我认为当输入值为 0 时 mask_zero=True 将输出 0，因此后续层可以跳过计算或其他操作。

mask_zero 是如何工作的？

示例:

data_in = np.array([
  [1, 2, 0, 0]
])
data_in.shape
>>> (1, 4)

# model
x = Input(shape=(4,))
e = Embedding(5, 5, mask_zero=True)(x)

m = Model(inputs=x, outputs=e)
p = m.predict(data_in)
print(p.shape)
print(p)

实际输出是:(数字是随机的)

(1, 4, 5)
[[[ 0.02499047  0.04617121  0.01586803  0.0338897   0.009652  ]
  [ 0.04782704 -0.04035913 -0.0341589   0.03020919 -0.01157228]
  [ 0.00451764 -0.01433611  0.02606953  0.00328832  0.02650392]
  [ 0.00451764 -0.01433611  0.02606953  0.00328832  0.02650392]]]

但是，我认为输出将是:

[[[ 0.02499047  0.04617121  0.01586803  0.0338897   0.009652  ]
  [ 0.04782704 -0.04035913 -0.0341589   0.03020919 -0.01157228]
  [ 0 0 0 0 0]
  [ 0 0 0 0 0]]]

最佳答案

实际上，为嵌入层设置 mask_zero=True 不会导致返回零向量。相反，嵌入层的行为不会改变，它会返回索引为零的嵌入向量。您可以通过检查嵌入层权重来确认这一点(即在您提到的示例中它是 m.layers[0].get_weights())。相反，它会影响后续层(例如 RNN 层)的行为。

如果您检查嵌入层的源代码，您会看到一个名为 compute_mask 的方法:

def compute_mask(self, inputs, mask=None):
    if not self.mask_zero:
        return None
    output_mask = K.not_equal(inputs, 0)
    return output_mask

此输出掩码将作为 mask 参数传递给支持掩码的后续层。这已在 __call__ 中实现基础层的方法，Layer:

# Handle mask propagation.
previous_mask = _collect_previous_mask(inputs)
user_kwargs = copy.copy(kwargs)
if not is_all_none(previous_mask):
    # The previous layer generated a mask.
    if has_arg(self.call, 'mask'):
        if 'mask' not in kwargs:
            # If mask is explicitly passed to __call__,
            # we should override the default mask.
            kwargs['mask'] = previous_mask

这使得后面的层忽略(即不在计算中考虑)这个输入步骤。这是一个最小的例子:

data_in = np.array([
  [1, 0, 2, 0]
])

x = Input(shape=(4,))
e = Embedding(5, 5, mask_zero=True)(x)
rnn = LSTM(3, return_sequences=True)(e)

m = Model(inputs=x, outputs=rnn)
m.predict(data_in)

array([[[-0.00084503, -0.00413611,  0.00049972],
        [-0.00084503, -0.00413611,  0.00049972],
        [-0.00144554, -0.00115775, -0.00293898],
        [-0.00144554, -0.00115775, -0.00293898]]], dtype=float32)

如您所见，LSTM 层的第二个和第四个时间步的输出分别与第一个和第三个时间步的输出相同。这意味着这些时间步长已被屏蔽。

更新:计算损失时也将考虑掩码，因为损失函数在内部增强以支持使用 weighted_masked_objective 的掩码:

def weighted_masked_objective(fn):
    """Adds support for masking and sample-weighting to an objective function.
    It transforms an objective function `fn(y_true, y_pred)`
    into a sample-weighted, cost-masked objective function
    `fn(y_true, y_pred, weights, mask)`.
    # Arguments
        fn: The objective function to wrap,
            with signature `fn(y_true, y_pred)`.
    # Returns
        A function with signature `fn(y_true, y_pred, weights, mask)`.
    """

when compiling the model :

weighted_losses = [weighted_masked_objective(fn) for fn in loss_functions]

您可以使用以下示例验证这一点:

data_in = np.array([[1, 2, 0, 0]])
data_out = np.arange(12).reshape(1,4,3)

x = Input(shape=(4,))
e = Embedding(5, 5, mask_zero=True)(x)
d = Dense(3)(e)

m = Model(inputs=x, outputs=d)
m.compile(loss='mse', optimizer='adam')
preds = m.predict(data_in)
loss = m.evaluate(data_in, data_out, verbose=0)
print(preds)
print('Computed Loss:', loss)

[[[ 0.009682    0.02505393 -0.00632722]
  [ 0.01756451  0.05928303  0.0153951 ]
  [-0.00146054 -0.02064196 -0.04356086]
  [-0.00146054 -0.02064196 -0.04356086]]]
Computed Loss: 9.041069030761719

# verify that only the first two outputs 
# have been considered in the computation of loss
print(np.square(preds[0,0:2] - data_out[0,0:2]).mean())

9.041070036475277

关于python - Keras Embedding 层中的 mask_zero 是如何工作的？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47485216/