How efficient/intelligent is Theano in computing gradients? - python-2.7

Suppose I have an artificial neural networks with 5 hidden layers. For the moment, forget about the details of the neural network model such as biases, the activation functions used, type of data and so on ... . Of course, the activation functions are differentiable.
With symbolic differentiation, the following computes the gradients of the objective function with respect to the layers' weights:
w1_grad = T.grad(lost, [w1])
w2_grad = T.grad(lost, [w2])
w3_grad = T.grad(lost, [w3])
w4_grad = T.grad(lost, [w4])
w5_grad = T.grad(lost, [w5])
w_output_grad = T.grad(lost, [w_output])
This way, to compute the gradients w.r.t w1 the gradients w.r.t w2, w3, w4 and w5 must first be computed. Similarly to compute the gradients w.r.t w2 the gradients w.r.t w3, w4 and w5 must be computed first.
However, I could the following code also computes the gradients w.r.t to each weight matrix:
w1_grad, w2_grad, w3_grad, w4_grad, w5_grad, w_output_grad = T.grad(lost, [w1, w2, w3, w4, w5, w_output])
I was wondering, is there any difference between these two methods in terms of performance? Is Theano intelligent enough to avoid re-computing the gradients using the second method? By intelligent I mean to compute w3_grad, Theano should [preferably] use the pre-computed gradients of w_output_grad, w5_grad and w4_grad instead of computing them again.

Well it turns out Theano does not take the previously-computed gradients to compute the gradients in lower layers of a computational graph. Here's a dummy example of a neural network with 3 hidden layers and an output layer. However, it's not going to be a big deal at all since computing the gradients is a once-in-a-life-time operation unless you have to compute the gradient on each iteration. Theano returns a symbolic expression for the derivatives as a computational graph and you can simply use it as a function from that point on. From that point on we simply use the function derived by Theano to compute numerical values and update the weights using those.
import theano.tensor as T
import time
import numpy as np
class neuralNet(object):
def __init__(self, examples, num_features, num_classes):
self.w = shared(np.random.random((16384, 5000)).astype(T.config.floatX), borrow = True, name = 'w')
self.w2 = shared(np.random.random((5000, 3000)).astype(T.config.floatX), borrow = True, name = 'w2')
self.w3 = shared(np.random.random((3000, 512)).astype(T.config.floatX), borrow = True, name = 'w3')
self.w4 = shared(np.random.random((512, 40)).astype(T.config.floatX), borrow = True, name = 'w4')
self.b = shared(np.ones(5000, dtype=T.config.floatX), borrow = True, name = 'b')
self.b2 = shared(np.ones(3000, dtype=T.config.floatX), borrow = True, name = 'b2')
self.b3 = shared(np.ones(512, dtype=T.config.floatX), borrow = True, name = 'b3')
self.b4 = shared(np.ones(40, dtype=T.config.floatX), borrow = True, name = 'b4')
self.x = examples
L1 = T.nnet.sigmoid(T.dot(self.x, self.w) + self.b)
L2 = T.nnet.sigmoid(T.dot(L1, self.w2) + self.b2)
L3 = T.nnet.sigmoid(T.dot(L2, self.w3) + self.b3)
L4 = T.dot(L3, self.w4) + self.b4
self.forwardProp = T.nnet.softmax(L4)
self.predict = T.argmax(self.forwardProp, axis = 1)
def loss(self, y):
return -T.mean(T.log(self.forwardProp)[T.arange(y.shape[0]), y])
x = T.matrix('x')
y = T.ivector('y')
nnet = neuralNet(x)
loss = nnet.loss(y)
diffrentiationTime = []
for i in range(100):
t1 = time.time()
gw, gw2, gw3, gw4, gb, gb2, gb3, gb4 = T.grad(loss, [nnet.w, nnet.w2, logReg.w3, nnet.w4, nnet.b, nnet.b2, nnet.b3, nnet.b4])
diffrentiationTime.append(time.time() - t1)
print 'Efficient Method: Took %f seconds with std %f' % (np.mean(diffrentiationTime), np.std(diffrentiationTime))
diffrentiationTime = []
for i in range(100):
t1 = time.time()
gw = T.grad(loss, [nnet.w])
gw2 = T.grad(loss, [nnet.w2])
gw3 = T.grad(loss, [nnet.w3])
gw4 = T.grad(loss, [nnet.w4])
gb = T.grad(loss, [nnet.b])
gb2 = T.grad(loss, [nnet.b2])
gb3 = T.grad(loss, [nnet.b3])
gb4 = T.grad(loss, [nnet.b4])
diffrentiationTime.append(time.time() - t1)
print 'Inefficient Method: Took %f seconds with std %f' % (np.mean(diffrentiationTime), np.std(diffrentiationTime))
This will print out the followings:
Efficient Method: Took 0.061056 seconds with std 0.013217
Inefficient Method: Took 0.305081 seconds with std 0.026024
This shows that Theano uses a dynamic-programming approach to compute gradients for the efficient method.

Related

Plot variables out of differential equations system function

I have a 4-4 differential equations system in a function (subsystem4) and I solved it with odeint funtion. I managed to plot the results of the system. My problem is that I want to plot and some other equations (e.g. x,y,vcxdot...) which are included in the same function (subsystem4) but I get NameError: name 'vcxdot' is not defined. Also, I want to use some of these equations (not only the results of the equation's system) as inputs in a following differential equations system and plot all the equations in the same period of time (t). I have done this using Matlab-Simulink but it was much easier because of Simulink blocks. How can I have access to and plot all the equations of a function (subsystem4) and use them as input in a following system? I am new in python and I use Python 2.7.12. Thank you in advance!
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
def subsystem4(u,t):
added_mass_x = 0.03 # kg
added_mass_y = 0.04
mb = 0.3 # kg
m1 = mb-added_mass_x
m2 = mb-added_mass_y
l1 = 0.07 # m
l2 = 0.05 # m
J = 0.00050797 # kgm^2
Sa = 0.0110 # m^2
Cd = 2.44
Cl = 3.41
Kd = 0.000655 # kgm^2
r = 1000 # kg/m^3
f = 2 # Hz
c1 = 0.5*r*Sa*Cd
c2 = 0.5*r*Sa*Cl
c3 = 0.5*mb*(l1**2)
c4 = Kd/J
c5 = (1/(2*J))*(l1**2)*mb*l2
c6 = (1/(3*J))*(l1**3)*mb
vcx = u[0]
vcy = u[1]
psi = u[2]
wz = u[3]
x = 3 + 0.3*np.cos(t)
y = 0.5 + 0.3*np.sin(t)
xdot = -0.3*np.sin(t)
ydot = 0.3*np.cos(t)
xdotdot = -0.3*np.cos(t)
ydotdot = -0.3*np.sin(t)
vcx = xdot*np.cos(psi)-ydot*np.sin(psi)
vcy = ydot*np.cos(psi)+xdot*np.sin(psi)
psidot = wz
vcxdot = xdotdot*np.cos(psi)-xdot*np.sin(psi)*psidot-ydotdot*np.sin(psi)-ydot*np.cos(psi)*psidot
vcydot = ydotdot*np.cos(psi)-ydot*np.sin(psi)*psidot+xdotdot*np.sin(psi)+xdot*np.cos(psi)*psidot
g1 = -(m1/c3)*vcxdot+(m2/c3)*vcy*wz-(c1/c3)*vcx*np.sqrt((vcx**2)+(vcy**2))+(c2/c3)*vcy*np.sqrt((vcx**2)+(vcy**2))*np.arctan2(vcy,vcx)
g2 = (m2/c3)*vcydot+(m1/c3)*vcx*wz+(c1/c3)*vcy*np.sqrt((vcx**2)+(vcy**2))+(c2/c3)*vcx*np.sqrt((vcx**2)+(vcy**2))*np.arctan2(vcy,vcx)
A = 12*np.sin(2*np.pi*f*t+np.pi)
if A>=0.1:
wzdot = ((m1-m2)/J)*vcx*vcy-c4*wz**2*np.sign(wz)-c5*g2-c6*np.sqrt((g1**2)+(g2**2))
elif A<-0.1:
wzdot = ((m1-m2)/J)*vcx*vcy-c4*wz**2*np.sign(wz)-c5*g2+c6*np.sqrt((g1**2)+(g2**2))
else:
wzdot = ((m1-m2)/J)*vcx*vcy-c4*wz**2*np.sign(wz)-c5*g2
return [vcxdot,vcydot,psidot,wzdot]
u0 = [0,0,0,0]
t = np.linspace(0,15,1000)
u = odeint(subsystem4,u0,t)
vcx = u[:,0]
vcy = u[:,1]
psi = u[:,2]
wz = u[:,3]
plt.figure(1)
plt.subplot(211)
plt.plot(t,vcx,'r-',linewidth=2,label='vcx')
plt.plot(t,vcy,'b--',linewidth=2,label='vcy')
plt.plot(t,psi,'g:',linewidth=2,label='psi')
plt.plot(t,wz,'c',linewidth=2,label='wz')
plt.xlabel('time')
plt.legend()
plt.show()
To the immediate question of plotting the derivatives, you can get the velocities by directly calling the ODE function again on the solution,
u = odeint(subsystem4,u0,t)
udot = subsystem4(u.T,t)
and get the separate velocity arrays via
vcxdot,vcydot,psidot,wzdot = udot
In this case the function involves branching, which is not very friendly to vectorized calls of it. There are ways to vectorize branching, but the easiest work-around is to loop manually through the solution points, which is slower than a working vectorized implementation. This will again procude a list of tuples like odeint, so the result has to be transposed as a tuple of lists for "easy" assignment to the single array variables.
udot = [ subsystem4(uk, tk) for uk, tk in zip(u,t) ];
vcxdot,vcydot,psidot,wzdot = np.asarray(udot).T
This may appear to double somewhat the computation, but not really, as the solution points are usually interpolated from the internal step points of the solver. The evaluation of the ODE function during integration will usually happen at points that are different from the solution points.
For the other variables, extract the computation of position and velocities into functions to have the constant and composition in one place only:
def xy_pos(t): return 3 + 0.3*np.cos(t), 0.5 + 0.3*np.sin(t)
def xy_vel(t): return -0.3*np.sin(t), 0.3*np.cos(t)
def xy_acc(t): return -0.3*np.cos(t), -0.3*np.sin(t)
or similar that you can then use both inside the ODE function and in preparing the plots.
What Simulink most likely does is to collect all the equations of all the blocks and form this into one big ODE system which is then solved for the whole state at once. You will need to implement something similar. One big state vector, and each subsystem knows its slice of the state resp. derivatives vector to get its specific state variables from and write the derivatives to. The computation of the derivatives can then use values communicated among the subsystems.
What you are trying to do, solving the subsystems separately, works only for resp. will likely result in a order 1 integration method. All higher order methods need to be able to simultaneously shift the state in some direction computed from previous stages of the method, and evaluate the whole system there.

Use Chi-Squared statistic in pymc3

I am trying to use PyMC3 to fit a model to some observed data. This model is based on external code (interfaced via theano.ops.as_op), and depends on multiple parameters that should be fit by the MCMC process. Since the gradient of the external code cannot be determined, I use the Metropolis-Hastings sampler.
I have established Uniform priors for my inputs, and generate a model using my custom code. However, I want to compare the simulated data to my observations (a 3D np.ndarray) using the chi-squared statistic (sum of the squares of data-model/sigma^2) to obtain a log-likelihood. When the MCMC samples are drawn, this should lead to the trace converging on the best values of each parameter.
My model is explained in the following semi-pseudocode (if that's even a word):
import pymc3 as pm
#Some stuff setting up the data, preparing some functions etc.
#theano.compile.ops.as_op(itypes=[input types],otypes = [output types])
def make_model(inputs):
#Wrapper to external code to generate simulated data
return simulated data
model = pm.model()
with model:
#priors for 13 input parameters
simData = make_model(inputs)
I now want to obtain the chi-squared logLikelihood for this model versus the data, which I think can be done using pm.ChiSquared, however I do not see how to combine the data, model and this distribution together to cause the sampler to perform correctly. I would guess it might look something like:
chiSq = pm.ChiSquared(nu=data.size, observed = (data-simData)**2/err**2)
trace = pm.sample(1000)
Is this correct? In running previous tests, I have found the samples appear to be simply drawn from the priors.
Thanks in advance.
Taking aloctavodia's advice, I was able to get parameter estimates for some toy exponential data using a pm.Normal likelihood. Using a pm.ChiSquared likelihood as the OP suggested, the model converged to correct values, but the posteriors on the parameters were roughly three times as broad. Here's the code for the model; I first generated data and then fit with PyMC3.
# Draw `nPoints` observed data points `y_obs` from the function
# 3. + 18. * numpy.exp(-.2 * x)
# with the points evaluated at `x_obs`
# x_obs = numpy.linspace(0, 100, nPoints)
# Add Normal(mu=0,sd=`cov`) noise to each point in `y_obs`
# Then instantiate PyMC3 model for fit:
def YModel(x, c, a, l):
# exponential model expected to describe the data
mu = c + a * pm.math.exp(-l * x)
return mu
def logp(y_mod, y_obs):
# Normal distribution likelihood
return pm.Normal.dist(mu = y_mod, sd = cov).logp(y_obs)
# Chi squared likelihood (to use, comment preceding line & uncomment next 2 lines)
#chi2 = chi2 = pm.math.sum( ((y_mod - y_obs)/cov)**2 )
#return pm.ChiSquared.dist(nu = nPoints).logp(chi2)
with pm.Model() as model:
c = pm.Uniform('constant', lower = 0., upper = 10., testval = 5.)
a = pm.Uniform('amplitude', lower = 0., upper = 50., testval = 25.)
l = pm.Uniform('lambda', lower = 0., upper = 10., testval = 5.)
y_mod = YModel(x_obs, c, a, l)
L = pm.DensityDist('L', logp, observed = {'y_mod': y_mod, 'y_obs': y_obs}, testval = {'y_mod': y_mod, 'y_obs': y_obs})
step = pm.Metropolis([c, a, l])
trace = pm.sample(draws = 10000, step = step)
The above model converged, but I found that success was sensitive to the bounds on the priors and the initial guesses on those parameters.
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
c 3.184397 0.111933 0.002563 2.958383 3.397741 1834.0 1.000260
a 18.276887 0.747706 0.019857 16.882025 19.762849 1343.0 1.000411
l 0.200201 0.013486 0.000361 0.174800 0.226480 1282.0 0.999991
(Edited: I had forgotten to sum the squares of the normalized residuals for chi2)

Tensorflow RNN training won't execute?

I am currently trying to train this RNN network, but seem to be running into weird errors, which I am not able to decode.
The input to my rnn network is digital sampled audio files. As the audio file can be of different length, will the vector of the sampled audio also have different lengths.
The output or the target of the neural network is to recreate a 14 dimensional vector, containing certain information of the audio files. I've already know the target, by manually calculating it, but need to make it work with a neural network.
I am currently using tensorflow as framework.
My network setup looks like this:
def last_relevant(output):
max_length = int(output.get_shape()[1])
relevant = tf.reduce_sum(tf.mul(output, tf.expand_dims(tf.one_hot(length, max_length), -1)), 1)
return relevant
def length(sequence): ##Zero padding to fit the max lenght... Question whether that is a good idea.
used = tf.sign(tf.reduce_max(tf.abs(sequence), reduction_indices=2))
length = tf.reduce_sum(used, reduction_indices=1)
length = tf.cast(length, tf.int32)
return length
def cost(output, target):
# Compute cross entropy for each frame.
cross_entropy = target * tf.log(output)
cross_entropy = -tf.reduce_sum(cross_entropy, reduction_indices=2)
mask = tf.sign(tf.reduce_max(tf.abs(target), reduction_indices=2))
cross_entropy *= mask
# Average over actual sequence lengths.
cross_entropy = tf.reduce_sum(cross_entropy, reduction_indices=1)
cross_entropy /= tf.reduce_sum(mask, reduction_indices=1)
return tf.reduce_mean(cross_entropy)
#----------------------------------------------------------------------#
#----------------------------Main--------------------------------------#
### Tensorflow neural network setup
batch_size = None
sequence_length_max = max_length
input_dimension=1
data = tf.placeholder(tf.float32,[batch_size,sequence_length_max,input_dimension])
target = tf.placeholder(tf.float32,[None,14])
num_hidden = 24 ## Hidden layer
cell = tf.nn.rnn_cell.LSTMCell(num_hidden,state_is_tuple=True) ## Long short term memory
output, state = tf.nn.dynamic_rnn(cell, data, dtype=tf.float32,sequence_length = length(data)) ## Creates the Rnn skeleton
last = last_relevant(output)#tf.gather(val, int(val.get_shape()[0]) - 1) ## Appedning as last
weight = tf.Variable(tf.truncated_normal([num_hidden, int(target.get_shape()[1])]))
bias = tf.Variable(tf.constant(0.1, shape=[target.get_shape()[1]]))
prediction = tf.nn.softmax(tf.matmul(last, weight) + bias)
cross_entropy = cost(output,target)# How far am I from correct value?
optimizer = tf.train.AdamOptimizer() ## TensorflowOptimizer
minimize = optimizer.minimize(cross_entropy)
mistakes = tf.not_equal(tf.argmax(target, 1), tf.argmax(prediction, 1))
error = tf.reduce_mean(tf.cast(mistakes, tf.float32))
## Training ##
init_op = tf.initialize_all_variables()
sess = tf.Session()
sess.run(init_op)
batch_size = 1000
no_of_batches = int(len(train_data)/batch_size)
epoch = 5000
for i in range(epoch):
ptr = 0
for j in range(no_of_batches):
inp, out = train_data[ptr:ptr+batch_size], train_output[ptr:ptr+batch_size]
ptr+=batch_size
sess.run(minimize,{data: inp, target: out})
print "Epoch - ",str(i)
incorrect = sess.run(error,{data: test_data, target: test_output})
print('Epoch {:2d} error {:3.1f}%'.format(i + 1, 100 * incorrect))
sess.close()
The error seem to be the usage of the function last_relevant, which should take the output, and feed it back.
This is the error message:
TypeError: Expected binary or unicode string, got <function length at 0x7f846594dde8>
Anyway to tell what could be wrong here?
I tried to build your code in my local.
There is a fundamental mistake in the code which is that you call tf.one_hot but what you pass don't really fit with what is expected:
Read documentation here:
https://github.com/tensorflow/tensorflow/blob/master/tensorflow/g3doc/api_docs/python/functions_and_classes/shard6/tf.one_hot.md
tf.one_hot(indices, depth, on_value=None, off_value=None, axis=None, dtype=None, name=None)
However, you are passing a function pointer ("length" is a function in your code, I recommend naming your function in a meaningful manner by refraining yourself from using common keywords) instead of the first parameter.
For a wild guide, you can put your indices as first param (instead of my placeholder empty list) and it will be fixed
relevant = tf.reduce_sum(
tf.mul(output, tf.expand_dims(tf.one_hot([], max_length), -1)), 1)

Creating Emax model in pymc3

I am trying to build an Emax model using pymc3 based on the data and model in this video..
(about 40mins in)
https://www.youtube.com/watch?v=U9Nf-ZYHRQA&feature=youtu.be&list=PLvLDbH2lpyXNGV8mpBdF7EFK9LQJzGL-Y
Here is a screen shot showing the model...
My code is here...
pkpd_model = Model()
with pkpd_model:
# Hyperparameter Priors
mu_e0 = Normal('mu_e0', mu=0, tau =1000)
tau_e0 = Uniform('tau_e0', lower=0, upper =100)
mu_emax = Normal('mu_emax', mu=0, tau =1000)
tau_emax = Uniform('tau_emax', lower=0, upper =100)
e0 = Lognormal('e0', mu = mu_e0, tau=tau_e0, shape =n_studies)
emax= Lognormal('emax', mu = mu_emax, tau =tau_emax, shape =n_studies)
ed50 = Lognormal('ed50', mu=1, tau = 1000)
# Normalise sigma for sample size
sigma = np.sqrt(np.square(Uniform('sigma', lower = 0, upper = 1000 ))/n)
# Expected value of outcome
resp_median = e0[study] + (emax[study]*dose)/(ed50+dose)
# Likelihood (sampling distribution) of observations
resp = Lognormal('resp', mu=resp_median, tau =sigma, observed =mean_response)
resp_pred = Lognormal('resp_pred', mu=resp_median, tau =sigma, shape =len(dose))
The model runs and fits okay, its just that the posterior estimates for the model parameter are not near what I am expecting. for example my emax estimate is around 2 but you can clearly see that from the data it should be around 10. So I can only assume I have made an error in building the model, but I cannot for the the life of me see what it is.
Can you help?
Thanks
Mark
After reexamining the model specs I saw that the resp_median is actually the log of that expression. So...
esp_median = np.log(e0[study] + (emax[study]*dose)/(ed50+dose))
improves things. However I am still not sure that I have specified the distributions correctly. Any pointers anyone?

Error using scipy.optimize nonlinear solvers

I am trying to solve a set of M simultaneous eqns with M variables. I input a M X 2 matrix in as an initial guess to my function and it returns a M X 2 matrix, where each entry would equal zero if my guess was correct. Thus my function can be represented as f_k(u1,u2,...uN) = 0 for k=1,2,...N. Below is the code for my function, (for simplicities sake I have left out the modules that go with this code, i.e. p. or phi. for instance. I was more wondering if anyone else has had this error before)
M = len(p.x_lat)
def main(u_A):
## unpack u_A
u_P = u_total[:,0]
u_W = u_total[:,1]
## calculate phi_A for all monomeric species
G_W = exp(-u_W)
phi_W = zeros(M)
phi_W[1:] = p.phi_Wb * G_W[1:]
## calculate phi_A for all polymeric species
G_P = exp(-u_P)
G_P[0] = 0.
G_fwd = phi.fwd_propagator(G_P,p.Np,0) #(function that takes G_P and propagates outward)
G_bkwd = phi.bkwd_propagator(G_P,p.Np,0) #(function that takes G_P and propagates inward)
phi_P = phi.phi_P(G_fwd,G_bkwd,p.norm_graft_density,p.Np) #(function that takes the two propagators and combines them to calculate a segment density at each point)
## calculate u_A components
u_intW = en.u_int_AB(p.chi_PW,phi_P,p.phi_Pb) + en.u_int_AB(p.chi_SW,p.phi_S,p.phi_Sb) #(fxn that calculates new potential from the new segment densities)
u_intW[0] = 0.
u_Wprime = u_W - u_intW
u_intP = en.u_int_AB(p.chi_PW,phi_W,p.phi_Wb) + en.u_int_AB(p.chi_PS,p.phi_S,p.phi_Sb) #(fxn that calculates new potential from the new segment densities)
u_intP[0] = 0.
u_Pprime = u_P - u_intP
## calculate f_A
phi_total = p.phi_S + phi_W + phi_P
u_prime = 0.5 * (u_Wprime + u_Pprime)
f_total = zeros( (M, 2) )
f_total[:,0] = 1. - 1./phi_total + u_prime - u_Wprime
f_total[:,1] = 1. - 1./phi_total + u_prime - u_Pprime
return f_total
I researched ways of solving nonlinear equations such as this one using python. I came across the scipy.optimize library with the several options for solvers http://docs.scipy.org/doc/scipy-0.13.0/reference/optimize.nonlin.html. I first tried to use the newton_krylov solver and received the following error message:
ValueError: Jacobian inversion yielded zero vector. This indicates a bug in the Jacobian approximation.
I also tried broyden1 solver and it never converged but simply stayed stagnant. Code for implementation of both below:
sol = newton_krylov(main, guess, verbose=1, f_tol=10e-7)
sol = broyden1(main, guess, verbose=1, f_tol=10e-7)
My initial guess is given below here:
## first guess of u_A(x)
u_P = zeros(M)
u_P[1] = -0.0001
u_P[M-1] = 0.0001
u_W = zeros(M)
u_W[1] = 0.0001
u_W[M-1] = -0.0001
u_total = zeros( (M,2) )
u_total[:,0] = u_P
u_total[:,1] = u_W
guess = u_total
Any help would be greatly appreciated!