问题
The docs for an Embedding Layer in Keras say:
Turns positive integers (indexes) into dense vectors of fixed size. eg.
[[4], [20]]
->[[0.25, 0.1], [0.6, -0.2]]
I believe this could also be achieved by encoding the inputs as one-hot vectors of length vocabulary_size
, and feeding them into a Dense Layer.
Is an Embedding Layer merely a convenience for this two-step process, or is something fancier going on under the hood?
回答1:
Mathematically, the difference is this:
An embedding layer performs select operation. In keras, this layer is equivalent to:
K.gather(self.embeddings, inputs) # just one matrix
A dense layer performs dot-product operation, plus an optional activation:
outputs = matmul(inputs, self.kernel) # a kernel matrix outputs = bias_add(outputs, self.bias) # a bias vector return self.activation(outputs) # an activation function
You can emulate an embedding layer with fully-connected layer via one-hot encoding, but the whole point of dense embedding is to avoid one-hot representation. In NLP, the word vocabulary size can be of the order 100k (sometimes even a million). On top of that, it's often needed to process the sequences of words in a batch. Processing the batch of sequences of word indices would be much more efficient than the batch of sequences of one-hot vectors. In addition, gather
operation itself is faster than matrix dot-product, both in forward and backward pass.
回答2:
An embedding layer is faster, because it is essentially the equivalent of a dense layer that makes simplifying assumptions.
Imagine a word-to-embedding layer with these weights:
w = [[0.1, 0.2, 0.3, 0.4],
[0.5, 0.6, 0.7, 0.8],
[0.9, 0.0, 0.1, 0.2]]
A Dense
layer will treat these like actual weights with which to perform matrix multiplication. An embedding layer will simply treat these weights as a list of vectors, each vector representing one word; the 0th word in the vocabulary is w[0]
, 1st is w[1]
, etc.
For an example, use the weights above and this sentence:
[0, 2, 1, 2]
A naive Dense
-based net needs to convert that sentence to a 1-hot encoding
[[1, 0, 0],
[0, 0, 1],
[0, 1, 0],
[0, 0, 1]]
then do a matrix multiplication
[[1 * 0.1 + 0 * 0.5 + 0 * 0.9, 1 * 0.2 + 0 * 0.6 + 0 * 0.0, 1 * 0.3 + 0 * 0.7 + 0 * 0.1, 1 * 0.4 + 0 * 0.8 + 0 * 0.2],
[0 * 0.1 + 0 * 0.5 + 1 * 0.9, 0 * 0.2 + 0 * 0.6 + 1 * 0.0, 0 * 0.3 + 0 * 0.7 + 1 * 0.1, 0 * 0.4 + 0 * 0.8 + 1 * 0.2],
[0 * 0.1 + 1 * 0.5 + 0 * 0.9, 0 * 0.2 + 1 * 0.6 + 0 * 0.0, 0 * 0.3 + 1 * 0.7 + 0 * 0.1, 0 * 0.4 + 1 * 0.8 + 0 * 0.2],
[0 * 0.1 + 0 * 0.5 + 1 * 0.9, 0 * 0.2 + 0 * 0.6 + 1 * 0.0, 0 * 0.3 + 0 * 0.7 + 1 * 0.1, 0 * 0.4 + 0 * 0.8 + 1 * 0.2]]
=
[[0.1, 0.2, 0.3, 0.4],
[0.9, 0.0, 0.1, 0.2],
[0.5, 0.6, 0.7, 0.8],
[0.9, 0.0, 0.1, 0.2]]
However, an Embedding
layer simply looks at [0, 2, 1, 2]
and takes the weights of the layer at indices zero, two, one, and two to immediately get
[w[0],
w[2],
w[1],
w[2]]
=
[[0.1, 0.2, 0.3, 0.4],
[0.9, 0.0, 0.1, 0.2],
[0.5, 0.6, 0.7, 0.8],
[0.9, 0.0, 0.1, 0.2]]
So it's the same result, just obtained in a hopefully faster way.
The Embedding
layer does have limitations:
- The input needs to be integers in [0, vocab_length).
- No bias.
- No activation.
However, none of those limitations should matter if you just want to convert an integer-encoded word into an embedding.
来源:https://stackoverflow.com/questions/47868265/what-is-the-difference-between-an-embedding-layer-and-a-dense-layer