I want to do max pooling in my network, like this:
My network is 1D (you can think the above example is one sentence with 6 words while every word has 3 embedding )I don't know the length of feature(not every sentence has the same length), so I can't set the pool_size in tf.layers.MaxPooling1D(https://www.tensorflow.org/api_docs/python/tf/layers/MaxPooling1D)
I just want to pooling every half features(or half sentence), Is there any function or method to do that?
(Note: My previous answer had an error that would have result in incorrect pooling windows. This one should be fine.)
Here is one possible way written in "low level" tensorflow. You might need to wrap this in a keras layer (or just use Lambda) to integrate it into your model.
x = ... # input, shape batch x n_words x features
x = tf.reshape(x, [batch, 2, n_words//2, features]) # need to get these dimensions, can get them from tf.shape(x) as well
x = tf.reduce_max(x, axis=2)
This would implement max pooling; you could also use reduce_mean for average pooling, for example.
This has one limitation, namely it's not going to work if n_words is odd. In that case, you might have to check whether it is and use tf.pad to add one element in the word axis to make it even.
Related
I am using 1x1 convolution in the deep network to reduce a feature x: Bx2CxHxW to BxCxHxW. I have three options:
x -> Conv (1x1) -> Batchnorm-->ReLU. Code will be output = ReLU(BN(Conv(x))). Reference resnet
x -> BN -> ReLU-> Conv. So the code will be output = Conv(ReLU(BN(x))) . Reference densenet
x-> Conv. The code is output = Conv(x)
Which one is most using for feature reduction? Why?
Since you are going to train your net end-to-end, whatever configuration you are using - the weights will be trained to accommodate them.
BatchNorm?
I guess the first question you need to ask yourself is do you want to use BatchNorm? If your net is deep and you are concerned with covariate shifts then you probably should have a BatchNorm -- and here goes option no. 3
BatchNorm first?
If your x is the output of another conv layer, than there's actually no difference between your first and second alternatives: your net is a cascade of ...-conv-bn-ReLU-conv-BN-ReLU-conv-... so it's only an "artificial" partitioning of the net into triplets of functions conv, bn, relu and up to the very first and last functions you can split things however you wish. Moreover, since Batch norm is a linear operation (scale + bias) it can be "folded" into an adjacent conv layer without changing the net, so you basically left with conv-relu pairs.
So, there's not really a big difference between the first two options you highlighted.
What else to consider?
Do you really need ReLU when changing dimension of features? You can think of the reducing dimensions as a linear mapping - decomposing the weights mapping to x into a lower rank matrix that ultimately maps into c dimensional space instead of 2c space. If you consider a linear mapping, then you might omit the ReLU altogether.
See fast RCNN SVD trick for an example.
Saying I have a 2000x100 matrix, I put it into 10 dimension embedding layer, which gives me 2000x100x10 tensor. so it's 2000 examples and each example has a 100x10 matrix. and then, I pass it to a conv1d and KMaxpolling to get 2000x24 matrix, which is 2000 examples and each example has a 24 dimension vector. and now, I would like to recombine those examples before I apply another layer. I would like to combine the first 10 examples together, and such and such, so I get a tuple. and then I pass that tuple to the next layer.
My question is, Can I do that with Keras? and any idea on how to do it?
The idea of using "samples" is that these samples should be unique and not relate to each other.
This is something Keras will demand from your model: if it started with 2000 samples, it must end with 2000 samples. Ideally, these samples do not talk to each other, but you can use custom layers to hack this, but only in the middle. You will need to end with 2000 samples anyway.
I believe you're going to end your model with 200 groups, so maybe you should already start with shape (200,10,100) and use TimeDistributed wrappers:
inputs = Input((10,100)) #shape (200,10,100)
out = TimeDistributed(Embedding(....))(inputs) #shape (200,10,100,10)
out = TimeDistributed(Conv1D(...))(out) #shape (200,10,len,filters)
#here, you use your layer that will work on the groups without TimeDistributed.
To reshape a tensor without changing the batch size, use the Reshape(newShape) layer, where newShape does not include the first dimension (batch size).
To reshape a tensor including the batch size, use a Lambda(lambda x: K.reshape(x,newShape)) layer, where newShape includes the first dimension (batch size) - Here you must remember the warning above: somewhere you will need to undo this change so you end up with the same batch size as the input.
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 want to create a network that has specific fixed connections between layers.
For example,
Sparsely connected neural network
I tried looking into functions in Tensorflow, but I only found dense networks with regularizers, which doesn't function as I want.
If it's not possible in tensorflow, then please suggest some other library that can be used. Thanks!
You can always find a workaround. Let's say a layer does y = xW (Wx is also correct) but you want some of the entries in W always be zeros. You can do it column-wise:
For column i (or element i since y is a vector) of the output, y_i = x * D_i * W_i. The matrix D_i is a constant diagonal matrix (tf.constant, tf.diag) that controls what element would be zeros.
Then you can use tf.concat to combine all y_i to matrix Y.
You can abstract this into a function whose signature may look like def sparse_layer(input_layer, gates_matrix, activation_f, ...) which returns the output layer.