    =================================================================
    Total params: 1,306,112
    Trainable params: 1,306,112
    Non-trainable params: 0

``

当我试图适应它时......

``

    history = model.fit(X_train, y_train,
                        batch_size=batch_size,
                        epochs=epochs,
                        verbose=1,
                        validation_data=(X_test, y_test))

``我得到一个错误:

``

  ValueError: Error when checking target: expected dense_21 to have shape (512,) but got array with shape (1,)                    

``

python keras-layer
1个回答
0
投票

修改

model.add(Dense(512, activation='relu'))

model.add(Dense(1, activation='relu'))

输出形状的大小为1,与y_train.shape [1]相同。

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