I am currently stuck trying to utilize the Optim package in Julia in an attempt to minimize a cost function. The cost function is the cost function for an L2 regularised logistic regression. It is constructed as follows;
using Optim
function regularised_cost(X, y, θ, λ)
m = length(y)
# Sigmoid predictions
h = sigmoid(X * θ)
# left side of the cost function
positive_class_cost = ((-y)' * log.(h))
# right side of the cost function
negative_class_cost = ((1 .- y)' * log.(1 .- h))
# lambda effect
lambda_regularization = (λ/(2*m) * sum(θ[2 : end] .^ 2))
# Current batch cost
𝐉 = (1/m) * (positive_class_cost - negative_class_cost) + lambda_regularization
# Gradients for all the theta members with regularization except the constant
∇𝐉 = (1/m) * (X') * (h-y) + ((1/m) * (λ * θ))
∇𝐉[1] = (1/m) * (X[:, 1])' * (h-y) # Exclude the constant
return (𝐉, ∇𝐉)
end
I would like to use LBFGS algorithm as a solver to find the best weights that minimize this function based on my training examples and labels which are defined as:
opt_train = [ones(size(X_train_scaled, 1)) X_train_scaled] # added intercept
initial_theta = zeros(size(opt_train, 2))
Having read the documentation, here's my current implementation which is currently not working:
cost, gradient! = regularised_cost(opt_train, y_train, initial_theta, 0.01)
res = optimize(regularised_cost,
gradient!,
initial_theta,
LBFGS(),
Optim.Options(g_tol = 1e-12,
iterations = 1000,
store_trace = true,
show_trace = true))
How do I pass my training examples and labels along with the gradients so that the solver (LBFGS) can find me the best weights for theta?
You need to close over your train data and make a loss function that only takes the parameters as inputs.
As per the docs on dealing with constant parameterised
It should be so.wthing like:
loss_and_grad(theta) = regularised_cost(opt_train, y_train, theta, 0.01)
loss(theta) = first(loss_and_grad(theta))
res = optimize(loss, initial_theta)
I will leave it to you to see how to hook the gradient in.
A reminder though: don't use non-const globals.
They are slow, in particular the way they are used in the loss_and_grad function I wrote will be slow.
So you should declare opt_train and y_train as const.
Or make a function that takes them and returns a loss function etc
Related
My script is failing due to too high memory usage. When I reduce the batch size it works.
#tf.function(autograph=not DEBUG)
def step(prev_state, input_b):
input_b = tf.reshape(input_b, shape=[1,input_b.shape[0]])
state = FastALIFStateTuple(v=prev_state[0], z=prev_state[1], b=prev_state[2], r=prev_state[3])
new_b = self.decay_b * state.b + (tf.ones(shape=[self.units],dtype=tf.float32) - self.decay_b) * state.z
thr = self.thr + new_b * self.beta
z = state.z
i_in = tf.matmul(input_b, W_in)
i_rec = tf.matmul(z, W_rec)
i_t = i_in + i_rec
I_reset = z * thr * self.dt
new_v = self._decay * state.v + (1 - self._decay) * i_t - I_reset
# Spike generation
is_refractory = tf.greater(state.r, .1)
zeros_like_spikes = tf.zeros_like(z)
new_z = tf.where(is_refractory, zeros_like_spikes, self.compute_z(new_v, thr))
new_r = tf.clip_by_value(state.r + self.n_refractory * new_z - 1,
0., float(self.n_refractory))
return [new_v, new_z, new_b, new_r]
#tf.function(autograph=not DEBUG)
def evolve_single(inputs):
accumulated_state = tf.scan(step, inputs, initializer=state0)
Z = tf.squeeze(accumulated_state[1]) # -> [T,units]
if self.model_settings['avg_spikes']:
Z = tf.reshape(tf.reduce_mean(Z, axis=0), shape=(1,-1))
out = tf.matmul(Z, W_out) + b_out
return out # - [BS,Num_labels]
# # - Using a simple loop
# out_store = []
# for i in range(fingerprint_3d.shape[0]):
# out_store.append(tf.squeeze(evolve_single(fingerprint_3d[i,:,:])))
# return tf.reshape(out_store, shape=[fingerprint_3d.shape[0],self.d_out])
final_out = tf.squeeze(tf.map_fn(evolve_single, fingerprint_3d)) # -> [BS,T,self.units]
return final_out
This code snippet is inside a tf.function, but I omitted it since I don't think it's relevant.
As can be seen, I run the code on fingerprint_3d, a tensor that has the dimension [BatchSize,Time,InputDimension], e.g. [50,100,20]. When I run this with BatchSize < 10 everything works fine, although tf.scan already uses a lot of memory for that.
When I now execute the code on a batch of size 50, suddenly I get an OOM, even though I am executing it in an iterative matter (here commented out).
How should I execute this code so that the Batch Size doesn't matter?
Is tensorflow maybe parallelizing my for loop so that it executed over multiple batches at once?
Another unrelated question is the following: What function instead of tf.scan should I use if I only want to accumulate one state variable, compared to the case for tf.scan where it just accumulates all the state variables? Or is that possible with tf.scan?
As mentioned in the discussions here, tf.foldl, tf.foldr, and tf.scan all require keeping track of all values for all iterations, which is necessary for computations like gradients. I am not aware of any ways to mitigate this issue; still, I would also be interested if anyone has a better answer than mine.
When I used
#tf.function
def get_loss_and_gradients():
with tf.GradientTape(persistent=False) as tape:
logits, spikes = rnn.call(fingerprint_input=graz_dict["train_input"], W_in=W_in, W_rec=W_rec, W_out=W_out, b_out=b_out)
loss = loss_normal(tf.cast(graz_dict["train_groundtruth"],dtype=tf.int32), logits)
gradients = tape.gradient(loss, [W_in,W_rec,W_out,b_out])
return loss, logits, spikes, gradients
it works.
When I remove the #tf.function decorator the memory blows up. So it really seems important that tensorflow can create a graph for you computations.
I'm trying to make LSTM in tensorflow 2.1 from scratch, without using the one already supplied with keras (tf.keras.layers.LSTM), just to learn and code something. To do so, I've defined a class "Model" that when called (like with model(input)) it computes the matrix multiplications of the LSTM. I'm pasting here part of my code, the other parts are on github (link)
class Model(object):
[...]
def __call__(self, inputs):
assert inputs.shape == (vocab_size, T_steps)
outputs = []
for time_step in range(T_steps):
x = inputs[:,time_step]
x = tf.expand_dims(x,axis=1)
z = tf.concat([self.h_prev,x],axis=0)
f = tf.matmul(self.W_f, z) + self.b_f
f = tf.sigmoid(f)
i = tf.matmul(self.W_i, z) + self.b_i
i = tf.sigmoid(i)
o = tf.matmul(self.W_o, z) + self.b_o
o = tf.sigmoid(o)
C_bar = tf.matmul(self.W_C, z) + self.b_C
C_bar = tf.tanh(C_bar)
C = (f * self.C_prev) + (i * C_bar)
h = o * tf.tanh(C)
v = tf.matmul(self.W_v, h) + self.b_v
v = tf.sigmoid(v)
y = tf.math.softmax(v, axis=0)
self.h_prev = h
self.C_prev = C
outputs.append(y)
outputs = tf.squeeze(tf.stack(outputs,axis=1))
return outputs
But this neural netoworks has three problems:
1) it is way slow during training. In comparison a model that uses tf.keras.layers.LSTM() is trained more than 10 times faster. Why is this? Maybe because I didn't use a minibatch training, but a stochastic one?
2) the NN seems to not learn anything at all. After just some (very few!) training examples, the loss seems to settle down and it won't decrease anymore, but rather it oscillates around the reached value. After training, I tested the NN making it generate some text, but it just outputs non-sense gibberish. Why isn't learning anything?
3) the loss function outputs very high values. I've coded a categorical cross-entropy loss function but, with 100 characters long sequence, the value of the function is over 370 per training example. Shouldn't it be way lower than this?
I've wrote the loss function like this:
def compute_loss(predictions, desired_outputs):
l = 0
for i in range(T_steps):
l -= tf.math.log(predictions[desired_outputs[i], i])
return l
I know they're open questions, but unfortunately I can't make it works. So any answer, even a short answer that help me to make myself solve the problem, is fine :)
I am currently trying to create my own loss function for Keras (using Tensorflow backend). This is a simple categorical crossentropy but I am applying a factor on the 1st column to penalize more loss from the 1st class.
Yet I am new to Keras and I can't figure out how to translate my function (below) as I have to use symbolic expressions and it seems I can't go element-wise:
def custom_categorical_crossentropy(y_true, y_pred):
y_pred = np.clip(y_pred, _EPSILON, 1.0-_EPSILON)
out = np.zeros(y_true.shape).astype('float32')
for i in range(0,y_true.shape[0]):
for j in range (0,y_true.shape[1]):
#penalize more all elements on class 1 so that loss takes its low proportion in the dataset into account
if(j==0):
out[i][j] = -(prop_database*(y_true[i][j] * np.log(y_pred[i][j]) + (1.0 - y_true[i][j]) * np.log(1.0 - y_pred[i][j])))
else:
out[i][j] = -(y_true[i][j] * np.log(y_pred[i][j]) + (1.0 - y_true[i][j]) * np.log(1.0 - y_pred[i][j]))
out = np.mean(out.astype('float32'), axis=-1)
return tf.convert_to_tensor(out,
dtype=tf.float32,
name='custom_loss')
Can someone help me?
Many thanks!
You can use class_weight in the fit method to penalize classes without creating functions:
weights = {
0:2,
1:1,
2:1,
3:1,
...
}
model.compile(optimizer=chooseOne, loss='categorical_crossentropy')
model.fit(......., class_weight = weights)
This will make the first class be twice as important as the others.
I am trying to train an autoencoder NN (3 layers - 2 visible, 1 hidden) using numpy and scipy for the MNIST digits images dataset. The implementation is based on the notation given here Below is my code:
def autoencoder_cost_and_grad(theta, visible_size, hidden_size, lambda_, data):
"""
The input theta is a 1-dimensional array because scipy.optimize.minimize expects
the parameters being optimized to be a 1d array.
First convert theta from a 1d array to the (W1, W2, b1, b2)
matrix/vector format, so that this follows the notation convention of the
lecture notes and tutorial.
You must compute the:
cost : scalar representing the overall cost J(theta)
grad : array representing the corresponding gradient of each element of theta
"""
training_size = data.shape[1]
# unroll theta to get (W1,W2,b1,b2) #
W1 = theta[0:hidden_size*visible_size]
W1 = W1.reshape(hidden_size,visible_size)
W2 = theta[hidden_size*visible_size:2*hidden_size*visible_size]
W2 = W2.reshape(visible_size,hidden_size)
b1 = theta[2*hidden_size*visible_size:2*hidden_size*visible_size + hidden_size]
b2 = theta[2*hidden_size*visible_size + hidden_size: 2*hidden_size*visible_size + hidden_size + visible_size]
#feedforward pass
a_l1 = data
z_l2 = W1.dot(a_l1) + numpy.tile(b1,(training_size,1)).T
a_l2 = sigmoid(z_l2)
z_l3 = W2.dot(a_l2) + numpy.tile(b2,(training_size,1)).T
a_l3 = sigmoid(z_l3)
#backprop
delta_l3 = numpy.multiply(-(data-a_l3),numpy.multiply(a_l3,1-a_l3))
delta_l2 = numpy.multiply(W2.T.dot(delta_l3),
numpy.multiply(a_l2, 1 - a_l2))
b2_derivative = numpy.sum(delta_l3,axis=1)/training_size
b1_derivative = numpy.sum(delta_l2,axis=1)/training_size
W2_derivative = numpy.dot(delta_l3,a_l2.T)/training_size + lambda_*W2
#print(W2_derivative.shape)
W1_derivative = numpy.dot(delta_l2,a_l1.T)/training_size + lambda_*W1
W1_derivative = W1_derivative.reshape(hidden_size*visible_size)
W2_derivative = W2_derivative.reshape(visible_size*hidden_size)
b1_derivative = b1_derivative.reshape(hidden_size)
b2_derivative = b2_derivative.reshape(visible_size)
grad = numpy.concatenate((W1_derivative,W2_derivative,b1_derivative,b2_derivative))
cost = 0.5*numpy.sum((data-a_l3)**2)/training_size + 0.5*lambda_*(numpy.sum(W1**2) + numpy.sum(W2**2))
return cost,grad
I have also implemented a function to estimate the numerical gradient and verify the correctness of my implementation (below).
def compute_gradient_numerical_estimate(J, theta, epsilon=0.0001):
"""
:param J: a loss (cost) function that computes the real-valued loss given parameters and data
:param theta: array of parameters
:param epsilon: amount to vary each parameter in order to estimate
the gradient by numerical difference
:return: array of numerical gradient estimate
"""
gradient = numpy.zeros(theta.shape)
eps_vector = numpy.zeros(theta.shape)
for i in range(0,theta.size):
eps_vector[i] = epsilon
cost1,grad1 = J(theta+eps_vector)
cost2,grad2 = J(theta-eps_vector)
gradient[i] = (cost1 - cost2)/(2*epsilon)
eps_vector[i] = 0
return gradient
The norm of the difference between the numerical estimate and the one computed by the function is around 6.87165125021e-09 which seems to be acceptable. My main problem seems to be to get the gradient descent algorithm "L-BGFGS-B" working using the scipy.optimize.minimize function as below:
# theta is the 1-D array of(W1,W2,b1,b2)
J = lambda x: utils.autoencoder_cost_and_grad(theta, visible_size, hidden_size, lambda_, patches_train)
options_ = {'maxiter': 4000, 'disp': False}
result = scipy.optimize.minimize(J, theta, method='L-BFGS-B', jac=True, options=options_)
I get the below output from this:
scipy.optimize.minimize() details:
fun: 90.802022224079778
hess_inv: <16474x16474 LbfgsInvHessProduct with dtype=float64>
jac: array([ -6.83667742e-06, -2.74886002e-06, -3.23531941e-06, ...,
1.22425735e-01, 1.23425062e-01, 1.28091250e-01])
message: b'ABNORMAL_TERMINATION_IN_LNSRCH'
nfev: 21
nit: 0
status: 2
success: False
x: array([-0.06836677, -0.0274886 , -0.03235319, ..., 0. ,
0. , 0. ])
Now, this post seems to indicate that the error could mean that the gradient function implementation could be wrong? But my numerical gradient estimate seems to confirm that my implementation is correct. I have tried varying the initial weights by using a uniform distribution as specified here but the problem still persists. Is there anything wrong with my backprop implementation?
Turns out the issue was a syntax error (very silly) with this line:
J = lambda x: utils.autoencoder_cost_and_grad(theta, visible_size, hidden_size, lambda_, patches_train)
I don't even have the lambda parameter x in the function declaration. So the theta array wasn't even being passed whenever J was being invoked.
This fixed it:
J = lambda x: utils.autoencoder_cost_and_grad(x, visible_size, hidden_size, lambda_, patches_train)
I'm using tensorflow batch normalization in my deep neural network successfully. I'm doing it the following way:
if apply_bn:
with tf.variable_scope('bn'):
beta = tf.Variable(tf.constant(0.0, shape=[out_size]), name='beta', trainable=True)
gamma = tf.Variable(tf.constant(1.0, shape=[out_size]), name='gamma', trainable=True)
batch_mean, batch_var = tf.nn.moments(z, [0], name='moments')
ema = tf.train.ExponentialMovingAverage(decay=0.5)
def mean_var_with_update():
ema_apply_op = ema.apply([batch_mean, batch_var])
with tf.control_dependencies([ema_apply_op]):
return tf.identity(batch_mean), tf.identity(batch_var)
mean, var = tf.cond(self.phase_train,
mean_var_with_update,
lambda: (ema.average(batch_mean), ema.average(batch_var)))
self.z_prebn.append(z)
z = tf.nn.batch_normalization(z, mean, var, beta, gamma, 1e-3)
self.z.append(z)
self.bn.append((mean, var, beta, gamma))
And it works fine both for training and testing phases.
However I encounter problems when I try to use the computed neural network parameters in my another project, where I need to compute all the matrix multiplications and stuff by myself. The problem is that I can't reproduce the behavior of the tf.nn.batch_normalization function:
feed_dict = {
self.tf_x: np.array([range(self.x_cnt)]) / 100,
self.keep_prob: 1,
self.phase_train: False
}
for i in range(len(self.z)):
# print 0 layer's 1 value of arrays
print(self.sess.run([
self.z_prebn[i][0][1], # before bn
self.bn[i][0][1], # mean
self.bn[i][1][1], # var
self.bn[i][2][1], # offset
self.bn[i][3][1], # scale
self.z[i][0][1], # after bn
], feed_dict=feed_dict))
# prints
# [-0.077417567, -0.089603029, 0.000436493, -0.016652612, 1.0055743, 0.30664611]
According to the formula on the page https://www.tensorflow.org/versions/r1.2/api_docs/python/tf/nn/batch_normalization:
bn = scale * (x - mean) / (sqrt(var) + 1e-3) + offset
But as we can see,
1.0055743 * (-0.077417567 - -0.089603029)/(0.000436493^0.5 + 1e-3) + -0.016652612
= 0.543057
Which differs from the value 0.30664611, computed by Tensorflow itself.
So what am I doing wrong here and why I can't just calculate batch normalized value myself?
Thanks in advance!
The formula used is slightly different from:
bn = scale * (x - mean) / (sqrt(var) + 1e-3) + offset
It should be:
bn = scale * (x - mean) / (sqrt(var + 1e-3)) + offset
The variance_epsilon variable is supposed to scale with the variance, not with sigma, which is the square-root of variance.
After the correction, the formula yields the correct value:
1.0055743 * (-0.077417567 - -0.089603029)/((0.000436493 + 1e-3)**0.5) + -0.016652612
# 0.30664642276945747