I have implemented Autoencoder using Keras that takes 112*112*3 neurons as input and 100 neurons as the compressed/encoded state. I want to find the neurons out of these 100 that learns the important features. So far i have calculated eigen values(e) and eigen vectors(v) using the following steps. And i found out that around first 30 values of (e) is greater than 0. Does that mean the first 30 modes are the important ones? Is there any other method that could find the important neurons?
Thanks in Advance
x_enc = enc_model.predict(x_train, batch_size=BATCH_SIZE) # shape (3156,100)
x_mean = np.mean(x_enc, axis=0) # shape (100,)
x_stds = np.std(x_enc, axis=0) # shape (100,)
x_cov = np.cov((x_enc - x_mean).T) # shape (100,100)
e, v = np.linalg.eig(x_cov) # shape (100,) and (100,100) respectively
I don't know if the approach you are using will actually give you any useful results since the way the network learns and what it exactly learns aren't known, I suggest you use a different kind of autoencoder, that automatically learns disentangled representations of the data in a latent space, this way you can be sure that all the parameters you find are actually contributing to the representation of your data. check this article
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Denote a[2, 3] to be a matrix of dimension 2x3. Say there are 10 elements in each input and the network is a two-element classifier (cat or dog, for example). Say there is just one dense layer. For now I am ignoring the bias vector. I know this is an over-simplified neural net, but it is just for this example. Each output in a dense layer of a neural net can be calculated as
output = matmul(input, weights)
Where weights is a weight matrix 10x2, input is an input vector 1x10, and output is an output vector 1x2.
My question is this: Can an entire series of inputs be computed at the same time with a single matrix multiplication? It seems like you could compute
output = matmul(input, weights)
Where there are 100 inputs total, and input is 100x10, weights is 10x2, and output is 100x2.
In back propagation, you could do something similar:
input_err = matmul(output_err, transpose(weights))
weights_err = matmul(transpose(input), output_err)
weights -= learning_rate*weights_err
Where weights is the same, output_err is 100x2, and input is 100x10.
However, I tried to implement a neural network in this way from scratch and I am currently unsuccessful. I am wondering if I have some other error or if my approach is fundamentally wrong.
Thanks!
If anyone else is wondering, I found the answer to my question. This does not in fact work, for a few reasons. Essentially, computing all inputs in this way is like running a network with a batch size equal to the number of inputs. The weights do not get updated between inputs, but rather all at once. And so while it seems that calculating together would be valid, it makes it so that each input does not individually influence the training step by step. However, with a reasonable batch size, you can do 2d matrix multiplications, where the input is batch_size by input_size in order to speed up training.
In addition, if predicting on many inputs (in the test stage, for example), since no weights are updated, an entire matrix multiplication of num_inputs by input_size can be run to compute all inputs in parallel.
I am trying to produce a mathematical operation selection nn model, which is based on the scalar input. The operation is selected based on the softmax result which is produce by the nn. Then this operation has to be applied to the scalar input in order to produce the final output. So far I’ve come up with applying argmax and onehot on the softmax output in order to produce a mask which then is applied on the concated values matrix from all the possible operations to be performed (as show in the pseudo code below). The issue is that neither argmax nor onehot appears to be differentiable. I am new to this, so any would be highly appreciated. Thanks in advance.
#perform softmax
logits = tf.matmul(current_input, W) + b
softmax = tf.nn.softmax(logits)
#perform all possible operations on the input
op_1_val = tf_op_1(current_input)
op_2_val = tf_op_2(current_input)
op_3_val = tf_op_2(current_input)
values = tf.concat([op_1_val, op_2_val, op_3_val], 1)
#create a mask
argmax = tf.argmax(softmax, 1)
mask = tf.one_hot(argmax, num_of_operations)
#produce the input, by masking out those operation results which have not been selected
output = values * mask
I believe that this is not possible. This is similar to Hard Attention described in this paper. Hard attention is used in Image captioning to allow the model to focus only on a certain part of the image at each step. Hard attention is not differentiable but there are 2 ways to go around this:
1- Use Reinforcement Learning (RL): RL is made to train models that makes decisions. Even though, the loss function won't back-propagate any gradients to the softmax used for the decision, you can use RL techniques to optimize the decision. For a simplified example, you can consider the loss as penalty, and send to the node, with the maximum value in the softmax layer, a policy gradient proportional to the penalty in order to decrease the score of the decision if it was bad (results in a high loss).
2- Use something like soft attention: instead of picking only one operation, mix them with weights based on the softmax. so instead of:
output = values * mask
Use:
output = values * softmax
Now, the operations will converge down to zero based on how much the softmax will not select them. This is easier to train compared to RL but it won't work if you must completely remove the non-selected operations from the final result (set them to zero completely).
This is another answer that talks about Hard and Soft attention that you may find helpful: https://stackoverflow.com/a/35852153/6938290
I'm pretty sure this is a silly question but I can't find it anywhere else so I'm going to ask it here.
I'm doing semantic image segmentation using a cnn (unet) in keras with 7 labels. So my label for each image is (7,n_rows,n_cols) using the theano backend. So across the 7 layers for each pixel, it's one-hot encoded. In this case, is the correct error function to use categorical cross-entropy? It seems that way to me but the network seems to learn better with binary cross-entropy loss. Can someone shed some light on why that would be and what the principled objective is?
Binary cross-entropy loss should be used with sigmod activation in the last layer and it severely penalizes opposite predictions. It does not take into account that the output is a one-hot coded and the sum of the predictions should be 1. But as mis-predictions are severely penalizing the model somewhat learns to classify properly.
Now to enforce the prior of one-hot code is to use softmax activation with categorical cross-entropy. This is what you should use.
Now the problem is using the softmax in your case as Keras don't support softmax on each pixel.
The easiest way to go about it is permute the dimensions to (n_rows,n_cols,7) using Permute layer and then reshape it to (n_rows*n_cols,7) using Reshape layer. Then you can added the softmax activation layer and use crossentopy loss. The data should also be reshaped accordingly.
The other way of doing so will be to implement depth-softmax :
def depth_softmax(matrix):
sigmoid = lambda x: 1 / (1 + K.exp(-x))
sigmoided_matrix = sigmoid(matrix)
softmax_matrix = sigmoided_matrix / K.sum(sigmoided_matrix, axis=0)
return softmax_matrix
and use it as a lambda layer:
model.add(Deconvolution2D(7, 1, 1, border_mode='same', output_shape=(7,n_rows,n_cols)))
model.add(Permute(2,3,1))
model.add(BatchNormalization())
model.add(Lambda(depth_softmax))
If tf image_dim_ordering is used then you can do way with the Permute layers.
For more reference check here.
I tested the solution of #indraforyou and think that the proposed method has some mistakes. As the commentsection does not allow for proper code segments, here is what I think would be the fixed version:
def depth_softmax(matrix):
from keras import backend as K
exp_matrix = K.exp(matrix)
softmax_matrix = exp_matrix / K.expand_dims(K.sum(exp_matrix, axis=-1), axis=-1)
return softmax_matrix
This method will expect the ordering of the matrix to be (height, width, channels).
I can't get TensorFlow RELU activations (neither tf.nn.relu nor tf.nn.relu6) working without NaN values for activations and weights killing my training runs.
I believe I'm following all the right general advice. For example I initialize my weights with
weights = tf.Variable(tf.truncated_normal(w_dims, stddev=0.1))
biases = tf.Variable(tf.constant(0.1 if neuron_fn in [tf.nn.relu, tf.nn.relu6] else 0.0, shape=b_dims))
and use a slow training rate, e.g.,
tf.train.MomentumOptimizer(0.02, momentum=0.5).minimize(cross_entropy_loss)
But any network of appreciable depth results in NaN for cost and and at least some weights (at least in the summary histograms for them). In fact, the cost is often NaN right from the start (before training).
I seem to have these issues even when I use L2 (about 0.001) regularization, and dropout (about 50%).
Is there some parameter or setting that I should adjust to avoid these issues? I'm at a loss as to where to even begin looking, so any suggestions would be appreciated!
Following He et. al (as suggested in lejlot's comment), initializing the weights of the l-th layer to a zero-mean Gaussian distribution with standard deviation
where nl is the flattened length of the the input vector or
stddev=np.sqrt(2 / np.prod(input_tensor.get_shape().as_list()[1:]))
results in weights that generally do not diverge.
If you use a softmax classifier at the top of your network, try to make the initial weights of the layer just below the softmax very small (e.g. std=1e-4). This makes the initial distribution of outputs of the network very soft (high temperature), and helps ensure that the first few steps of your optimization are not too large and numerically unstable.
Have you tried gradient clipping and/or a smaller learning rate?
Basically, you will need to process your gradients before applying them, as follows (from tf docs, mostly):
# Replace this with what follows
# opt = tf.train.MomentumOptimizer(0.02, momentum=0.5).minimize(cross_entropy_loss)
# Create an optimizer.
opt = tf.train.MomentumOptimizer(learning_rate=0.001, momentum=0.5)
# Compute the gradients for a list of variables.
grads_and_vars = opt.compute_gradients(cross_entropy_loss, tf.trainable_variables())
# grads_and_vars is a list of tuples (gradient, variable). Do whatever you
# need to the 'gradient' part, for example cap them, etc.
capped_grads_and_vars = [(tf.clip_by_value(gv[0], -5., 5.), gv[1]) for gv in grads_and_vars]
# Ask the optimizer to apply the capped gradients.
opt = opt.apply_gradients(capped_grads_and_vars)
Also, the discussion in this question might help.
I am trying to use RBFNN for point cloud to surface reconstruction but I couldn't understand what would be my feature vectors in RBFNN.
Can any one please help me to understand this one.
A goal to get to this:
From inputs like this:
An RBF network essentially involves fitting data with a linear combination of functions that obey a set of core properties -- chief among these is radial symmetry. The parameters of each of these functions is learned by incremental adjustment based on errors generated through repeated presentation of inputs.
If I understand (it's been a very long time since I used one of these networks), your question pertains to preprocessing of the data in the point cloud. I believe that each of the points in your point cloud should serve as one input. If I understand properly, the features are your three dimensions, and as such each point can already be considered a "feature vector."
You have other choices that remain, namely the number of radial basis neurons in your hidden layer, and the radial basis functions to use (a Gaussian is a popular first choice). The training of the network and the surface reconstruction can be done in a number of ways but I believe this is beyond the scope of the question.
I don't know if it will help, but here's a simple python implementation of an RBF network performing function approximation, with one-dimensional inputs:
import numpy as np
import matplotlib.pyplot as plt
def fit_me(x):
return (x-2) * (2*x+1) / (1+x**2)
def rbf(x, mu, sigma=1.5):
return np.exp( -(x-mu)**2 / (2*sigma**2));
# Core parameters including number of training
# and testing points, minimum and maximum x values
# for training and testing points, and the number
# of rbf (hidden) nodes to use
num_points = 100 # number of inputs (each 1D)
num_rbfs = 20.0 # number of centers
x_min = -5
x_max = 10
# Training data, evenly spaced points
x_train = np.linspace(x_min, x_max, num_points)
y_train = fit_me(x_train)
# Testing data, more evenly spaced points
x_test = np.linspace(x_min, x_max, num_points*3)
y_test = fit_me(x_test)
# Centers of each of the rbf nodes
centers = np.linspace(-5, 10, num_rbfs)
# Everything is in place to train the network
# and attempt to approximate the function 'fit_me'.
# Start by creating a matrix G in which each row
# corresponds to an x value within the domain and each
# column i contains the values of rbf_i(x).
center_cols, x_rows = np.meshgrid(centers, x_train)
G = rbf(center_cols, x_rows)
plt.plot(G)
plt.title('Radial Basis Functions')
plt.show()
# Simple training in this case: use pseudoinverse to get weights
weights = np.dot(np.linalg.pinv(G), y_train)
# To test, create meshgrid for test points
center_cols, x_rows = np.meshgrid(centers, x_test)
G_test = rbf(center_cols, x_rows)
# apply weights to G_test
y_predict = np.dot(G_test, weights)
plt.plot(y_predict)
plt.title('Predicted function')
plt.show()
error = y_predict - y_test
plt.plot(error)
plt.title('Function approximation error')
plt.show()
First, you can explore the way in which inputs are provided to the network and how the RBF nodes are used. This should extend to 2D inputs in a straightforward way, though training may get a bit more involved.
To do proper surface reconstruction you'll likely need a representation of the surface that is altogether different than the representation of the function that's learned here. Not sure how to take this last step.