I asked similar question on CrossValidation for the image interpretation. I'm moving my detailed question here to include some code details.
The results I'm having are not fully desirable So maybe you have faced this issue before and you can help me find it out.
It is fully convolution neural network "no fully connected part".
Training part
first the images are transposed to match the convolution function. (batch_no,img_channels,width,height)
input.transpose(0, 3, 1, 2)
Learning optimized using learning rate:3e-6, Hu_uniform initialization and nestrove for 500 epochs until this convergence.
Training cost: 1.602449
Training loss: 4.610442
validation error: 5.126761
Test loss: 5.885714
Backward part
Loading Image
jpgfile = np.array(Image.open(join(testing_folder,img_name)))
Reshape to one batch
batch = jpgfile.reshape(1, jpgfile.shape[0], jpgfile.shape[1], 3)
Run the model to extract first feature map after activation using Relu
output = classifier.layer0.output
Test_model = theano.function(
inputs=[x],
outputs=output,
)
layer_Fmaps = Test_model(test_set_x)
Apply backwork model to reconstruct the image using the only activated
neurons
bch, ch, row, col = layer_Fmaps.shape
output_grad_reshaped = layer_Fmaps.reshape((-1, 1, row, col))
output_grad_reshaped = output_grad_reshaped[0].reshape(1,1,row,col)
input_shape = (1, 3, 226, 226)
W = classifier.layer0.W.get_value()[0].reshape(1,3,7,7)
kernel = theano.shared(W)
inp = T.tensor4('inp')
deconv_out = T.nnet.abstract_conv.conv2d_grad_wrt_inputs(
output_grad = inp,
filters=kernel,
input_shape= input_shape,
filter_shape=(1,3,7,7),
border_mode=(0,0),
subsample=(1,1)
)
f = theano.function(
inputs = [inp],
outputs= deconv_out)
f_out = f(output_grad_reshaped)
deconved_relu = T.nnet.relu(f_out)[0].transpose(1,2,0)
deconved = f_out[0].transpose(1,2,0)
Here we have two images results, the first is the transposed image without activation and the second with relu since kernels might have some negative weights.
It is clear from the transposed convolution image that this kernel is learn to detect some useful feature related to this image. But the reconstructing part is breaking the image color scheme during the transpose convolution. It might be because the pixels values are small float numbers. Do you see where is the problem here ?
Related
I've been implementing VAE and IWAE models on the caltech silhouettes dataset and am having an issue where the VAE outperforms IWAE by a modest margin (test LL ~120 for VAE, ~133 for IWAE!). I don't believe this should be the case, according to both theory and experiments produced here.
I'm hoping someone can find some issue in how I'm implementing that's causing this to be the case.
The network I'm using to approximate q and p is the same as that detailed in the appendix of the paper above. The calculation part of the model is below:
data_k_vec = data.repeat_interleave(K,0) # Generate K samples (in my case K=50 is producing this behavior)
mu, log_std = model.encode(data_k_vec)
z = model.reparameterize(mu, log_std) # z = mu + torch.exp(log_std)*epsilon (epsilon ~ N(0,1))
decoded = model.decode(z) # this is the sigmoid output of the model
log_prior_z = torch.sum(-0.5 * z ** 2, 1)-.5*z.shape[1]*T.log(torch.tensor(2*np.pi))
log_q_z = compute_log_probability_gaussian(z, mu, log_std) # Definitions below
log_p_x = compute_log_probability_bernoulli(decoded,data_k_vec)
if model_type == 'iwae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, K)
elif model_type =='vae':
log_w_matrix = (log_prior_z + log_p_x - log_q_z).view(-1, 1)*1/K
log_w_minus_max = log_w_matrix - torch.max(log_w_matrix, 1, keepdim=True)[0]
ws_matrix = torch.exp(log_w_minus_max)
ws_norm = ws_matrix / torch.sum(ws_matrix, 1, keepdim=True)
ws_sum_per_datapoint = torch.sum(log_w_matrix * ws_norm, 1)
loss = -torch.sum(ws_sum_per_datapoint) # value of loss that gets returned to training function. loss.backward() will get called on this value
Here are the likelihood functions. I had to fuss with the bernoulli LL in order to not get nan during training
def compute_log_probability_gaussian(obs, mu, logstd, axis=1):
return torch.sum(-0.5 * ((obs-mu) / torch.exp(logstd)) ** 2 - logstd, axis)-.5*obs.shape[1]*T.log(torch.tensor(2*np.pi))
def compute_log_probability_bernoulli(theta, obs, axis=1): # Add 1e-18 to avoid nan appearances in training
return torch.sum(obs*torch.log(theta+1e-18) + (1-obs)*torch.log(1-theta+1e-18), axis)
In this code there's a "shortcut" being used in that the row-wise importance weights are being calculated in the model_type=='iwae' case for the K=50 samples in each row, while in the model_type=='vae' case the importance weights are being calculated for the single value left in each row, so that it just ends up calculating a weight of 1. Maybe this is the issue?
Any and all help is huge - I thought that addressing the nan issue would permanently get me out of the weeds but now I have this new problem.
EDIT:
Should add that the training scheme is the same as that in the paper linked above. That is, for each of i=0....7 rounds train for 2**i epochs with a learning rate of 1e-4 * 10**(-i/7)
The K-sample importance weighted ELBO is
$$ \textrm{IW-ELBO}(x,K) = \log \sum_{k=1}^K \frac{p(x \vert z_k) p(z_k)}{q(z_k;x)}$$
For the IWAE there are K samples originating from each datapoint x, so you want to have the same latent statistics mu_z, Sigma_z obtained through the amortized inference network, but sample multiple z K times for each x.
So its computationally wasteful to compute the forward pass for data_k_vec = data.repeat_interleave(K,0), you should compute the forward pass once for each original datapoint, then repeat the statistics output by the inference network for sampling:
mu = torch.repeat_interleave(mu,K,0)
log_std = torch.repeat_interleave(log_std,K,0)
Then sample z_k. And now repeat your datapoints data_k_vec = data.repeat_interleave(K,0), and use the resulting tensor to efficiently evaluate the conditional p(x |z_k) for each importance sample z_k.
Note you may also want to use the logsumexp operation when calculating the IW-ELBO for numerical stability. I can't quite figure out what's going on with the log_w_matrix calculation in your post, but this is what I would do:
log_pz = ...
log_qzCx = ....
log_pxCz = ...
log_iw = log_pxCz + log_pz - log_qzCx
log_iw = log_iw.reshape(-1, K)
iwelbo = torch.logsumexp(log_iw, dim=1) - np.log(K)
EDIT: Actually after thinking about it a bit and using the score function identity, you can interpret the IWAE gradient as an importance weighted estimate of the standard single-sample gradient, so the method in the OP for calculation of the importance weights is equivalent (if a bit wasteful), provided you place a stop_gradient operator around the normalized importance weights, which you call w_norm. So I the main problem is the absence of this stop_gradient operator.
If I have a string, say "abc" and target of that string in reverse, say "cba".
Can a neural network, in particular an encoder-decoder model, learn this mapping? If so, what is the best model to accomplish this.
I ask, as this is a structural translation rather than a simple character mapping as in normal machine translation
If your network is an old-fashioned encoder-decoder model (without attention), then, as #Prune said, it has memory bottleneck (encoder dimensionality). Thus, such a network cannot learn to reverse strings of arbitrary size. However, you can train such an RNN to reverse strings of limited size. For example, the following toy seq2seq LSTM is able to reverse sequences of digits with length up to 10. Here is how you train it:
from keras.models import Model
from keras.layers import Input, LSTM, Dense, Embedding
import numpy as np
emb_dim = 20
latent_dim = 100 # Latent dimensionality of the encoding space.
vocab_size = 12 # digits 0-9, 10 is for start token, 11 for end token
encoder_inputs = Input(shape=(None, ), name='enc_inp')
common_emb = Embedding(input_dim=vocab_size, output_dim=emb_dim)
encoder_emb = common_emb(encoder_inputs)
encoder = LSTM(latent_dim, return_state=True)
encoder_outputs, state_h, state_c = encoder(encoder_emb)
encoder_states = [state_h, state_c]
decoder_inputs = Input(shape=(None,), name='dec_inp')
decoder_emb = common_emb(decoder_inputs)
decoder_lstm = LSTM(latent_dim, return_sequences=True, return_state=True)
decoder_outputs, _, _ = decoder_lstm(decoder_emb, initial_state=encoder_states)
decoder_dense = Dense(vocab_size, activation='softmax')
decoder_outputs = decoder_dense(decoder_outputs)
model = Model([encoder_inputs, decoder_inputs], decoder_outputs)
def generate_batch(length=4, batch_size=64):
x = np.random.randint(low=0, high=10, size=(batch_size, length))
y = x[:, ::-1]
start = np.ones((batch_size, 1), dtype=int) * 10
end = np.ones((batch_size, 1), dtype=int) * 11
enc_x = np.concatenate([start, x], axis=1)
dec_x = np.concatenate([start, y], axis=1)
dec_y = np.concatenate([y, end], axis=1)
dec_y_onehot = np.zeros(shape=(batch_size, length+1, vocab_size), dtype=int)
for row in range(batch_size):
for col in range(length+1):
dec_y_onehot[row, col, dec_y[row, col]] = 1
return [enc_x, dec_x], dec_y_onehot
def generate_batches(batch_size=64, max_length=10):
while True:
length = np.random.randint(low=1, high=max_length)
yield generate_batch(length=length, batch_size=batch_size)
model.compile(optimizer='rmsprop', loss='categorical_crossentropy', metrics=['categorical_accuracy'])
model.fit_generator(generate_batches(), steps_per_epoch=1000, epochs=20)
Now you can apply it to reverse a sequence (my decoder is very inefficient, but it does illustrate the principle)
input_seq = np.array([[10, 2, 1, 2, 8, 5, 0, 6]])
result = np.array([[10]])
next_digit = -1
for i in range(100):
next_digit = model.predict([input_seq, result])[0][-1].argmax()
if next_digit == 11:
break
result = np.concatenate([result, [[next_digit]]], axis=1)
print(result[0][1:])
Hoorray, it prints [6 0 5 8 2 1 2] !
Generally, you can think of such a model as a weird autoencoder (with a reversal side-effect), and choose architecture and training procedure suitable for autoencoders. And there is quite a vast literature about text autoencoders.
Moreover, if you make an encoder-decoder model with attention, then, it will have no memory bottleneck, so, in principle, it is possible to reverse a sequence of any length with a neural network. However, attention requires quadratic computational time, so in practice even neural networks with attention will be very inefficient for long sequences.
I doubt that a NN will learn the abstract structural transformation. Since the string is of unbounded input length, the finite NN won't have the info necessary. NLP processes generally work with identifying small blocks and simple context-sensitive shifts. I don't think they'd identify the end-to-end swaps needed.
However, I expect that an image processor, adapted to a single dimension, would learn this quite quickly. Some can learn how to rotate a sub-image.
I wrote a vanilla autoencoder using only Dense layer.
Below is my code:
iLayer = Input ((784,))
layer1 = Dense(128, activation='relu' ) (iLayer)
layer2 = Dense(64, activation='relu') (layer1)
layer3 = Dense(28, activation ='relu') (layer2)
layer4 = Dense(64, activation='relu') (layer3)
layer5 = Dense(128, activation='relu' ) (layer4)
layer6 = Dense(784, activation='softmax' ) (layer5)
model = Model (iLayer, layer6)
model.compile(loss='binary_crossentropy', optimizer='adam')
(trainX, trainY), (testX, testY) = mnist.load_data()
print ("shape of the trainX", trainX.shape)
trainX = trainX.reshape(trainX.shape[0], trainX.shape[1]* trainX.shape[2])
print ("shape of the trainX", trainX.shape)
model.fit (trainX, trainX, epochs=5, batch_size=100)
Questions:
1) softmax provides probability distribution. Understood. This means, I would have a vector of 784 values with probability between 0 and 1. For example [ 0.02, 0.03..... upto 784 items], summing all 784 elements provides 1.
2) I don't understand how the binary crossentropy works with these values. Binary cross entropy is for two values of output, right?
In the context of autoencoders the input and output of the model is the same. So, if the input values are in the range [0,1] then it is acceptable to use sigmoid as the activation function of last layer. Otherwise, you need to use an appropriate activation function for the last layer (e.g. linear which is the default one).
As for the loss function, it comes back to the values of input data again. If the input data are only between zeros and ones (and not the values between them), then binary_crossentropy is acceptable as the loss function. Otherwise, you need to use other loss functions such as 'mse' (i.e. mean squared error) or 'mae' (i.e. mean absolute error). Note that in the case of input values in range [0,1] you can use binary_crossentropy, as it is usually used (e.g. Keras autoencoder tutorial and this paper). However, don't expect that the loss value becomes zero since binary_crossentropy does not return zero when both prediction and label are not either zero or one (no matter they are equal or not). Here is a video from Hugo Larochelle where he explains the loss functions used in autoencoders (the part about using binary_crossentropy with inputs in range [0,1] starts at 5:30)
Concretely, in your example, you are using the MNIST dataset. So by default the values of MNIST are integers in the range [0, 255]. Usually you need to normalize them first:
trainX = trainX.astype('float32')
trainX /= 255.
Now the values would be in range [0,1]. So sigmoid can be used as the activation function and either of binary_crossentropy or mse as the loss function.
Why binary_crossentropy can be used even when the true label values (i.e. ground-truth) are in the range [0,1]?
Note that we are trying to minimize the loss function in training. So if the loss function we have used reaches its minimum value (which may not be necessarily equal to zero) when prediction is equal to true label, then it is an acceptable choice. Let's verify this is the case for binray cross-entropy which is defined as follows:
bce_loss = -y*log(p) - (1-y)*log(1-p)
where y is the true label and p is the predicted value. Let's consider y as fixed and see what value of p minimizes this function: we need to take the derivative with respect to p (I have assumed the log is the natural logarithm function for simplicity of calculations):
bce_loss_derivative = -y*(1/p) - (1-y)*(-1/(1-p)) = 0 =>
-y/p + (1-y)/(1-p) = 0 =>
-y*(1-p) + (1-y)*p = 0 =>
-y + y*p + p - y*p = 0 =>
p - y = 0 => y = p
As you can see binary cross-entropy have the minimum value when y=p, i.e. when the true label is equal to predicted label and this is exactly what we are looking for.
I am trying to create a list based on my neural network outputs and use it in Tensorflow as a loss function.
Assume that results is list of size [1, batch_size] that is output by a neural network. I check to see whether the first value of this list is in a specific range passed in as a placeholder called valid_range, and if it is add 1 to a list. If it is not, add -1. The goal is to make all predictions of the network in the range, so the correct predictions is a tensor of all 1, which I call correct_predictions.
values_list = []
for j in range(batch_size):
a = results[0, j] >= valid_range[0]
b = result[0, j] <= valid_range[1]
c = tf.logical_and(a, b)
if (c == 1):
values_list.append(1)
else:
values_list.append(-1.)
values_list_tensor = tf.convert_to_tensor(values_list)
correct_predictions = tf.ones([batch_size, ], tf.float32)
Now, I want to use this as a loss function in my network, so that I can force all the predictions to be in the specified range. I try to train like this:
loss = tf.reduce_mean(tf.squared_difference(values_list_tensor, correct_predictions))
optimizer = tf.train.AdamOptimizer(learning_rate=learning_rate)
gradients, variables = zip(*optimizer.compute_gradients(loss))
gradients, _ = tf.clip_by_global_norm(gradients, gradient_clip_threshold)
optimize = optimizer.apply_gradients(zip(gradients, variables))
This, however, has a problem and throws an error on the last optimize line, saying:
ValueError: No gradients provided for any variable: ['<tensorflow.python.training.optimizer._RefVariableProcessor object at 0x7f0245d4afd0>',
'<tensorflow.python.training.optimizer._RefVariableProcessor object at 0x7f0245d66050>'
...
I tried to debug this in Tensorboard, and I notice that the list I am creating does not appear in the graph, so basically the x part of the loss function is not part of the network itself. Is there some way to accurately create a list based on the predictions of a neural network and use it in the loss function in Tensorflow to train the network?
Please help, I have been stuck on this for a few days now.
Edit:
Following what was suggested in the comments, I decided to use a l2 loss function, multiplying it by the binary vector I had from before values_list_tensor. The binary vector now has values 1 and 0 instead of 1 and -1. This way when the prediction is in the range the loss is 0, else it is the normal l2 loss. As I am unable to see the values of the tensors, I am not sure if this is correct. However, I can view the final loss and it is always 0, so something is wrong here. I am unsure if the multiplication is being done correctly and if values_list_tensor is calculated accurately? Can someone help and tell me what could be wrong?
loss = tf.reduce_mean(tf.nn.l2_loss(tf.matmul(tf.transpose(tf.expand_dims(values_list_tensor, 1)), tf.expand_dims(result[0, :], 1))))
Thanks
To answer the question in the comment. One way to write a piece-wise function is using tf.cond. For example, here is a function that returns 0 in [-1, 1] and x everywhere else:
sess = tf.InteractiveSession()
x = tf.placeholder(tf.float32)
y = tf.cond(tf.logical_or(tf.greater(x, 1.0), tf.less(x, -1.0)), lambda : x, lambda : 0.0)
y.eval({x: 1.5}) # prints 1.5
y.eval({x: 0.5}) # prints 0.0
Could someone give a clear explanation of backpropagation for LSTM RNNs?
This is the type structure I am working with. My question is not posed at what is back propagation, I understand it is a reverse order method of calculating the error of the hypothesis and output used for adjusting the weights of neural networks. My question is how LSTM backpropagation is different then regular neural networks.
I am unsure of how to find the initial error of each gates. Do you use the first error (calculated by hypothesis minus output) for each gate? Or do you adjust the error for each gate through some calculation? I am unsure how the cell state plays a role in the backprop of LSTMs if it does at all. I have looked thoroughly for a good source for LSTMs but have yet to find any.
That's a good question. You certainly should take a look at suggested posts for details, but a complete example here would be helpful too.
RNN Backpropagaion
I think it makes sense to talk about an ordinary RNN first (because LSTM diagram is particularly confusing) and understand its backpropagation.
When it comes to backpropagation, the key idea is network unrolling, which is way to transform the recursion in RNN into a feed-forward sequence (like on the picture above). Note that abstract RNN is eternal (can be arbitrarily large), but each particular implementation is limited because the memory is limited. As a result, the unrolled network really is a long feed-forward network, with few complications, e.g. the weights in different layers are shared.
Let's take a look at a classic example, char-rnn by Andrej Karpathy. Here each RNN cell produces two outputs h[t] (the state which is fed into the next cell) and y[t] (the output on this step) by the following formulas, where Wxh, Whh and Why are the shared parameters:
In the code, it's simply three matrices and two bias vectors:
# model parameters
Wxh = np.random.randn(hidden_size, vocab_size)*0.01 # input to hidden
Whh = np.random.randn(hidden_size, hidden_size)*0.01 # hidden to hidden
Why = np.random.randn(vocab_size, hidden_size)*0.01 # hidden to output
bh = np.zeros((hidden_size, 1)) # hidden bias
by = np.zeros((vocab_size, 1)) # output bias
The forward pass is pretty straightforward, this example uses softmax and cross-entropy loss. Note each iteration uses the same W* and h* arrays, but the output and hidden state are different:
# forward pass
for t in xrange(len(inputs)):
xs[t] = np.zeros((vocab_size,1)) # encode in 1-of-k representation
xs[t][inputs[t]] = 1
hs[t] = np.tanh(np.dot(Wxh, xs[t]) + np.dot(Whh, hs[t-1]) + bh) # hidden state
ys[t] = np.dot(Why, hs[t]) + by # unnormalized log probabilities for next chars
ps[t] = np.exp(ys[t]) / np.sum(np.exp(ys[t])) # probabilities for next chars
loss += -np.log(ps[t][targets[t],0]) # softmax (cross-entropy loss)
Now, the backward pass is performed exactly as if it was a feed-forward network, but the gradient of W* and h* arrays accumulates the gradients in all cells:
for t in reversed(xrange(len(inputs))):
dy = np.copy(ps[t])
dy[targets[t]] -= 1
dWhy += np.dot(dy, hs[t].T)
dby += dy
dh = np.dot(Why.T, dy) + dhnext # backprop into h
dhraw = (1 - hs[t] * hs[t]) * dh # backprop through tanh nonlinearity
dbh += dhraw
dWxh += np.dot(dhraw, xs[t].T)
dWhh += np.dot(dhraw, hs[t-1].T)
dhnext = np.dot(Whh.T, dhraw)
Both passes above are done in chunks of size len(inputs), which corresponds to the size of the unrolled RNN. You might want to make it bigger to capture longer dependencies in the input, but you pay for it by storing all outputs and gradients per each cell.
What's different in LSTMs
LSTM picture and formulas look intimidating, but once you coded plain vanilla RNN, the implementation of LSTM is pretty much same. For example, here is the backward pass:
# Loop over all cells, like before
d_h_next_t = np.zeros((N, H))
d_c_next_t = np.zeros((N, H))
for t in reversed(xrange(T)):
d_x_t, d_h_prev_t, d_c_prev_t, d_Wx_t, d_Wh_t, d_b_t = lstm_step_backward(d_h_next_t + d_h[:,t,:], d_c_next_t, cache[t])
d_c_next_t = d_c_prev_t
d_h_next_t = d_h_prev_t
d_x[:,t,:] = d_x_t
d_h0 = d_h_prev_t
d_Wx += d_Wx_t
d_Wh += d_Wh_t
d_b += d_b_t
# The step in each cell
# Captures all LSTM complexity in few formulas.
def lstm_step_backward(d_next_h, d_next_c, cache):
"""
Backward pass for a single timestep of an LSTM.
Inputs:
- dnext_h: Gradients of next hidden state, of shape (N, H)
- dnext_c: Gradients of next cell state, of shape (N, H)
- cache: Values from the forward pass
Returns a tuple of:
- dx: Gradient of input data, of shape (N, D)
- dprev_h: Gradient of previous hidden state, of shape (N, H)
- dprev_c: Gradient of previous cell state, of shape (N, H)
- dWx: Gradient of input-to-hidden weights, of shape (D, 4H)
- dWh: Gradient of hidden-to-hidden weights, of shape (H, 4H)
- db: Gradient of biases, of shape (4H,)
"""
x, prev_h, prev_c, Wx, Wh, a, i, f, o, g, next_c, z, next_h = cache
d_z = o * d_next_h
d_o = z * d_next_h
d_next_c += (1 - z * z) * d_z
d_f = d_next_c * prev_c
d_prev_c = d_next_c * f
d_i = d_next_c * g
d_g = d_next_c * i
d_a_g = (1 - g * g) * d_g
d_a_o = o * (1 - o) * d_o
d_a_f = f * (1 - f) * d_f
d_a_i = i * (1 - i) * d_i
d_a = np.concatenate((d_a_i, d_a_f, d_a_o, d_a_g), axis=1)
d_prev_h = d_a.dot(Wh.T)
d_Wh = prev_h.T.dot(d_a)
d_x = d_a.dot(Wx.T)
d_Wx = x.T.dot(d_a)
d_b = np.sum(d_a, axis=0)
return d_x, d_prev_h, d_prev_c, d_Wx, d_Wh, d_b
Summary
Now, back to your questions.
My question is how is LSTM backpropagation different then regular Neural Networks
The are shared weights in different layers, and few more additional variables (states) that you need to pay attention to. Other than this, no difference at all.
Do you use the first error (calculated by hypothesis minus output) for each gate? Or do you adjust the error for each gate through some calculation?
First up, the loss function is not necessarily L2. In the example above it's a cross-entropy loss, so initial error signal gets its gradient:
# remember that ps is the probability distribution from the forward pass
dy = np.copy(ps[t])
dy[targets[t]] -= 1
Note that it's the same error signal as in ordinary feed-forward neural network. If you use L2 loss, the signal indeed equals to ground-truth minus actual output.
In case of LSTM, it's slightly more complicated: d_next_h = d_h_next_t + d_h[:,t,:], where d_h is the upstream gradient the loss function, which means that error signal of each cell gets accumulated. But once again, if you unroll LSTM, you'll see a direct correspondence with the network wiring.
I think your questions could not be answered in a short response. Nico's simple LSTM has a link to a great paper from Lipton et.al., please read this. Also his simple python code sample helps to answer most of your questions.
If you understand Nico's last sentence
ds = self.state.o * top_diff_h + top_diff_s
in detail, please give me a feed back. At the moment I have a final problem with his "Putting all this s and h derivations together".