I'm trying to build a model that will learn features of a 3D space. Unlike image processing, the values of the 3D matrix are not continuous; they represent some discrete value of what "material" can be found at that specific coordinate (grass with value 1 or stairs with value 2 for example).
Is it possible to train a model to learn the features of the space without interpolating in-between values? For example, I don't want the neural net to deduce 1.5 to be some kind of grass stairs.
You'll want to use one-hot encoding, which represents categorical values as arrays of zeroes with a single value set to one. This means that grass (id = 1) would be [0, 1, 0, 0, ...] and stairs (id = 2) would be [0, 0, 1, 0, ...]. To perform one-hot encoding, look into keras' to_categorical function.
Further reading:
one-hot encoding tutorial
one-hot preprocessing using to_categorical
one-hot on the fly using an embedding layer
As any categorical model, this should be a "one-hot" data.
The "channels" dimension of your data should have a size of n-materials.
Values = 0 mean there is no presence of that material
Values = 1 mean there is presence of that material
So, your input shape will be something like (samples, spatial1, spatial2, spatial3, materials). If your data is currently shaped as (samples, s1, s2, s3) and has the materias as integers as you described, you can use to_categorical to transform the integers to "one-hot".
Although I am not sure if this is what you are asking for, I would imagine that t after the bottleneck of the convolutional network, one would typically use a flatten layer and then the output goes to a dense layer. The output layer, if using sigmoid activation will give you probabilities for each of the classes which have to be one-hot encoded, as others have suggested.
If you want the output of the network itself to be in discreet values, I suppose you can use some sort of step-wise activation function in the output layer. However you have to take care that your loss remains differentiable throughout the network (which is why such activation functions are not available in keras). This might be of interest: https://github.com/keras-team/keras/issues/7370
Related
Tensoflow Embedding Layer (https://www.tensorflow.org/api_docs/python/tf/keras/layers/Embedding) is easy to use,
and there are massive articles talking about
"how to use" Embedding (https://machinelearningmastery.com/what-are-word-embeddings/, https://www.sciencedirect.com/topics/computer-science/embedding-method)
.
However, I want to know the Implemention of the very "Embedding Layer" in Tensorflow or Pytorch.
Is it a word2vec?
Is it a Cbow?
Is it a special Dense Layer?
Structure wise, both Dense layer and Embedding layer are hidden layers with neurons in it. The difference is in the way they operate on the given inputs and weight matrix.
A Dense layer performs operations on the weight matrix given to it by multiplying inputs to it ,adding biases to it and applying activation function to it. Whereas Embedding layer uses the weight matrix as a look-up dictionary.
The Embedding layer is best understood as a dictionary that maps integer indices (which stand for specific words) to dense vectors. It takes integers as input, it looks up these integers in an internal dictionary, and it returns the associated vectors. It’s effectively a dictionary lookup.
from keras.layers import Embedding
embedding_layer = Embedding(1000, 64)
Here 1000 means the number of words in the dictionary and 64 means the dimensions of those words. Intuitively, embedding layer just like any other layer will try to find vector (real numbers) of 64 dimensions [ n1, n2, ..., n64] for any word. This vector will represent the semantic meaning of that particular word. It will learn this vector while training using backpropagation just like any other layer.
When you instantiate an Embedding layer, its weights (its internal dictionary of token vectors) are initially random, just as with any other layer. During training, these word vectors are gradually adjusted via backpropagation, structuring the space into something the downstream model can exploit. Once fully trained, the embedding space will show a lot of structure—a kind of structure specialized for the specific problem for which you’re training your model.
-- Deep Learning with Python by F. Chollet
Edit - How "Backpropagation" is used to train the look-up matrix of the Embedding Layer ?
Embedding layer is similar to the linear layer without any activation function. Theoretically, Embedding layer also performs matrix multiplication but doesn't add any non-linearity to it by using any kind of activation function. So backpropagation in the Embedding layer is similar to as of any linear layer. But practically, we don't do any matrix multiplication in the embedding layer because the inputs are generally one hot encoded and the matrix multiplication of weights by a one-hot encoded vector is as easy as a look-up.
In the keras documentation it states that the embedding layer "can only be used as the first layer in a model." This makes no sense to me, I might want to do a reshape/flatten on an input before passing it to the embedding layer, but this is not allowed. Why must the embedding layer be used only as the first layer?
"can only be used as the first layer in a model." This makes no sense
to me
Generally, an embedding layer maps discrete values to continues values. In the subsequence layers, we have continues vector representation that means there is no need to convert the vectors again.
I might want to do a reshape/flatten on input before passing it to
the embedding layer
Of course, you can reshape or flatten an input but in most cases is meaningless. For example, assume we have sentences with a length of 30 and want to flatten them before passed them to embedding:
input_layer = Input(shape=(30))
flatten = Flatten()(input_layer)
embedd = Embedding(1000, 100)(flatten)
In the above example, flatten layer has no effect at all. Before and after flatten our vector size is [batch, 30].
Let look at another example, assume our inputs vector our 2D with the shape of [batch, 30, 2]. After flatting the input, the vectors have the size of [batch, 60]. We can feed them into Embedding layer but in most of the scenarios, it has no meaning. In fact, we destroy the logical relationship between features.
input_layer = Input(shape=(30, 2))
flatten = Flatten()(input_layer)
embedd = Embedding(1000, 100)(flatten)
I have a model that outputs a Softmax, and I would like to develop a custom loss function. The desired behaviour would be:
1) Softmax to one-hot (normally I do numpy.argmax(softmax_vector) and set that index to 1 in a null vector, but this is not allowed in a loss function).
2) Multiply the resulting one-hot vector by my embedding matrix to get an embedding vector (in my context: the word-vector that is associated to a given word, where words have been tokenized and assigned to indices, or classes for the Softmax output).
3) Compare this vector with the target (this could be a normal Keras loss function).
I know how to write a custom loss function in general, but not to do this. I found this closely related question (unanswered), but my case is a bit different, since I would like to preserve my softmax output.
It is possible to mix tensorflow and keras in you customer loss function. Once you can access to all Tensorflow function, things become very easy. I just give you a example of how this function could be imlement.
import tensorflow as tf
def custom_loss(target, softmax):
max_indices = tf.argmax(softmax, -1)
# Get the embedding matrix. In Tensorflow, this can be directly done
# with tf.nn.embedding_lookup
embedding_vectors = tf.nn.embedding_lookup(you_embedding_matrix, max_indices)
# Do anything you want with normal keras loss function
loss = some_keras_loss_function(target, embedding_vectors)
loss = tf.reduce_mean(loss)
return loss
Fan Luo's answer points in the right direction, but ultimately will not work because it involves non-derivable operations. Note such operations are acceptable for the real value (a loss function takes a real value and a predicted value, non-derivable operations are only fine for the real value).
To be fair, that was what I was asking in the first place. It is not possible to do what I wanted, but we can get a similar and derivable behaviour:
1) Element-wise power of the softmax values. This makes smaller values much smaller. For example, with a power of 4 [0.5, 0.2, 0.7] becomes [0.0625, 0.0016, 0.2400]. Note that 0.2 is comparable to 0.7, but 0.0016 is negligible with respect to 0.24. The higher my_power is, the more similar to a one-hot the final result will be.
soft_extreme = Lambda(lambda x: x ** my_power)(softmax)
2) Importantly, both softmax and one-hot vectors are normalized, but not our "soft_extreme". First, find the sum of the array:
norm = tf.reduce_sum(soft_extreme, 1)
3) Normalize soft_extreme:
almost_one_hot = Lambda(lambda x: x / norm)(soft_extreme)
Note: Setting my_power too high in 1) will result in NaNs. If you need a better softmax to one-hot conversion, then you may do steps 1 to 3 two or more times in a row.
4) Finally we want the vector from the dictionary. Lookup is forbidden, but we can take the average vector using matrix multiplication. Because our soft_normalized is similar to one-hot encoding this average will be similar to the vector associated to the highest argument (original intended behaviour). The higher my_power is in (1), the truer this will be:
target_vectors = tf.tensordot(almost_one_hot, embedding_matrix, axes=[[1], [0]])
Note: This will not work directly using batches! In my case, I reshaped my "one hot" (from [batch, dictionary_length] to [batch, 1, dictionary_length] using tf.reshape. Then tiled my embedding_matrix batch times and finally used:
predicted_vectors = tf.matmul(reshaped_one_hot, tiled_embedding)
There may be more elegant solutions (or less memory-hungry, if tiling the embedding matrix is not an option), so feel free to explore more.
I'm working on a problem using Keras that has been presenting me with issues:
My X data is all of shape (num_samples, 8192, 8), but my Y data is of shape (num_samples, 4), where 4 is a one-hot encoded vector.
Both X and Y data will be run through LSTM layers, but the layers are rejecting the Y data because it doesn't match the shape of the X data.
Is padding the Y data with 0s so that it matches the dimensions of the X data unreasonable? What kind of effects would that have? Is there a better solution?
Edited for clarification:
As requested, here is more information:
My Y data represents the expected output of passing the X data through my model. This is my first time working with LSTMs, so I don't have an architecture in mind, but I'd like to use an architecture that works well with classifying long (8192-length) sequences of words into one of several categories. Additionally, the dataset that I have is of an immense size when fed through an LSTM, so I'm currently using batch-training.
Technologies being used:
Keras (Tensorflow Backend)
TL;DR Is padding one tensor with zeroes in all dimensions to match another tensor's shape a bad idea? What could be a better approach?
First of all, let's make sure your representation is actually what you think it is; the input to an LSTM (or any recurrent layer, for that matter) must be of dimensionality: (timesteps, shape), i.e. if you have 1000 training samples, each consisting of 100 timesteps, with each timestep having 10 values, your input shape will be (100,10,). Therefore I assume from your question that each input sample in your X set has 8192 steps and 8 values per step. Great; a single LSTM layer can iterate over these and produce 4-dimensional representations with absolutely no problem, just like so:
myLongInput = Input(shape=(8192,8,))
myRecurrentFunction = LSTM(4)
myShortOutput = myRecurrentFunction(myLongInput)
myShortOutput.shape
TensorShape([Dimension(None), Dimension(4)])
I assume your problem stems from trying to apply yet another LSTM on top of the first one; the next LSTM expects a tensor that has a time dimension, but your output has none. If that is the case, you'll need to let your first LSTM also output the intermediate representations at each time step, like so:
myNewRecurrentFunction=LSTM(4, return_sequences=True)
myLongOutput = myNewRecurrentFunction(myLongInput)
myLongOutput.shape
TensorShape([Dimension(None), Dimension(None), Dimension(4)])
As you can see the new output is now a 3rd order tensor, with the second dimension now being the (yet unassigned) timesteps. You can repeat this process until your final output, where you usually don't need the intermediate representations but rather only the last one. (Sidenote: make sure to set the activation of your last layer to a softmax if your output is in one-hot format)
On to your original question, zero-padding has very little negative impact on your network. The network will strain itself a bit in the beginning trying to figure out the concept of the additional values you have just thrown at it, but will very soon be able to learn they're meaningless. This comes at a cost of a larger parameter space (therefore more time and memory complexity), but doesn't really affect predictive power most of the time.
I hope that was helpful.
I'm following udacity MNIST tutorial and MNIST data is originally 28*28 matrix. However right before feeding that data, they flatten the data into 1d array with 784 columns (784 = 28 * 28).
For example,
original training set shape was (200000, 28, 28).
200000 rows (data). Each data is 28*28 matrix
They converted this into the training set whose shape is (200000, 784)
Can someone explain why they flatten the data out before feeding to tensorflow?
Because when you're adding a fully connected layer, you always want your data to be a (1 or) 2 dimensional matrix, where each row is the vector representing your data. That way, the fully connected layer is just a matrix multiplication between your input (of size (batch_size, n_features)) and the weights (of shape (n_features, n_outputs)) (plus the bias and the activation function), and you get an output of shape (batch_size, n_outputs). Plus, you really don't need the original shape information in a fully connected layer, so it's OK to lose it.
It would be more complicated and less efficient to get the same result without reshaping first, that's why we always do it before a fully connected layer. For a convolutional layer, on the opposite, you'll want to keep the data in original format (width, height).
That is a convention with fully connected layers. Fully connected layers connect every node in the previous layer with every node in the successive layer so locality is not an issue for this type of layer.
Additionally by defining the layer like this we can efficiently calculate the next step by calculating the formula: f(Wx + b) = y. This would not be as easily possible with multidimensional input and reshaping the input is low cost and easy to accomplish.