I was performing semantic segmentation using PyTorch. There are a total of 103 different classes in the dataset and the targets are RGB images with only the Red channel containing the labels. I was using nn.CrossEntropyLoss as my loss function. For sanity, I wanted to check if using nn.CrossEntropyLoss is correct for this problem and whether it has the expected behaviour
I pick a random mask from my dataset and create a categorical version of it using this custom transform
class ToCategorical:
def __init__(self, n_classes: int) -> None:
self.n_classes = n_classes
def __call__(self, sample: torch.Tensor):
mask = sample.permute(1, 2, 0)
categories = torch.unique(mask).tolist()[1:] # get all categories other than 0
# build a tensor with `n_classes` channels
one_hot_image = torch.zeros(self.n_classes, *mask.shape[:-1])
for category in categories:
# get spacial locs where the categ is present
rows, cols, _ = torch.where(mask == category)
# in same spacial loc but in `categ` channel fill 1
one_hot_image[category, rows, cols] = 1
return one_hot_image
And then I send this image as the output (prediction) and use the ground truth mask as the target to the loss function.
import torch.nn as nn
mask = T.PILToTensor()(Image.open("path_to_image").convert("RGB"))
categorical_mask = ToCategorical(103)(mask).unsqueeze(0)
mask = mask[0].unsqueeze(0) # get only the red channel, add fake batch_dim
loss_fn = nn.CrossEntropyLoss()
target = mask
output = categorical_mask
print(output.shape, target.shape)
print(loss_fn(output, target.to(torch.long)))
I expected the loss to be zero but to my surprise, the output is as follows
torch.Size([1, 103, 600, 800]) torch.Size([1, 600, 800])
tensor(4.2836)
I verified with other samples in the dataset and I obtained similar values for other masks as well. Am I doing something wrong? I expect the loss to be = 0 when the output is the same as the target.
PS. I also know that nn.CrossEntropyLoss is the same as using log_softmax followed by nn.NLLLoss() but even I obtained the same value by using nllloss as well
For Reference
Dataset used: UECFoodPixComplete
I would like to adress this:
I expect the loss to be = 0 when the output is the same as the target.
If the prediction matches the target, i.e. the prediction corresponds to a one-hot-encoding of the labels contained in the dense target tensor, but the loss itself is not supposed to equal to zero. Actually, it can never be equal to zero because the nn.CrossEntropyLoss function is always positive by definition.
Let us take a minimal example with number of #C classes and a target y_pred and a prediction y_pred consisting of prefect predictions:
As a quick reminder:
The softmax is applied on the logits (q_i) as p_i = log(exp(q_i)/sum_j(exp(q_j)):
>>> p = F.softmax(y_pred, 1)
Similarly if you are using the log-softmax, defined as logp_i = log(p_i):
>>> logp = F.log_softmax(y_pred, 1)
Then comes the negative likelihood function computed between x the input and y the target: -y*x. In association with the softmax, it comes down to -y*p, or -y*logp respectively. In any case, whether you apply the log or not, only the predictions corresponding to the true classes will remain since the others ones are zeroed-out.
That being said, applying the NLLLoss on y_pred would indeed result with a 0 as you expected in your question. However, here we apply it on the probability distribution or log-probability: p, or logp respectively!
In our specific case, p_i = 1 for the true class and p_i = 0 for all other classes (there are #C - 1 of those). This means the softmax of the logit associated with the true class will equal to exp(1)/sum_i(p_i). And since sum_i(p_i) = (#C-1)*exp(0) + exp(1). We therefore have:
softmax(p) = e / (#C - 1 + e)
Similarly for log-softmax:
log-softmax(p) = log(e / (#C-1 + e)) = 1 - log(#C - 1 + e)
If we proceed by applying the negative likelihood function we simply get cross-entropy(y_pred, y_true) = (nllloss o log-softmax)(y_pred, y_true). This results in:
loss = - (1 - log(#C - 1 + e)) = log(#C - 1 + e) - 1
This effectively corresponds to the minimum of the nn.CrossEntropyLoss function.
Regarding your specific case where #C = 103, you may have an issue in your code... since the average loss should equal to log(102 + e) - 1 i.e. around 3.65.
>>> y_true = torch.randint(0,103,(1,1,2,5))
>>> y_pred = torch.zeros(1,103,2,5).scatter(1, y_true, value=1)
You can see for yourself with one of the provided methods:
the builtin function nn.functional.cross_entropy:
>>> F.cross_entropy(y_pred, y_true[:,0])
tensor(3.6513)
manually computing the quantity:
>>> logp = F.log_softmax(y_pred, 1)
>>> -logp.gather(1, y_true).mean()
tensor(3.6513)
analytical result:
>>> log(102 + e) - 1
3.6513
I have a concern in understanding the Cartpole code as an example for Deep Q Learning. The DQL Agent part of the code as follow:
class DQLAgent:
def __init__(self, env):
# parameter / hyperparameter
self.state_size = env.observation_space.shape[0]
self.action_size = env.action_space.n
self.gamma = 0.95
self.learning_rate = 0.001
self.epsilon = 1 # explore
self.epsilon_decay = 0.995
self.epsilon_min = 0.01
self.memory = deque(maxlen = 1000)
self.model = self.build_model()
def build_model(self):
# neural network for deep q learning
model = Sequential()
model.add(Dense(48, input_dim = self.state_size, activation = "tanh"))
model.add(Dense(self.action_size,activation = "linear"))
model.compile(loss = "mse", optimizer = Adam(lr = self.learning_rate))
return model
def remember(self, state, action, reward, next_state, done):
# storage
self.memory.append((state, action, reward, next_state, done))
def act(self, state):
# acting: explore or exploit
if random.uniform(0,1) <= self.epsilon:
return env.action_space.sample()
else:
act_values = self.model.predict(state)
return np.argmax(act_values[0])
def replay(self, batch_size):
# training
if len(self.memory) < batch_size:
return
minibatch = random.sample(self.memory,batch_size)
for state, action, reward, next_state, done in minibatch:
if done:
target = reward
else:
target = reward + self.gamma*np.amax(self.model.predict(next_state)[0])
train_target = self.model.predict(state)
train_target[0][action] = target
self.model.fit(state,train_target, verbose = 0)
def adaptiveEGreedy(self):
if self.epsilon > self.epsilon_min:
self.epsilon *= self.epsilon_decay
In the training section, we found our target and train_target. So why did we set train_target[0][action] = target here?
Every predict made while learning is not correct, but thanks to error calculation and backpropagation, the predict made at the end of the network will get closer and closer, but when we make train_target[0][action] = target here the error becomes 0, and in this case, how will the learning be?
self.model.predict(state) will return a tensor of shape of (1, 2) containing the estimated Q values for each action (in cartpole the action space is {0,1}).
As you know the Q value is a measure of the expected reward.
By setting self.model.predict(state)[0][action] = target (where target is the expected sum of rewards) it is creating a target Q value on which to train the model. By then calling model.fit(state, train_target) it is using the target Q value to train said model to approximate better Q values for each state.
I don't understand why you are saying that the loss becomes 0: the target is set to the discounted sum of rewards plus the current reward
target = reward + self.gamma*np.amax(self.model.predict(next_state)[0])
while the network prediction for the highest Q value is
np.amax(self.model.predict(next_state)[0])
The loss between the target and the predicted values is what is used to train the model.
Edit - more detailed explaination
(you can ignore the [0] to the predicted values, as it is just to access the right column and unimportant in the understanding)
The target variable is set to the sum between the current reward and the estimated sum of future rewards, or the Q value. Note that this variable is called target but it is not the target of the network, but the target Q value for the chosen action.
The train_target variable is used as what you call the "dataset". It represents the target of the network.
train_target = self.model.predict(state)
train_target[0][action] = target
You can clearly see that:
train_target[<taken action>] = reward + self.gamma*np.amax(self.model.predict(next_state)[0])
train_target[<any other action>] = <prediction from the model>
the loss (mean squared error):
prediction = self.model.predict(state)
loss = (train_target - prediction)^2
For any line of the that is not the the loss is 0. For the one line that has been set the loss is
(target - prediction[action])^2
or
((reward + self.gamma*np.amax(self.model.predict(next_state)[0])) - self.model.predict(state)[0][action])^2
which is clearly different from 0.
Note that this agent is not ideal. I would strongly recommend the use of a target model instead of creating target Q values that way. Check out this answer as for why.
I designed the Graph Attention Network.
However, during the operations inside the layer, the values of features becoming equal.
class GraphAttentionLayer(nn.Module):
## in_features = out_features = 1024
def __init__(self, in_features, out_features, dropout):
super(GraphAttentionLayer, self).__init__()
self.dropout = dropout
self.in_features = in_features
self.out_features = out_features
self.W = nn.Parameter(torch.zeros(size=(in_features, out_features)))
self.a1 = nn.Parameter(torch.zeros(size=(out_features, 1)))
self.a2 = nn.Parameter(torch.zeros(size=(out_features, 1)))
nn.init.xavier_normal_(self.W.data, gain=1.414)
nn.init.xavier_normal_(self.a1.data, gain=1.414)
nn.init.xavier_normal_(self.a2.data, gain=1.414)
self.leakyrelu = nn.LeakyReLU()
def forward(self, input, adj):
h = torch.mm(input, self.W)
a_input1 = torch.mm(h, self.a1)
a_input2 = torch.mm(h, self.a2)
a_input = torch.mm(a_input1, a_input2.transpose(1, 0))
e = self.leakyrelu(a_input)
zero_vec = torch.zeros_like(e)
attention = torch.where(adj > 0, e, zero_vec) # most of values is close to 0
attention = F.softmax(attention, dim=1) # all values are 0.0014 which is 1/707 (707^2 is the dimension of attention)
attention = F.dropout(attention, self.dropout)
return attention
The dimension of 'attention' is (707 x 707) and I observed the value of attention is near 0 before the softmax.
After the softmax, all values are 0.0014 which is 1/707.
I wonder how to keep the values normalized and prevent this situation.
Thanks
Since you say this happens during training I would assume it is at the start. With random initialization you often get near identical values at the end of the network during the start of the training process.
When all values are more or less equal the output of the softmax will be 1/num_elements for every element, so they sum up to 1 over the dimension you chose. So in your case you get 1/707 as all the values, which just sounds to me your weights are freshly initialized and the outputs are mostly random at this stage.
I would let it train for a while and observe if this changes.
I would like to estimate the parameters of a mixture model of normal distributions in OpenTURNS (that is, the distribution of a weighted sum of Gaussian random variables). OpenTURNS can create such a mixture, but it cannot estimate its parameters. Moreover, I need to create the mixture as an OpenTURNS distribution in order to propagate uncertainty through a function.
For example, I know how to create a mixture of two normal distributions:
import openturns as ot
mu1 = 1.0
sigma1 = 0.5
mu2 = 3.0
sigma2 = 2.0
weights = [0.3, 0.7]
n1 = ot.Normal(mu1, sigma1)
n2 = ot.Normal(mu2, sigma2)
m = ot.Mixture([n1, n2], weights)
In this example, I would like to estimate mu1, sigma1, mu2, sigma2 on a given sample. In order to create a working example, it is easy to generate a sample by simulation.
s = m.getSample(100)
You can rely on scikit-learn's GaussianMixture to estimate the parameters and then use them to define a Mixture model in OpenTURNS.
The script hereafter contains a Python class MixtureFactory that estimates the parameters of a scikitlearn GaussianMixture and outputs an OpenTURNS Mixture distribution:
from sklearn.mixture import GaussianMixture
from sklearn.utils.validation import check_is_fitted
import openturns as ot
import numpy as np
class MixtureFactory(GaussianMixture):
"""
Representation of a Gaussian mixture model probability distribution.
This class allows to estimate the parameters of a Gaussian mixture
distribution using scikit algorithms & provides openturns Mixture object.
Read more in scikit learn user guide & openturns theory.
Parameters:
-----------
n_components : int, defaults to 1.
The number of mixture components.
covariance_type : {'full' (default), 'tied', 'diag', 'spherical'}
String describing the type of covariance parameters to use.
Must be one of:
'full'
each component has its own general covariance matrix
'tied'
all components share the same general covariance matrix
'diag'
each component has its own diagonal covariance matrix
'spherical'
each component has its own single variance
tol : float, defaults to 1e-3.
The convergence threshold. EM iterations will stop when the
lower bound average gain is below this threshold.
reg_covar : float, defaults to 1e-6.
Non-negative regularization added to the diagonal of covariance.
Allows to assure that the covariance matrices are all positive.
max_iter : int, defaults to 100.
The number of EM iterations to perform.
n_init : int, defaults to 1.
The number of initializations to perform. The best results are kept.
init_params : {'kmeans', 'random'}, defaults to 'kmeans'.
The method used to initialize the weights, the means and the
precisions.
Must be one of::
'kmeans' : responsibilities are initialized using kmeans.
'random' : responsibilities are initialized randomly.
weights_init : array-like, shape (n_components, ), optional
The user-provided initial weights, defaults to None.
If it None, weights are initialized using the `init_params` method.
means_init : array-like, shape (n_components, n_features), optional
The user-provided initial means, defaults to None,
If it None, means are initialized using the `init_params` method.
precisions_init : array-like, optional.
The user-provided initial precisions (inverse of the covariance
matrices), defaults to None.
If it None, precisions are initialized using the 'init_params' method.
The shape depends on 'covariance_type'::
(n_components,) if 'spherical',
(n_features, n_features) if 'tied',
(n_components, n_features) if 'diag',
(n_components, n_features, n_features) if 'full'
random_state : int, RandomState instance or None, optional (default=None)
If int, random_state is the seed used by the random number generator;
If RandomState instance, random_state is the random number generator;
If None, the random number generator is the RandomState instance used
by `np.random`.
warm_start : bool, default to False.
If 'warm_start' is True, the solution of the last fitting is used as
initialization for the next call of fit(). This can speed up
convergence when fit is called several times on similar problems.
In that case, 'n_init' is ignored and only a single initialization
occurs upon the first call.
See :term:`the Glossary <warm_start>`.
verbose : int, default to 0.
Enable verbose output. If 1 then it prints the current
initialization and each iteration step. If greater than 1 then
it prints also the log probability and the time needed
for each step.
verbose_interval : int, default to 10.
Number of iteration done before the next print.
"""
def __init__(self, n_components=2, covariance_type='full', tol=1e-6,
reg_covar=1e-6, max_iter=1000, n_init=1, init_params='kmeans',
weights_init=None, means_init=None, precisions_init=None,
random_state=41, warm_start=False,
verbose=0, verbose_interval=10):
super().__init__(n_components, covariance_type, tol, reg_covar,
max_iter, n_init, init_params, weights_init, means_init,
precisions_init, random_state, warm_start, verbose, verbose_interval)
def fit(self, X):
"""
Fit the mixture model parameters.
EM algorithm is applied here to estimate the model parameters and build a
Mixture distribution (see openturns mixture).
The method fits the model ``n_init`` times and sets the parameters with
which the model has the largest likelihood or lower bound. Within each
trial, the method iterates between E-step and M-step for ``max_iter``
times until the change of likelihood or lower bound is less than
``tol``, otherwise, a ``ConvergenceWarning`` is raised.
If ``warm_start`` is ``True``, then ``n_init`` is ignored and a single
initialization is performed upon the first call. Upon consecutive
calls, training starts where it left off.
Parameters
----------
X : array-like, shape (n_samples, n_features)
List of n_features-dimensional data points. Each row
corresponds to a single data point.
Returns
-------
"""
data = np.array(X)
# Evaluate the model parameters.
super().fit(data)
# openturns mixture
# n_components ==> weight of size n_components
weights = self.weights_
n_components = len(weights)
# Create ot distribution
collection = n_components * [0]
# Covariance matrices
cov = self.covariances_
mu = self.means_
# means : n_components x n_features
n_components, n_features = mu.shape
# Following the type of covariance, we define the collection of gaussians
# Spherical : C_k = Identity * sigma_k
if self.covariance_type is 'spherical':
c = ot.CorrelationMatrix(n_features)
for l in range(n_components):
sigma = np.sqrt(cov[l])
collection[l] = ot.Normal(list(mu[l]), [ sigma ] * n_features , c)
elif self.covariance_type is 'diag' :
for l in range(n_components):
c = ot.CovarianceMatrix(n_features)
for i in range(n_features):
c[i,i] = cov[l, i]
collection[l] = ot.Normal(list(mu[l]), c)
elif self.covariance_type == 'tied':
# Same covariance for all clusters
c = ot.CovarianceMatrix(n_features)
for i in range(n_features):
for j in range(0, i+1):
c[i,j] = cov[i,j]
# Define the collection with the same covariance
for l in range(n_components):
collection[l] = ot.Normal(list(mu[l]), c)
else:
n_features = cov.shape[1]
for l in range(n_components):
c = ot.CovarianceMatrix(n_features)
for i in range(n_features):
for j in range(0, i+1):
c[i,j] = cov[l][i,j]
collection[l] = ot.Normal(list(mu[l]), c)
self._mixture = ot.Mixture(collection, weights)
return self
def get_mixture(self):
"""
Returns the Mixture object
"""
check_is_fitted(self)
return self._mixture
if __name__ == "__main__":
mu1 = 1.0
sigma1 = 0.5
mu2 = 3.0
sigma2 = 2.0
weights = [0.3, 0.7]
n1 = ot.Normal(mu1, sigma1)
n2 = ot.Normal(mu2, sigma2)
m = ot.Mixture([n1, n2], weights)
x = m.getSample(1000)
est_dist = MixtureFactory(random_state=1)
est_dist.fit(x)
print(est_dist.get_mixture())
I have actually tried this method and it works perfectly. On top of that the fit of the model through the SciKit GMM and the ulterior adjustment thanks to OpenTurns are very fast. I recommend future users to test several numbers of components and covariance matrix structures, as it will not take a lot of time and can substantially improve the goodness of fit to the data.
Thanks for the answer.
Here is a pure OpenTURNS solution. It is probably slower than the scikit-learn-based method, but it is more generic: you could use it to estimate the parameters of any mixture model, not necessarily a mixture of normal distributions.
The idea is to retrieve the log-likelihood function from the Mixture object and minimize it.
In the following, let us assume that s is the sample we want to fit the mixture on.
First, we need to build the mixture we want to estimate the parameters of. We can specify any valid set of parameters, it does not matter. In your example, you want a mixture of 2 normal distributions.
mixture = ot.Mixture([ot.Normal()]*2, [0.5]*2)
There is a small hurdle. All weights sum to 1, thus one of them is determined by the others: the solver must not be allowed to freely set it. The order of the parameters of an OpenTURNS Mixture is as follows:
weight of the first distribution;
parameters of the first distribution;
weight of the second distribution;
parameters of the second distribution:
...
You can view all parameters with mixture.getParameter() and their names with mixture.getParameterDescription(). The following is a helper function that:
takes as input the list containing of all mixture parameters except the weight of its first distribution;
outputs a Point containing all parameters including the weight of the first distribution.
def full(params):
"""
Point of all mixture parameters from a list that omits the first weight.
"""
params = ot.Point(params)
aux_mixture = ot.Mixture(mixture)
dist_number = aux_mixture.getDistributionCollection().getSize()
index = aux_mixture.getDistributionCollection()[0].getParameter().getSize()
list_weights = []
for num in range(1, dist_number):
list_weights.append(params[index])
index += 1 + aux_mixture.getDistributionCollection()[num].getParameter().getSize()
complementary_weight = ot.Point([abs(1.0 - sum(list_weights))])
complementary_weight.add(params)
return complementary_weight
The next function computes the opposite of the log-likelihood of a given list of parameters (except the first weight).
For the sake of numerical stability, it divides this value by the number of observations.
We will minimize this function in order to find the Maximum Likelihood Estimate.
def minus_log_pdf(params):
"""
- log-likelihood of a list of parameters excepting the first weight
divided by the number of observations
"""
aux_mixture = ot.Mixture(mixture)
full_params = full(params)
try:
aux_mixture.setParameter(full_params)
except TypeError:
# case where the proposed parameters are invalid:
# return a huge value
return [ot.SpecFunc.LogMaxScalar]
res = - aux_mixture.computeLogPDF(s).computeMean()
return res
To use OpenTURNS optimization facilities, we need to turn this function into a PythonFunction object.
OT_minus_log_pdf = ot.PythonFunction(mixture.getParameter().getSize()-1, 1, minus_log_pdf)
Cobyla is usually good at likelihood optimization.
problem = ot.OptimizationProblem(OT_minus_log_pdf)
algo = ot.Cobyla(problem)
In order to decrease chances of Cobyla being stuck on a local minimum, we are going to use MultiStart. We pick a starting set of parameters and randomly change the weights. The following helper function makes it easy:
def random_weights(params, nb):
"""
List of nb Points representing mixture parameters with randomly varying weights.
"""
aux_mixture = ot.Mixture(mixture)
full_params = full(params)
aux_mixture.setParameter(full_params)
list_params = []
for num in range(nb):
dirichlet = ot.Dirichlet([1.0] * aux_mixture.getDistributionCollection().getSize()).getRealization()
dirichlet.add(1.0 - sum(dirichlet))
aux_mixture.setWeights(dirichlet)
list_params.append(aux_mixture.getParameter()[1:])
return list_params
We pick 10 starting points and increase the number of maximum evaluations of the log-likelihood from 100 (by default) to 10000.
init = mixture.getParameter()[1:]
starting_points = random_weights(init, 10)
algo_multistart = ot.MultiStart(algo, starting_points)
algo_multistart.setMaximumEvaluationNumber(10000)
Let's run the solver and retrieve the result.
algo_multistart.run()
result = algo_multistart.getResult()
All that remains is to set the mixture's parameters to the optimal value.
We must not forget to add the first weight back!
optimal_parameters = result.getOptimalPoint()
mixture.setParameter(full(optimal_parameters))
Below is an alternative.
The first step creates a new GaussianMixture class, derived from PythonDistribution. The key point is to implement the computeLogPDF method and the set/getParameters methods. Notice that this parametrization of a mixture only has one single weight w.
class GaussianMixture(ot.PythonDistribution):
def __init__(self, mu1 = -5.0, sigma1 = 1.0, \
mu2 = 5.0, sigma2 = 1.0, \
w = 0.5):
super(GaussianMixture, self).__init__(1)
if w < 0.0 or w > 1.0:
raise ValueError('The weight is not in [0, 1]. w=%s.' % (w))
self.mu1 = mu2
self.sigma1 = sigma1
self.mu2 = mu2
self.sigma2 = sigma2
self.w = w
collDist = [ot.Normal(mu1, sigma1), ot.Normal(mu2, sigma2)]
weight = [w, 1.0 - w]
self.distribution = ot.Mixture(collDist, weight)
def computeCDF(self, x):
p = self.distribution.computeCDF(x)
return p
def computePDF(self, x):
p = self.distribution.computePDF(x)
return p
def computeQuantile(self, prob, tail = False):
quantile = self.distribution.computeQuantile(prob, tail)
return quantile
def getSample(self, size):
X = self.distribution.getSample(size)
return X
def getParameter(self):
parameter = ot.Point([self.mu1, self.sigma1, \
self.mu2, self.sigma2, \
self.w])
return parameter
def setParameter(self, parameter):
[mu1, sigma1, mu2, sigma2, w] = parameter
self.__init__(mu1, sigma1, mu2, sigma2, w)
return parameter
def computeLogPDF(self, sample):
logpdf = self.distribution.computeLogPDF(sample)
return logpdf
In order to create the distribution, we use the Distribution class:
gm = ot.Distribution(GaussianMixture())
Estimating the parameters of this distribution is straightforward with MaximumLikelihoodFactory. However, we must set the bounds, because sigma cannot be negative and that w is in (0, 1).
factory = ot.MaximumLikelihoodFactory(gm)
lowerBound = [0.0, 1.e-6, 0.0, 1.e-6, 0.01]
upperBound = [0.0, 0.0, 0.0, 0.0, 0.99]
finiteLowerBound = [False, True, False, True, True]
finiteUpperBound = [False, False, False, False, True]
bounds = ot.Interval(lowerBound, upperBound, finiteLowerBound, finiteUpperBound)
factory.setOptimizationBounds(bounds)
Then we configure the optimization solver.
solver = factory.getOptimizationAlgorithm()
startingPoint = [-4.0, 1.0, 7.0, 1.5, 0.3]
solver.setStartingPoint(startingPoint)
factory.setOptimizationAlgorithm(solver)
Estimating the parameters is based on the build method.
distribution = factory.build(sample)
There are two limitations with this implementation.
First, it is not as fast as it should be, because of a limitation of the PythonDistribution (see https://github.com/openturns/openturns/issues/1391).
Estimating the parameters may be difficult, because the problem may have local optimas that cannot be retrieved with the default algorithm in MaximumLikelihoodFactory. This kind of task is generally done with the EM algorithm.
I am trying to implement q-learning with an action-value approximation-function. I am using openai-gym and the "MountainCar-v0" enviroment to test my algorithm out. My problem is, it does not converge or find the goal at all.
Basically the approximator works like the following, you feed in the 2 features: position and velocity and one of the 3 actions in a one-hot encoding: 0 -> [1,0,0], 1 -> [0,1,0] and 2 -> [0,0,1]. The output is the action-value approximation Q_approx(s,a), for one specific action.
I know that usually, the input is the state (2 features) and the output layer contains 1 output for each action. The big difference that I see is that I have run the feed forward pass 3 times (one for each action) and take the max, while in the standard implementation you run it once and take the max over the output.
Maybe my implementation is just completely wrong and I am thinking wrong. Gonna paste the code here, it is a mess but I am just experimenting a bit:
import gym
import numpy as np
from keras.models import Sequential
from keras.layers import Dense, Activation
env = gym.make('MountainCar-v0')
# The mean reward over 20 episodes
mean_rewards = np.zeros(20)
# Feature numpy holder
features = np.zeros(5)
# Q_a value holder
qa_vals = np.zeros(3)
one_hot = {
0 : np.asarray([1,0,0]),
1 : np.asarray([0,1,0]),
2 : np.asarray([0,0,1])
}
model = Sequential()
model.add(Dense(20, activation="relu",input_dim=(5)))
model.add(Dense(10,activation="relu"))
model.add(Dense(1))
model.compile(optimizer='rmsprop',
loss='mse',
metrics=['accuracy'])
epsilon_greedy = 0.1
discount = 0.9
batch_size = 16
# Experience replay containing features and target
experience = np.ones((10*300,5+1))
# Ring buffer
def add_exp(features,target,index):
if index % experience.shape[0] == 0:
index = 0
global filled_once
filled_once = True
experience[index,0:5] = features
experience[index,5] = target
index += 1
return index
for e in range(0,100000):
obs = env.reset()
old_obs = None
new_obs = obs
rewards = 0
loss = 0
for i in range(0,300):
if old_obs is not None:
# Find q_a max for s_(t+1)
features[0:2] = new_obs
for i,pa in enumerate([0,1,2]):
features[2:5] = one_hot[pa]
qa_vals[i] = model.predict(features.reshape(-1,5))
rewards += reward
target = reward + discount*np.max(qa_vals)
features[0:2] = old_obs
features[2:5] = one_hot[a]
fill_index = add_exp(features,target,fill_index)
# Find new action
if np.random.random() < epsilon_greedy:
a = env.action_space.sample()
else:
a = np.argmax(qa_vals)
else:
a = env.action_space.sample()
obs, reward, done, info = env.step(a)
old_obs = new_obs
new_obs = obs
if done:
break
if filled_once:
samples_ids = np.random.choice(experience.shape[0],batch_size)
loss += model.train_on_batch(experience[samples_ids,0:5],experience[samples_ids,5].reshape(-1))[0]
mean_rewards[e%20] = rewards
print("e = {} and loss = {}".format(e,loss))
if e % 50 == 0:
print("e = {} and mean = {}".format(e,mean_rewards.mean()))
Thanks in advance!
There shouldn't be much difference between the actions as inputs to your network or as different outputs of your network. It does make a huge difference if your states are images for example. because Conv nets work very well with images and there would be no obvious way of integrating the actions to the input.
Have you tried the cartpole balancing environment? It is better to test if your model is working correctly.
Mountain climb is pretty hard. It has no reward until you reach the top, which often doesn't happen at all. The model will only start learning something useful once you get to the top once. If you are never getting to the top you should probably increase your time doing exploration. in other words take more random actions, a lot more...