I'm trying to implement the gradient descent with PyTorch according to this schema but can't figure out how to properly update the weights. It is just a toy example with 2 linear layers with 2 nodes in hidden layer and one output.
Learning rate = 0.05;
target output = 1
https://hmkcode.github.io/ai/backpropagation-step-by-step/
Forward
Backward
My code is as following:
import torch
import torch.nn as nn
import torch.nn.functional as F
import torch.optim as optim
class MyNet(nn.Module):
def __init__(self):
super(MyNet, self).__init__()
self.linear1 = nn.Linear(2, 2, bias=None)
self.linear1.weight = torch.nn.Parameter(torch.tensor([[0.11, 0.21], [0.12, 0.08]]))
self.linear2 = nn.Linear(2, 1, bias=None)
self.linear2.weight = torch.nn.Parameter(torch.tensor([[0.14, 0.15]]))
def forward(self, inputs):
out = self.linear1(inputs)
out = self.linear2(out)
return out
losses = []
loss_function = nn.L1Loss()
model = MyNet()
optimizer = optim.SGD(model.parameters(), lr=0.05)
input = torch.tensor([2.0,3.0])
print('weights before backpropagation = ', list(model.parameters()))
for epoch in range(1):
result = model(input )
loss = loss_function(result , torch.tensor([1.00],dtype=torch.float))
print('result = ', result)
print("loss = ", loss)
model.zero_grad()
loss.backward()
print('gradients =', [x.grad.data for x in model.parameters()] )
optimizer.step()
print('weights after backpropagation = ', list(model.parameters()))
The result is following :
weights before backpropagation = [Parameter containing:
tensor([[0.1100, 0.2100],
[0.1200, 0.0800]], requires_grad=True), Parameter containing:
tensor([[0.1400, 0.1500]], requires_grad=True)]
result = tensor([0.1910], grad_fn=<SqueezeBackward3>)
loss = tensor(0.8090, grad_fn=<L1LossBackward>)
gradients = [tensor([[-0.2800, -0.4200], [-0.3000, -0.4500]]),
tensor([[-0.8500, -0.4800]])]
weights after backpropagation = [Parameter containing:
tensor([[0.1240, 0.2310],
[0.1350, 0.1025]], requires_grad=True), Parameter containing:
tensor([[0.1825, 0.1740]], requires_grad=True)]
Forward pass values:
2x0.11 + 3*0.21=0.85 ->
2x0.12 + 3*0.08=0.48 -> 0.85x0.14 + 0.48*0.15=0.191 -> loss =0.191-1 = -0.809
Backward pass: let's calculate w5 and w6 (output node weights)
w = w - (prediction-target)x(gradient)x(output of previous node)x(learning rate)
w5= 0.14 -(0.191-1)*1*0.85*0.05= 0.14 + 0.034= 0.174
w6= 0.15 -(0.191-1)*1*0.48*0.05= 0.15 + 0.019= 0.169
In my example Torch doesn't multiply the loss by derivative so we get wrong weights after updating. For the output node we got new weights w5,w6 [0.1825, 0.1740] , when it should be [0.174, 0.169]
Moving backward to update the first weight of the output node (w5) we need to calculate: (prediction-target)x(gradient)x(output of previous node)x(learning rate)=-0.809*1*0.85*0.05=-0.034. Updated weight w5 = 0.14-(-0.034)=0.174. But instead pytorch calculated new weight = 0.1825. It forgot to multiply by (prediction-target)=-0.809. For the output node we got gradients -0.8500 and -0.4800. But we still need to multiply them by loss 0.809 and learning rate 0.05 before we can update the weights.
What is the proper way of doing this?
Should we pass 'loss' as an argument to backward() as following: loss.backward(loss) .
That seems to fix it. But I couldn't find any example on this in documentation.
You should use .zero_grad() with optimizer, so optimizer.zero_grad(), not loss or model as suggested in the comments (though model is fine, but it is not clear or readable IMO).
Except that your parameters are updated fine, so the error is not on PyTorch's side.
Based on gradient values you provided:
gradients = [tensor([[-0.2800, -0.4200], [-0.3000, -0.4500]]),
tensor([[-0.8500, -0.4800]])]
Let's multiply all of them by your learning rate (0.05):
gradients_times_lr = [tensor([[-0.014, -0.021], [-0.015, -0.0225]]),
tensor([[-0.0425, -0.024]])]
Finally, let's apply ordinary SGD (theta -= gradient * lr), to get exactly the same results as in PyTorch:
parameters = [tensor([[0.1240, 0.2310], [0.1350, 0.1025]]),
tensor([[0.1825, 0.1740]])]
What you have done is taken the gradients calculated by PyTorch and multiplied them with the output of previous node and that's not how it works!.
What you've done:
w5= 0.14 -(0.191-1)*1*0.85*0.05= 0.14 + 0.034= 0.174
What should of been done (using PyTorch's results):
w5 = 0.14 - (-0.85*0.05) = 0.1825
No multiplication of previous node, it's done behind the scenes (that's what .backprop() does - calculates correct gradients for all of the nodes), no need to multiply them by previous ones.
If you want to calculate them manually, you have to start at the loss (with delta being one) and backprop all the way down (do not use learning rate here, it's a different story!).
After all of them are calculated, you can multiply each weight by optimizers learning rate (or any other formula for that matter, e.g. Momentum) and after this you have your correct update.
How to calculate backprop
Learning rate is not part of backpropagation, leave it alone until you calculate all of the gradients (it confuses separate algorithms together, optimization procedures and backpropagation).
1. Derivative of total error w.r.t. output
Well, I don't know why you are using Mean Absolute Error (while in the tutorial it is Mean Squared Error), and that's why both those results vary. But let's go with your choice.
Derivative of | y_true - y_pred | w.r.t. to y_pred is 1, so IT IS NOT the same as loss. Change to MSE to get equal results (here, the derivative will be (1/2 * y_pred - y_true), but we usually multiply MSE by two in order to remove the first multiplication).
In MSE case you would multiply by the loss value, but it depends entirely on the loss function (it was a bit unfortunate that the tutorial you were using didn't point this out).
2. Derivative of total error w.r.t. w5
You could probably go from here, but... Derivative of total error w.r.t to w5 is the output of h1 (0.85 in this case). We multiply it by derivative of total error w.r.t. output (it is 1!) and obtain 0.85, as done in PyTorch. Same idea goes for w6.
I seriously advise you not to confuse learning rate with backprop, you are making your life harder (and it's not easy with backprop IMO, quite counterintuitive), and those are two separate things (can't stress that one enough).
This source is nice, more step-by-step, with a little more complicated network idea (activations included), so you can get a better grasp if you go through all of it.
Furthermore, if you are really keen (and you seem to be), to know more ins and outs of this, calculate the weight corrections for other optimizers (say, nesterov), so you know why we should keep those ideas separated.
Related
I am creating a basic, and also my first neural network on handwritten digit recognition without any framework (like Tensorflow, PyTorch...) using the Backpropagation algorithm.
My NN has 784 inputs and 10 outputs. So for the last layer, I have to use Softmax.
Because of some memory errors, I have right now my images in shape (300, 784) and my labels in shape (300, 10)
After that I am calculating loss from Categorical Cross-entropy.
Now we are getting to my problem. In Backpropagation, I need manually compute the first derivative of an activation function. I am doing it like this:
dAl = -(np.divide(Y, Al) - np.divide(1 - Y, 1 - Al))
#Y = test labels
#Al - Activation value from my last layer
And after that my Backpropagation can start, so the last layer is softmax.
def SoftmaxDerivative(dA, Z):
#Z is an output from np.dot(A_prev, W) + b
#Where A_prev is an activation value from previous layer
#W is weight and b is bias
#dA is the derivative of an activation function value
x = activation_functions.softmax(dA)
s = x.reshape(-1,1)
dZ = np.diagflat(s) - np.dot(s, s.T)
return dZ
1. Is this function working properly?
In the end, I would like to compute derivatives of weights and biases, So I am using this:
dW = (1/m)*np.dot(dZ, A_prev.T)
#m is A_prev.shape[1] -> 10
db = (1/m)*np.sum(dZ, axis = 1, keepdims = True)
BUT it fails on dW, because dZ.shape is (3000, 3000) (compare to A_prev.shape, which is (300,10))
So from this I assume, that there are only 3 possible outcomes.
My Softmax backward is wrong
dW is wrong
I have some other bug completely somewhere else
Any help would be really appreciated!
I faced the same problem recently. I'm not sure but maybe this question will help you: Softmax derivative in NumPy approaches 0 (implementation)
I have a task in which I input a 500x500x1 image and get out a 500x500x1 binary segmentation. When working, only a small fraction of the 500x500 should be triggered (small "targets"). I'm using a sigmoid activation at the output. Since such a small fraction is desired to be positive, the training tends to stall with all outputs at zero, or very close. I've written my own loss function that partially deals with it, but I'd like to use binary cross entropy with a class weighting if possible.
My question is in two parts:
If I naively apply binary_crossentropy as the loss to my 500x500x1 output, will it apply on a per pixel basis as desired?
Is there a way for keras to apply class weighting with the single sigmoid output per pixel?
To answer your questions.
Yes, binary_cross_entropy will work per-pixel based, provided you feed to your image segmentation neural network pairs of the form (500x500x1 image(grayscale image) + 500x500x1 (corresponding mask to your image).
By feeding the parameter 'class_weight' parameter in model.fit()
Suppose you have 2 classes with 90%-10% distribution. Then you may want to penalise your algorithm 9 times more when it makes a mistake for the less well represented class(the class with 10% in this case). Suppose you have 900 examples of class 1 and 100 examples of class 2.
Then your class weights dictionary(there are multiple ways to compute it, what is important is to assign a greater weight to the less well represented class),
class_weights = {0:1000/900,1:1000/100}
Example : model.fit(X_train, Y_train, epochs = 30, batch_size=32, class_weight=class_weight)
NOTE: This is available only on 2d cases(class_weight). For 3D or higher dimensional spaces, one should use 'sample_weights'. For segmentation purposes, you would rather use sample_weights parameter.
The biggest gain you will have is by means of other loss functions. Other losses, apart from binary_crossentropy and categorical_crossentropy, inherently perform better on unbalanced datasets. Dice Loss is such a loss function.
Keras implementation:
smooth = 1.
def dice_coef(y_true, y_pred):
y_true_f = K.flatten(y_true)
y_pred_f = K.flatten(y_pred)
intersection = K.sum(y_true_f * y_pred_f)
return (2. * intersection + smooth) / (K.sum(y_true_f) + K.sum(y_pred_f) + smooth)
def dice_coef_loss(y_true, y_pred):
return 1 - dice_coef(y_true, y_pred)
You can also use as a loss function the sum of binary_crossentropy
and other losses if it suits you : i.e. loss = dice_loss + bce
I have a 1000 classes in the network and they have multi-label outputs. For each training example, the number of positive output is same(i.e 10) but they can be assigned to any of the 1000 classes. So 10 classes have output 1 and rest 990 have output 0.
For the multi-label classification, I am using 'binary-cross entropy' as cost function and 'sigmoid' as the activation function. When I tried this rule of 0.5 as the cut-off for 1 or 0. All of them were 0. I understand this is a class imbalance problem. From this link, I understand that, I might have to create extra output labels.Unfortunately, I haven't been able to figure out how to incorporate that into a simple neural network in keras.
nclasses = 1000
# if we wanted to maximize an imbalance problem!
#class_weight = {k: len(Y_train)/(nclasses*(Y_train==k).sum()) for k in range(nclasses)}
inp = Input(shape=[X_train.shape[1]])
x = Dense(5000, activation='relu')(inp)
x = Dense(4000, activation='relu')(x)
x = Dense(3000, activation='relu')(x)
x = Dense(2000, activation='relu')(x)
x = Dense(nclasses, activation='sigmoid')(x)
model = Model(inputs=[inp], outputs=[x])
adam=keras.optimizers.adam(lr=0.00001)
model.compile('adam', 'binary_crossentropy')
history = model.fit(
X_train, Y_train, batch_size=32, epochs=50,verbose=0,shuffle=False)
Could anyone help me with the code here and I would also highly appreciate if you could suggest a good 'accuracy' metric for this problem?
Thanks a lot :) :)
I have a similar problem and unfortunately have no answer for most of the questions. Especially the class imbalance problem.
In terms of metric there are several possibilities: In my case I use the top 1/2/3/4/5 results and check if one of them is right. Because in your case you always have the same amount of labels=1 you could take your top 10 results and see how many percent of them are right and average this result over your batch size. I didn't find a possibility to include this algorithm as a keras metric. Instead, I wrote a callback, which calculates the metric on epoch end on my validation data set.
Also, if you predict the top n results on a test dataset, see how many times each class is predicted. The Counter Class is really convenient for this purpose.
Edit: If found a method to include class weights without splitting the output.
You need a numpy 2d array containing weights with shape [number classes to predict, 2 (background and signal)].
Such an array could be calculated with this function:
def calculating_class_weights(y_true):
from sklearn.utils.class_weight import compute_class_weight
number_dim = np.shape(y_true)[1]
weights = np.empty([number_dim, 2])
for i in range(number_dim):
weights[i] = compute_class_weight('balanced', [0.,1.], y_true[:, i])
return weights
The solution is now to build your own binary crossentropy loss function in which you multiply your weights yourself:
def get_weighted_loss(weights):
def weighted_loss(y_true, y_pred):
return K.mean((weights[:,0]**(1-y_true))*(weights[:,1]**(y_true))*K.binary_crossentropy(y_true, y_pred), axis=-1)
return weighted_loss
weights[:,0] is an array with all the background weights and weights[:,1] contains all the signal weights.
All that is left is to include this loss into the compile function:
model.compile(optimizer=Adam(), loss=get_weighted_loss(class_weights))
I have a linear regression model that seems to work. I first load the data into X and the target column into Y, after that I implement the following...
X_train, X_test, Y_train, Y_test = train_test_split(
X_data,
Y_data,
test_size=0.2
)
rng = np.random
n_rows = X_train.shape[0]
X = tf.placeholder("float")
Y = tf.placeholder("float")
W = tf.Variable(rng.randn(), name="weight")
b = tf.Variable(rng.randn(), name="bias")
pred = tf.add(tf.multiply(X, W), b)
cost = tf.reduce_sum(tf.pow(pred-Y, 2)/(2*n_rows))
optimizer = tf.train.GradientDescentOptimizer(FLAGS.learning_rate).minimize(cost)
init = tf.global_variables_initializer()
init_local = tf.local_variables_initializer()
with tf.Session() as sess:
sess.run([init, init_local])
for epoch in range(FLAGS.training_epochs):
avg_cost = 0
for (x, y) in zip(X_train, Y_train):
sess.run(optimizer, feed_dict={X:x, Y:y})
# display logs per epoch step
if (epoch + 1) % FLAGS.display_step == 0:
c = sess.run(
cost,
feed_dict={X:X_train, Y:Y_train}
)
print("Epoch:", '%04d' % (epoch + 1), "cost=", "{:.9f}".format(c))
print("Optimization Finished!")
accuracy, accuracy_op = tf.metrics.accuracy(labels=tf.argmax(Y_test, 0), predictions=tf.argmax(pred, 0))
print(sess.run(accuracy))
I cannot figure out how to print out the model's accuracy. For example, in sklearn, it is simple, if you have a model you just print model.score(X_test, Y_test). But I do not know how to do this in tensorflow or if it is even possible.
I think I'd be able to calculate the Mean Squared Error. Does this help in any way?
EDIT
I tried implementing tf.metrics.accuracy as suggested in the comments but I'm having an issue implementing it. The documentation says it takes 2 arguments, labels and predictions, so I tried the following...
accuracy, accuracy_op = tf.metrics.accuracy(labels=tf.argmax(Y_test, 0), predictions=tf.argmax(pred, 0))
print(sess.run(accuracy))
But this gives me an error...
FailedPreconditionError (see above for traceback): Attempting to use uninitialized value accuracy/count
[[Node: accuracy/count/read = IdentityT=DT_FLOAT, _class=["loc:#accuracy/count"], _device="/job:localhost/replica:0/task:0/device:CPU:0"]]
How exactly does one implement this?
Turns out, since this is a multi-class Linear Regression problem, and not a classification problem, that tf.metrics.accuracy is not the right approach.
Instead of displaying the accuracy of my model in terms of percentage, I instead focused on reducing the Mean Square Error (MSE) instead.
From looking at other examples, tf.metrics.accuracy is never used for Linear Regression, and only classification. Normally tf.metric.mean_squared_error is the right approach.
I implemented two ways of calculating the total MSE of my predictions to my testing data...
pred = tf.add(tf.matmul(X, W), b)
...
...
Y_pred = sess.run(pred, feed_dict={X:X_test})
mse = tf.reduce_mean(tf.square(Y_pred - Y_test))
OR
mse = tf.metrics.mean_squared_error(labels=Y_test, predictions=Y_pred)
They both do the same but obviously the second approach is more concise.
There's a good explanation of how to measure the accuracy of a Linear Regression model here.
I didn't think this was clear at all from the Tensorflow documentation, but you have to declare the accuracy operation, and then initialize all global and local variables, before you run the accuracy calculation:
accuracy, accuracy_op = tf.metrics.accuracy(labels=tf.argmax(Y_test, 0), predictions=tf.argmax(pred, 0))
# ...
init_global = tf.global_variables_initializer
init_local = tf.local_variables_initializer
sess.run([init_global, init_local])
# ...
# run accuracy calculation
I read something on Stack Overflow about the accuracy calculation using local variables, which is why the local variable initializer is necessary.
After reading the complete code you posted, I noticed a couple other things:
In your calculation of pred, you use
pred = tf.add(tf.multiply(X, W), b). tf.multiply performs element-wise multiplication, and will not give you the fully connected layers you need for a neural network (which I am assuming is what you are ultimately working toward, since you're using TensorFlow). To implement fully connected layers, where each layer i (including input and output layers) has ni nodes, you need separate weight and bias matrices for each pair of successive layers. The dimensions of the i-th weight matrix (the weights between the i-th layer and the i+1-th layer) should be (ni, ni + 1), and the i-th bias matrix should have dimensions (ni + 1, 1). Then, going back to the multiplication operation - replace tf.multiply with tf.matmul, and you're good to go. I assume that what you have is probably fine for a single-class linear regression problem, but this is definitely the way you want to go if you plan to solve a multiclass regression problem or implement a deeper network.
Your weight and bias tensors have a shape of (1, 1). You give the variables the initial value of np.random.randn(), which according to the documentation, generates a single floating point number when no arguments are given. The dimensions of your weight and bias tensors need to be supplied as arguments to np.random.randn(). Better yet, you can actually initialize these to random values in Tensorflow: W = tf.Variable(tf.random_normal([dim0, dim1], seed = seed) (I always initialize random variables with a seed value for reproducibility)
Just a note in case you don't know this already, but non-linear activation functions are required for neural networks to be effective. If all your activations are linear, then no matter how many layers you have, it will reduce to a simple linear regression in the end. Many people use relu activation for hidden layers. For the output layer, use softmax activation for multiclass classification problems where the output classes are exclusive (i.e., where only one class can be correct for any given input), and sigmoid activation for multiclass classification problems where the output classes are not exlclusive.
I have trained GAN on celebA dataset. After that i separate G and D. Then i pick one image from celebA training dataset say yTrue and now i want to find the closest image to yTrue that G can generate say yPred. So the loss at output of G is ||yTrue - yPred||_2^{2} and i minimized it w.r.t generator input(latent variable from normal distribution). Below is code that is giving good results. Now the problem is i want to also add prior loss (log(1-D(G(z))) 1 in first line but i am not getting how to do it as D is not connected to G now and if i directly add k.mean(k.log(1-D.predict(G.output))) in first line it returns numpy array not tensor that is not allowed.
`loss = K.mean(K.square(yTrue - gf.output))
grad = K.gradients(loss,[gf.input])[0]
fn = K.function([gf.input], [grad])
generator_input = np.random.normal(0,1,[1,100])
for i in range(5000):
grad1 = fn([generator_input])
generator_input -= grads[0]*.01
recovered = gf.predict(generator_input)`
In keras, you get the final output to create loss functions. Then, you will have to train the full network to achieve that loss. (Train G+D joined as a single model).
In the loss function, you will have y_true and y_pred, and you use them to compare:
PS: if MSE is not taking the output of the discriminator, please detail your questoin better.
import keras.backend as K
def customLoss(yTrue,yPred):
mse = K.mean(K.square(yTrue-yPred)
prior = K.mean(K.log(1-yPred))
return mse + prior
Pass this function when compiling the model
discriminator.compile(loss=customLoss,optimizer=.....)