I am learning to implement the Factorization Machine in Pytorch.
And there should be some feature crossing operations.
For example, I've got three features [A,B,C], after embedding, they are [vA,vB,vC], so the feature crossing is "[vA·vB], [vA·vC], [vB·vc]".
I know this operation can be simplified by the following:
It can be implemented by MATRIX OPERATIONS.
But this only gives a final result, say, a single value.
The question is, how to get all cross_vec in the following without doing FOR loop:
note: size of "feature_emb" is [batch_size x feature_len x embedding_size]
g_feature = 0
for i in range(self.featurn_len):
for j in range(self.featurn_len):
if j <= i: continue
cross_vec = feature_emb[:,i,:] * feature_emb[:,j,:]
g_feature += torch.sum(cross_vec, dim=1)
You can
cross_vec = (feature_emb[:, None, ...] * feature_emb[..., None, :]).sum(dim=-1)
This should give you corss_vec of shape (batch_size, feature_len, feature_len).
Alternatively, you can use torch.bmm
cross_vec = torch.bmm(feature_emb, feature_emb.transpose(1, 2))
Related
I'm doing an experiment using face images in PyTorch framework. The input x is the given face image of size 5 * 5 (height * width) and there are 192 channels.
Objective: To obtain patches of x of patch_size(given as argument).
I have obtained the required result with the help of two for loops. But I want a better-vectorized solution so that the computation cost will be very less than using two for loops.
Used: PyTorch 0.4.1, (12 GB) Nvidia TitanX GPU.
The following is my implementation using two for loops
def extractpatches( x, patch_size): # x is bsx192x5x5
patches = x.unfold( 2, patch_size , 1).unfold(3,patch_size,1)
bs,c,pi,pj, _, _ = patches.size() #bs,192,
cnt = 0
p = torch.empty((bs,pi*pj,c,patch_size,patch_size)).to(device)
s = torch.empty((bs,pi*pj, c*patch_size*patch_size)).to(device)
//Want a vectorized method instead of two for loops below
for i in range(pi):
for j in range(pj):
p[:,cnt,:,:,:] = patches[:,:,i,j,:,:]
s[:,cnt,:] = p[:,cnt,:,:,:].view(-1,c*patch_size*patch_size)
cnt = cnt+1
return s
Thanks for your help in advance.
I think you can try this as following. I used some parts of your code for my experiment and it worked for me. Here l and f are the lists of tensor patches
l = [patches[:,:,int(i/pi),i%pi,:,:] for i in range(pi * pi)]
f = [l[i].contiguous().view(-1,c*patch_size*patch_size) for i in range(pi * pi)]
You can verify the above code using toy input values.
Thanks.
I was following Siraj Raval's videos on logistic regression using gradient descent :
1) Link to longer video :
https://www.youtube.com/watch?v=XdM6ER7zTLk&t=2686s
2) Link to shorter video :
https://www.youtube.com/watch?v=xRJCOz3AfYY&list=PL2-dafEMk2A7mu0bSksCGMJEmeddU_H4D
In the videos he talks about using gradient descent to reduce the error for a set number of iterations so that the function converges(slope becomes zero).
He also illustrates the process via code. The following are the two main functions from the code :
def step_gradient(b_current, m_current, points, learningRate):
b_gradient = 0
m_gradient = 0
N = float(len(points))
for i in range(0, len(points)):
x = points[i, 0]
y = points[i, 1]
b_gradient += -(2/N) * (y - ((m_current * x) + b_current))
m_gradient += -(2/N) * x * (y - ((m_current * x) + b_current))
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
return [new_b, new_m]
def gradient_descent_runner(points, starting_b, starting_m, learning_rate, num_iterations):
b = starting_b
m = starting_m
for i in range(num_iterations):
b, m = step_gradient(b, m, array(points), learning_rate)
return [b, m]
#The above functions are called below:
learning_rate = 0.0001
initial_b = 0 # initial y-intercept guess
initial_m = 0 # initial slope guess
num_iterations = 1000
[b, m] = gradient_descent_runner(points, initial_b, initial_m, learning_rate, num_iterations)
# code taken from Siraj Raval's github page
Why does the value of b & m continue to update for all the iterations? After a certain number of iterations, the function will converge, when we find the values of b & m that give slope = 0.
So why do we continue iteration after that point and continue updating b & m ?
This way, aren't we losing the 'correct' b & m values? How is learning rate helping the convergence process if we continue to update values after converging? Thus, why is there no check for convergence, and so how is this actually working?
In practice, most likely you will not reach to slope 0 exactly. Thinking of your loss function as a bowl. If your learning rate is too high, it is possible to overshoot over the lowest point of the bowl. On the contrary, if the learning rate is too low, your learning will become too slow and won't reach the lowest point of the bowl before all iterations are done.
That's why in machine learning, the learning rate is an important hyperparameter to tune.
Actually, once we reach a slope 0; b_gradient and m_gradient will become 0;
thus, for :
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
new_b and new_m will remain the old correct values; as nothing will be subtracted from them.
I am trying to implement a custom Keras objective function:
in 'Direct Intrinsics: Learning Albedo-Shading Decomposition by Convolutional Regression', Narihira et al.
This is the sum of equations (4) and (6) from the previous picture. Y* is the ground truth, Y a prediction map and y = Y* - Y.
This is my code:
def custom_objective(y_true, y_pred):
#Eq. (4) Scale invariant L2 loss
y = y_true - y_pred
h = 0.5 # lambda
term1 = K.mean(K.sum(K.square(y)))
term2 = K.square(K.mean(K.sum(y)))
sca = term1-h*term2
#Eq. (6) Gradient L2 loss
gra = K.mean(K.sum((K.square(K.gradients(K.sum(y[:,1]), y)) + K.square(K.gradients(K.sum(y[1,:]), y)))))
return (sca + gra)
However, I suspect that the equation (6) is not correctly implemented because the results are not good. Am I computing this right?
Thank you!
Edit:
I am trying to approximate (6) convolving with Prewitt filters. It works when my input is a chunk of images i.e. y[batch_size, channels, row, cols], but not with y_true and y_pred (which are of type TensorType(float32, 4D)).
My code:
def cconv(image, g_kernel, batch_size):
g_kernel = theano.shared(g_kernel)
M = T.dtensor3()
conv = theano.function(
inputs=[M],
outputs=conv2d(M, g_kernel, border_mode='full'),
)
accum = 0
for curr_batch in range (batch_size):
accum = accum + conv(image[curr_batch])
return accum/batch_size
def gradient_loss(y_true, y_pred):
y = y_true - y_pred
batch_size = 40
# Direction i
pw_x = np.array([[-1,0,1],[-1,0,1],[-1,0,1]]).astype(np.float64)
g_x = cconv(y, pw_x, batch_size)
# Direction j
pw_y = np.array([[-1,-1,-1],[0,0,0],[1,1,1]]).astype(np.float64)
g_y = cconv(y, pw_y, batch_size)
gra_l2_loss = K.mean(K.square(g_x) + K.square(g_y))
return (gra_l2_loss)
The crash is produced in:
accum = accum + conv(image[curr_batch])
...and error description is the following one:
*** TypeError: ('Bad input argument to theano function with name "custom_models.py:836" at index 0 (0-based)', 'Expected an array-like
object, but found a Variable: maybe you are trying to call a function
on a (possibly shared) variable instead of a numeric array?')
How can I use y (y_true - y_pred) as a numpy array, or how can I solve this issue?
SIL2
term1 = K.mean(K.square(y))
term2 = K.square(K.mean(y))
[...]
One mistake spread across the code was that when you see (1/n * sum()) in the equations, it is a mean. Not the mean of a sum.
Gradient
After reading your comment and giving it more thought, I think there is a confusion about the gradient. At least I got confused.
There are two ways of interpreting the gradient symbol:
The gradient of a vector where y should be differentiated with respect to the parameters of your model (usually the weights of the neural net). In previous edits I started to write in this direction because that's the sort of approach used to trained the model (eg. gradient descent). But I think I was wrong.
The pixel intensity gradient in a picture, as you mentioned in your comment. The diff of each pixel with its neighbor in each direction. In which case I guess you have to translate the example you gave into Keras.
To sum up, K.gradients() and numpy.gradient() are not used in the same way. Because numpy implicitly considers (i, j) (the row and column indices) as the two input variables, while when you feed a 2D image to a neural net, every single pixel is an input variable. Hope I'm clear.
There are several experiments that rely on gradient ascent rather than gradient descent. I have looked into some approaches to using "cost" and the minimize function to simulate the "maximize" function, but I am still not certain I know how to properly implement a maximize() function. Also, in most of these cases, I would say they are closer to an unsupervised learning. So given this code concept for a cost function:
cost = (Yexpected - Ycalculated)^2
train_step = tf.train.AdamOptimizer(0.5).minimize(cost)
I would like to write something were I am following the positive gradient and there may not be a Yexpected value:
maxMe = Function(Ycalculated)
train_step = tf.train.AdamOptimizer(0.5).maximize(maxMe)
A good example of this need is "http://cs229.stanford.edu/proj2009/LvDuZhai.pdf" with Recurrent Reinforcement Learning.
I have read a few papers and references that state changing the sign will flip the direction of movement to increasing gradient, but given TensorFlow's internal calculation of the gradient, I am not sure if this will work to Maximize as I don't know of a way to validate the results:
maxMe = Function(Ycalculated)
train_step = tf.train.AdamOptimizer(0.5).minimize( -1 * maxMe )
The intuition is simple, the minimize() function keeps squashing the given value, for example, if you start with 5, then for every iteration (for example and depending on the learning rate), the value will become say, 4, then 3, then 2, 1, 0 and so on if possible to bring it down more. Now if you pass -5 at the beginning (which is in fact a +5 but you changed the sign explicitly), the gradient will try to change the parameters to bring the number down more, as for example, -5, -6, -7, -8, ...etc. But in fact, the function is increasing because we changed the sign, and the actual sign is (+). In other words, the gradient, in the latter case, is changing the parameters of the neural network in a way that maximizes the function, not minimizing it.
Toy example with arbitrary numbers:
The input x = 1.5, The weight parameter at time (t) w_t = 0.1,
The observed response y = 3.0, The learning rate lr = 0.1.
x * w = 0.15 (this is y predicted for the current w)
loss function = (3.0 - 0.15)^2 = 8.1
Applying gradient descent:
w_(t+1) = w_t - lr * (derivative of loss function with respect to w)
w_(t+1) = 0.1 - (0.1 * [1.5 * 2(0.15 - 3.0)]) = 0.1 - (-0.855) = 0.955
If we use the new w_(t+1) we will have:
1.5 * 0.955 = 1.49 (which is closer to the correct answer 3.0)
and the new loss is: (3.0 - 1.49)^2 = 2.27 (smaller error).
If we keep iterating, we will adjust w to a value that gives us the minimum cost possible.
Now lets repeat the same experiment but with the sign flipped to negative:
loss function = - (3.0 - 0.15)^2 = -8.1
Applying gradient descent:
w_(t+1) = w_t - lr * (derivative of loss function with respect to w)
w_(t+1) = 0.1 - (0.1 * [1.5 * -2(0.15 - 3.0)]) = 0.1 - 0.855 = −0.755
If we apply the new w_(t+1) we will have:
1.5 * −0.755 = −1.1325 and the new loss is: (3.0 - (-1.1325))^2 = 17.07
(the loss function is maximizing!).
That is also applicable to any differentiable function, but this is just a simple naive example to demonstrate the idea.
So, you can do, as you suggested already:
optimizer.minimize( -1 * value)
Or if you like, create a wrapper function (which in fact is needless, but just to mention it):
def maximize(optimizer, value, **kwargs):
return optimizer.minimize(-value, **kwargs)
I have this Backpropagation implementation in MATLAB, and have an issue with training it. Early on in the training phase, all of the outputs go to 1. I have normalized the input data(except the desired class, which is used to generate a binary target vector) to the interval [0, 1]. I have been referring to the implementation in Artificial Intelligence: A Modern Approach, Norvig et al.
Having checked the pseudocode against my code(and studying the algorithm for some time), I cannot spot the error. I have not been using MATLAB for that long, so have been trying to use the documentation where needed.
I have also tried different amounts of nodes in the hidden layer and different learning rates (ALPHA).
The target data encodings are as follows: when the target is to classify as, say 2, the target vector would be [0,1,0], say it were 1, [1, 0, 0] so on and so forth. I have also tried using different values for the target, such as (for class 1 for example) [0.5, 0, 0].
I noticed that some of my weights go above 1, resulting in large net values.
%Topological constants
NUM_HIDDEN = 8+1;%written as n+1 so is clear bias is used
NUM_OUT = 3;
%Training constants
ALPHA = 0.01;
TARG_ERR = 0.01;
MAX_EPOCH = 50000;
%Read and normalize data file.
X = normdata(dlmread('iris.data'));
X = shuffle(X);
%X_test = normdata(dlmread('iris2.data'));
%epocherrors = fopen('epocherrors.txt', 'w');
%Weight matrices.
%Features constitute size(X, 2)-1, however size is (X, 2) to allow for
%appending bias.
w_IH = rand(size(X, 2), NUM_HIDDEN)-(0.5*rand(size(X, 2), NUM_HIDDEN));
w_HO = rand(NUM_HIDDEN+1, NUM_OUT)-(0.5*rand(NUM_HIDDEN+1, NUM_OUT));%+1 for bias
%Layer nets
net_H = zeros(NUM_HIDDEN, 1);
net_O = zeros(NUM_OUT, 1);
%Layer outputs
out_H = zeros(NUM_HIDDEN, 1);
out_O = zeros(NUM_OUT, 1);
%Layer deltas
d_H = zeros(NUM_HIDDEN, 1);
d_O = zeros(NUM_OUT, 1);
%Control variables
error = inf;
epoch = 0;
%Run the algorithm.
while error > TARG_ERR && epoch < MAX_EPOCH
for n=1:size(X, 1)
x = [X(n, 1:size(X, 2)-1) 1]';%Add bias for hiddens & transpose to column vector.
o = X(n, size(X, 2));
%Forward propagate.
net_H = w_IH'*x;%Transposed w.
out_H = [sigmoid(net_H); 1]; %Append 1 for bias to outputs
net_O = w_HO'*out_H;
out_O = sigmoid(net_O); %Again, transposed w.
%Calculate output deltas.
d_O = ((targetVec(o, NUM_OUT)-out_O) .* (out_O .* (1-out_O)));
%Calculate hidden deltas.
for i=1:size(w_HO, 1);
delta_weight = 0;
for j=1:size(w_HO, 2)
delta_weight = delta_weight + d_O(j)*w_HO(i, j);
end
d_H(i) = (out_H(i)*(1-out_H(i)))*delta_weight;
end
%Update hidden-output weights
for i=1:size(w_HO, 1)
for j=1:size(w_HO, 2)
w_HO(i, j) = w_HO(i, j) + (ALPHA*out_H(i)*d_O(j));
end
end
%Update input-hidden weights.
for i=1:size(w_IH, 1)
for j=1:size(w_IH, 2)
w_IH(i, j) = w_IH(i, j) + (ALPHA*x(i)*d_H(j));
end
end
out_O
o
%out_H
%w_IH
%w_HO
%d_O
%d_H
end
end
function outs = sigmoid(nets)
outs = zeros(size(nets, 1), 1);
for i=1:size(nets, 1)
if nets(i) < -45
outs(i) = 0;
elseif nets(i) > 45
outs(i) = 1;
else
outs(i) = 1/1+exp(-nets(i));
end
end
end
From what we've established in the comments the only thing that comes in my mind are all recipes written down together in this great NN archive:
ftp://ftp.sas.com/pub/neural/FAQ2.html#questions
First things you could try are:
1) How to avoid overflow in the logistic function? Probably that's the problem - many times I've implemented NNs the problem was with such an overflow.
2) How should categories be encoded?
And more general:
3) How does ill-conditioning affect NN training?
4) Help! My NN won't learn! What should I do?
After the discussion it turns out the problem lies within the sigmoid function:
function outs = sigmoid(nets)
%...
outs(i) = 1/1+exp(-nets(i)); % parenthesis missing!!!!!!
%...
end
It should be:
function outs = sigmoid(nets)
%...
outs(i) = 1/(1+exp(-nets(i)));
%...
end
The lack of parenthesis caused that the sigmoid output was larger than 1 sometimes. That made the gradient calculation incorrect (because it wasn't a gradient of this function). This caused the gradient to be negative. And this caused that the delta for the output layer was most of the time in the wrong direction. After the fix (the after correctly maintaining the error variable - this seems to be missing in your code) all seems to work fine.
Beside that, there are two other main problems with this code:
1) No bias. Without the bias each neuron can only represent a line which crosses the origin. If data is normalized (i.e. values are between 0 and 1), some configurations are inseparable.
2) Lack of guarding against high gradient values (point 1 in my previous answer).