Given the standard illustrative feed-forward neural net model, with the dots as neurons and the lines as neuron-to-neuron connection, what part is the (unfold) LSTM cell (see picture)? Is it a neuron (a dot) or a layer?

LSTM cell


1 Answer 1


The diagram you show works at least partially for describing both individual neurons and layers of those neurons.

However, the "incoming" data lines on the left represent all inputs under consideration, typically a vector of all inputs to the cell. That includes all data from current time steps (from input layer or earlier LSTM or time-distributed layers) - the line coming up into the LSTM cell - and the full output and cell state vectors from the whole LSTM layer on previous time step - the horizontal lines inside the cell starting on the left. Technically the top left input line could be read as either be a single neuron cell state value, or the full vector of cell state from the previous time step, depending on whether you were viewing the diagram as describing a single neuron in the layer, or the whole layer respectively.

If you visualise this cell connecting to itself over time steps, then the data lines both in and out must be whole layer vectors. The diagram is then best thought of as representing a whole LSTM layer, which is composed of various sub-layers which get combined, such as the forget gate layer (the leftmost yellow box).

Each yellow box in the diagram can be implemented very similar to a single layer of a simple feed forward NN, with its own weights and biases. So the forget gate can be implemented as

$$\mathbf{f}_t = \sigma(\mathbf{W}_f [\mathbf{x}_t , \mathbf{y}_{t-1}] + \mathbf{b}_f)$$

where $\mathbf{W}_f$ is the weight matrix of the forget gate, $\mathbf{b}_f$ is the bias vector, $\mathbf{x}_t$ is layer input on current time step, $\mathbf{y}_{t-1}$ is layer output from previous time step, and the comma $,$ is showing concatenation of those two (column) vectors into one large column vector.

If the input $\mathbf{x}$ is an $n$-dimensional vector, and output $\mathbf{y}$ is an $m$-dimensional vector, then:

  • $\mathbf{c}$, the cell state (top line), and all interim layer outputs are $m$-dimensional vectors
  • $\mathbf{W}_f$, the forget gate weights, and the three other internal weights matrices are $m \times (n+m)$ matrices
  • $\mathbf{b}_f$, the forget gate bias, and the three other internal bias vectors are $m$-dimensional vectors

The other gates are near identical (using their own weights and biases, plus the third cell candidate value uses tanh instead of logistic sigmoid), and have their own weight matrices etc. The whole thing is constructed a lot like four separate feed-forward layers, each receiving identical input, that are then combined using element-wise operations to perform the functions of the LSTM cell. There is nothing stopping you implementing each yellow box as a more sophisticated and deeper NN in its own right, but that seems relatively rare in practice, and it is more common to stack LSTM layers to achieve that kind of depth.

  • 2
    $\begingroup$ Your explanations seem to be consistent with my knowledge. I think it would even be more helpful if you also provide the dimensionality of the vectors and matrices, so that the OP can understand that the LSTM layer can be conceptually divided into LSTM units (or blocks). However, I know that this can easily become cumbersome and tedious. Moroever, it's also worth emphasing that this view is useful because libraries usually implement LSTM layers and not LSTM units. See, for example, tf.keras.layers.LSTM. $\endgroup$
    – nbro
    Feb 9, 2020 at 1:20
  • $\begingroup$ @nbro: I added concrete dimensions for matrices and vectors. I am not sure if adding an equivalent break down to the "neuron" unit level would be worth it, other than to note that this does exist conceptually, which I think is already implied. $\endgroup$ Feb 9, 2020 at 11:38
  • $\begingroup$ @NeilSlater, nice answer, I've just up-voted you, but just for the sake of completeness can you add as well the equations for input and output gates as well? $\endgroup$
    – JVGD
    Feb 11, 2020 at 10:48
  • $\begingroup$ @JVGD: The input and output gates use the exact same equation, just different variable names as they have dedicated weights and biases. E.g. $\mathbf{o}_t = \sigma(\mathbf{W}_o [\mathbf{x}_t , \mathbf{y}_{t-1}] + \mathbf{b}_o)$ - instead of repeating myself I will make it clear just how similar the gates are. $\endgroup$ Feb 11, 2020 at 10:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .