I would like to code in tf.Keras a Neural Network with a couple of loss functions. One is a standard mse (mean squared error) with a factor loading, while the other is basically a regularization term on the output of a hidden layer. This second loss is added through self.add_loss() in a user-defined class inheriting from tf.keras.layers.Layer. I have a couple of questions (the first is more important though).
1) The error I get when trying to combine the two losses together is the following:
ValueError: Shapes must be equal rank, but are 0 and 1
From merging shape 0 with other shapes. for '{{node AddN}} = AddN[N=2, T=DT_FLOAT](loss/weighted_loss/value, model/new_layer/mul_1)' with input shapes: [], [100].
So it comes from the fact that the tensors which should add up to make one unique loss value have different shapes (and ranks). Still, when I try to print the losses during the training, I clearly see that the vectors returned as losses have shape batch_size and rank 1. Could it be that when the 2 losses are summed I have to provide them (or at least the loss of add_loss) as scalar? I know the mse is usually returned as a vector where each entry is the mse from one sample in the batch, hence having batch_size as shape. I think I tried to do the same with the "regularization" loss. Do you have an explanation for this behavio(u)r?
The sample code which gives me error is the following:
import numpy as np
import tensorflow as tf
from tensorflow.keras import backend as K
from tensorflow.keras.models import Model
from tensorflow.keras.layers import Dense, Input
def rate_mse(rate=1e5):
#tf.function # also needed for printing
def loss(y_true, y_pred):
tmp = rate*K.mean(K.square(y_pred - y_true), axis=-1)
# tf.print('shape %s and rank %s output in mse'%(K.shape(tmp), tf.rank(tmp)))
tf.print('shape and rank output in mse',[K.shape(tmp), tf.rank(tmp)])
tf.print('mse loss:',tmp) # print when I put tf.function
return tmp
return loss
class newLayer(tf.keras.layers.Layer):
def __init__(self, rate=5e-2, **kwargs):
super(newLayer, self).__init__(**kwargs)
self.rate = rate
# #tf.function # to be commented for NN training
def call(self, inputs):
tmp = self.rate*K.mean(inputs*inputs, axis=-1)
tf.print('shape and rank output in regularizer',[K.shape(tmp), tf.rank(tmp)])
tf.print('regularizer loss:',tmp)
self.add_loss(tmp, inputs=True)
return inputs
tot_n = 10000
xx = np.random.rand(tot_n,1)
yy = np.pi*xx
train_size = int(0.9*tot_n)
xx_train = xx[:train_size]; xx_val = xx[train_size:]
yy_train = yy[:train_size]; yy_val = yy[train_size:]
reg_layer = newLayer()
input_layer = Input(shape=(1,)) # input
hidden = Dense(20, activation='relu', input_shape=(2,))(input_layer) # hidden layer
hidden = reg_layer(hidden)
output_layer = Dense(1, activation='linear')(hidden)
model = Model(inputs=[input_layer], outputs=[output_layer])
model.compile(optimizer='Adam', loss=rate_mse(), experimental_run_tf_function=False)
#model.compile(optimizer='Adam', loss=None, experimental_run_tf_function=False)
model.fit(xx_train, yy_train, epochs=100, batch_size = 100,
validation_data=(xx_val,yy_val), verbose=1)
#new_xx = np.random.rand(10,1); new_yy = np.pi*new_xx
#model.evaluate(new_xx,new_yy)
print(model.predict(np.array([[1]])))
2) I would also have a secondary question related to this code. I noticed that printing with tf.print inside the function rate_mse only works with tf.function. Similarly, the call method of newLayer is only taken into consideration if the same decorator is commented during training. Can someone explain why this is the case or reference me to a possible solution?
Thanks in advance to whoever can provide me help. I am currently using Tensorflow 2.2.0 and keras version is 2.3.0-tf.
I stuck with the same problem for a few days. "Standard" loss is going to be a scalar at the moment when we add it to the loss from add_loss. The only way how I get it working is to add one more axis while calculating mean. So we will get a scalar, and it will work.
tmp = self.rate*K.mean(inputs*inputs, axis=[0, -1])
Related
I want to create an MLP based custom CNN model (multi-scaled) consists of several parallel small networks (capsules). These simple small networks are instantiated as a custom layer (conv2d->Flatten->Dense) for each convolution scale i.e. 3x3, 5x5. The purpose of these capsule networks is to generate intermediate loss consciousness to reduce overall global loss using the CNN model. I have written some sketchy codes but I'm not able to write the correct code for computing local loss using these capsules. Here's the code:
from tensorflow.keras import layers
import tensorflow as tf
from tensorflow.keras.datasets import mnist
from tensorflow.keras.layers import Layer
class capsule(tf.keras.layers.Layer):
def __init__(self):
super(capsule, self).__init__()
self.loss_fn = tf.keras.losses.CategoricalCrossentropy(from_logits=True)
self.Flatten = tf.keras.layers.Flatten()
self.conv2D = tf.keras.layers.Conv2D(3,3,(1,1),padding='same', activation='relu',name="LocalLoss3x3")
self.classifier = tf.keras.layers.Dense(10,activation='softmax', name='capsule3Output')
def call(self, inputs):
x=self.conv2D(inputs)
x=self.Flatten(x)
x=self.classifier(x)
pred=self(x_train)
loss=self.loss_fn(pred,y_train)
#self.add_loss(self.rate * tf.reduce_sum(tf.square(inputs)))
return loss, x
(x_train, y_train), (x_test, y_test)= mnist.load_data()
from tensorflow.keras import layers
class SparseMLP(tf.keras.models.Model):
def __init__(self, output_dim):
super(SparseMLP, self).__init__()
self.dense_1 = layers.Dense(1, activation=tf.nn.relu)
self.capsule = capsule()
self.dense_2 = layers.Dense(output_dim)
def call(self, inputs):
x = self.dense_1(inputs)
loss,x = self.capsule(inputs)
return self.dense_2(x)
mlp = SparseMLP(10)
#x_train=x_train.reshape(-1,28,28,1)
y = mlp(x_train)
To include a loss within a layer , you can use add_loss function of tf.keras.layers.Layer class. This fucntion takes a loss value and adds it up to the global loss function define in compile function.
you can call self.add_loss(loss_value) from inside the call method of a custom
layer.Losses added in this way get added to the "main" loss during training
(the one passed to compile()).
So to make ur model consider the losses from intermediate layer , you should uncomment the add_loss fn , and then train the model in usual way that you train.
Please mind that it is totally fine to not declare a "main" loss in the compile function as there already is a loss that ur defining in your layer class.
Note that when you pass losses via add_loss(), it becomes possible to call compile() without a loss function, since the model already has a loss to minimize.
Please note that call function of SparseMLP model , should look like this:
x = self.dense_1(inputs)
# i dunno if u desire to do this, that is pass inputs in capsule
# instead of x.Currently the output from dense_1 is not used at all .
# so keep in mind to make sure ur passing proper inputs to layers.
# and u do not have to call loss here as it will tracked internally by
# keras.
x = self.capsule(inputs)
return self.dense_2(x)
So running your model like below should do the trick:
model.compile(loss = "define ur main loss is there is" , metrics = "define ur metrics")
model.fit(x = train_inst , y = train_targets)
I am using the input gradient as feature important and want to compare the feature importance of a train datapoint with the human annotated feature importance. I would like to make this comparison differentiable such that it can be learned through backpropagation. For that, I am writing a custom loss function that in addition to the regular loss (e.g. m.s.e. on the prediction vs true labels) also checks whether the input gradient is correct (e.g. m.s.e. of the input gradient vs the human annotated feature importance).
With the following code I am able to get the input gradient:
from keras import backend as K
import numpy as np
from keras.models import Model
from keras.layers import Input, Dense
def normalize(x):
# utility function to normalize a tensor by its L2 norm
return x / (K.sqrt(K.mean(K.square(x))) + 1e-5)
# Amount of training samples
N = 1000
input_dim = 10
# Generate training set make the 1st and 2nd feature same as the target feature
X = np.random.standard_normal(size=(N, input_dim))
y = np.random.randint(low=0, high=2, size=(N, 1))
X[:, 1] = y[:, 0]
X[:, 2] = y[:, 0]
# Create simple model
inputs = Input(shape=(input_dim,))
x = Dense(10, name="dense1")(inputs)
output = Dense(1, activation='sigmoid')(x)
model = Model(input=[inputs], output=output)
# Compile and fit model
model.compile(optimizer='adam', loss="mse", metrics=['accuracy'])
model.fit([X], y, epochs=100, batch_size=64)
# Get function to get input gradients
gradients = K.gradients(model.output, model.input)[0]
gradient_function = K.function([model.input], [normalize(gradients)])
# Get input gradient values of the training-set
grads_val = gradient_function([X])[0]
print(grads_val[:2])
This prints the following (you can see that the 1st and the 2nd features have the highest importance):
[[ 1.2629046e-02 2.2765596e+00 2.1479919e+00 2.1558853e-02
4.5277486e-03 2.9851785e-03 9.5279224e-04 -1.0903150e-02
-1.2230731e-02 2.1960819e-02]
[ 1.1318034e-02 2.0402350e+00 1.9250139e+00 1.9320872e-02
4.0577268e-03 2.6752844e-03 8.5390132e-04 -9.7713526e-03
-1.0961102e-02 1.9681118e-02]]
How can I write a custom loss function in which the input gradients are differentiable?
I started with the following loss function.
from keras.losses import mean_squared_error
def custom_loss():
# human annotated feature importance
# Let's say that it says to only look at the second feature
human_feature_importance = []
for i in range(N):
human_feature_importance.append([0,0,1,0,0,0,0,0,0,0])
def loss(y_true, y_pred):
# Get regular loss
regular_loss_value = mean_squared_error(y_true, y_pred)
# Somehow get the input gradient of each training sample as a tensor
# It should be differential w.r.t. all of the weights
gradients = ??
feature_importance_loss_value = mean_squared_error(gradients, human_feature_importance)
# Combine the both losses
return regular_loss_value + feature_importance_loss_value
return loss
I also found an implementation in tensorflow to make the input gradient differentialble: https://github.com/dtak/rrr/blob/master/rrr/tensorflow_perceptron.py#L18
I want to build an autoencoder where each layer in the encoder has the same meaning as a correspondent layer in the decoder. So if the autoencoder is perfectly trained, the values of those layers should be roughly the same.
So lets say the autoencoder consists of e1 -> e2 -> e3 -> d2 -> d1, whereas e1 is the input and d1 the output. A normal autoencoder trains to have the same result in d1 as e1, but I want the additional constraint, that e2 and d2 are the same. Therefore I want an additional backpropagation path which leads from d2 to e2 and trains at the same time as the normal path from d1 to e1. (d stands for decoder, e for encoder).
I tried to use the error between e2 and d2 as a regularization term with the CustomRegularization layer from the first answer of this link https://github.com/keras-team/keras/issues/5563. I also use this for the error between e1 and d1 instead of the normal path.
The following code is written such that more than 1 intermediate layer can be handled and also uses 4 layers.
In the out commented code is a normal autoencoder which only propagates from start to end.
from keras.layers import Dense
import numpy as np
from keras.datasets import mnist
from keras.models import Model
from keras.engine.topology import Layer
from keras import objectives
from keras.layers import Input
import keras
import matplotlib.pyplot as plt
#A layer which can be given as an output to force a regularization term between two layers
class CustomRegularization(Layer):
def __init__(self, **kwargs):
super(CustomRegularization, self).__init__(**kwargs)
def call(self, x, mask=None):
ld=x[0]
rd=x[1]
bce = objectives.binary_crossentropy(ld, rd)
loss2 = keras.backend.sum(bce)
self.add_loss(loss2, x)
return bce
def get_output_shape_for(self, input_shape):
return (input_shape[0][0],1)
def zero_loss(y_true, y_pred):
return keras.backend.zeros_like(y_pred)
#Create regularization layer between two corresponding layers of encoder and decoder
def buildUpDownRegularization(layerNo, input, up_layers, down_layers):
for i in range(0, layerNo):
input = up_layers[i](input)
start = input
for i in range(layerNo, len(up_layers)):
input = up_layers[i](input)
for j in range(0, len(down_layers) - layerNo):
input = down_layers[j](input)
end = input
cr = CustomRegularization()([start, end])
return cr
# Define shape of the network, layers, some hyperparameters and training data
sizes = [784, 400, 200, 100, 50]
up_layers = []
down_layers = []
for i in range(1, len(sizes)):
layer = Dense(units=sizes[i], activation='sigmoid', input_dim=sizes[i-1])
up_layers.append(layer)
for i in range(len(sizes)-2, -1, -1):
layer = Dense(units=sizes[i], activation='sigmoid', input_dim=sizes[i+1])
down_layers.append(layer)
batch_size = 128
num_classes = 10
epochs = 100
(x_train, y_train), (x_test, y_test) = mnist.load_data()
x_train = x_train.astype('float32')
x_test = x_test.astype('float32')
x_train /= 255
x_test /= 255
x_train = x_train.reshape([x_train.shape[0], 28*28])
x_test = x_test.reshape([x_test.shape[0], 28*28])
y_train = x_train
y_test = x_test
optimizer = keras.optimizers.Adam(lr=0.001, beta_1=0.9, beta_2=0.999, epsilon=None, decay=0.0, amsgrad=False)
"""
### Normal autoencoder like in base mnist example
model = keras.models.Sequential()
for layer in up_layers:
model.add(layer)
for layer in down_layers:
model.add(layer)
model.compile(optimizer=optimizer, loss=keras.backend.binary_crossentropy)
model.fit(x_train, y_train, batch_size=batch_size, epochs=epochs)
score = model.evaluate(x_test, y_test, verbose=0)
#print('Test loss:', score[0])
#print('Test accuracy:', score[1])
decoded_imgs = model.predict(x_test)
n = 10 # how many digits we will display
plt.figure(figsize=(20, 4))
for i in range(n):
# display original
ax = plt.subplot(2, n, i + 1)
plt.imshow(x_test[i].reshape(28, 28))
plt.gray()
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
# display reconstruction
ax = plt.subplot(2, n, i + 1 + n)
plt.imshow(decoded_imgs[i].reshape(28, 28))
plt.gray()
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
plt.show()
"""
### My autoencoder where each subpart is also an autoencoder
#This part is only because the model needs a path from start to end, contentwise this should do nothing
output = input = Input(shape=(sizes[0],))
for i in range(0, len(up_layers)):
output = up_layers[i](output)
for i in range(0, len(down_layers)):
output = down_layers[i](output)
crs = [output]
losses = [zero_loss]
#Build the regularization layer
for i in range(len(up_layers)):
crs.append(buildUpDownRegularization(i, input, up_layers, down_layers))
losses.append(zero_loss)
#Create and train model with adapted training data
network = Model([input], crs)
optimizer = keras.optimizers.Adam(lr=0.0001, beta_1=0.9, beta_2=0.999, epsilon=None, decay=0.0, amsgrad=False)
network.compile(loss=losses, optimizer=optimizer)
dummy_train = np.zeros([y_train.shape[0], 1])
dummy_test = np.zeros([y_test.shape[0], 1])
training_data = [y_train]
test_data = [y_test]
for i in range(len(network.outputs)-1):
training_data.append(dummy_train)
test_data.append(dummy_test)
network.fit(x_train, training_data, batch_size=batch_size, epochs=epochs,verbose=1, validation_data=(x_test, test_data))
score = network.evaluate(x_test, test_data, verbose=0)
print('Test loss:', score[0])
print('Test accuracy:', score[1])
decoded_imgs = network.predict(x_test)
n = 10 # how many digits we will display
plt.figure(figsize=(20, 4))
for i in range(n):
# display original
ax = plt.subplot(2, n, i + 1)
plt.imshow(x_test[i].reshape(28, 28))
plt.gray()
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
# display reconstruction
ax = plt.subplot(2, n, i + 1 + n)
plt.imshow(decoded_imgs[0][i].reshape(28, 28))
plt.gray()
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
plt.show()
If you run the code as is it will show, that the reproduction ability is no longer given in my code.
I expect a similar behavior to the uncommented code, which shows a normal autoencoder.
Edit: As mentioned in the answers this works well with MSE instead of crossentropy and a lr of .01. 100 epochs with that setting produce really good results.
Edit 2: I would like that the backpropagation works as in this [image] (https://imgur.com/OOo757x). So the backpropagation of the loss of a certain layer stops at the corresponding layer. I think I didn't make this clear before and I don't know if the code currently does that.
Edit 3: Although this code runs and returns a good looking solution the CustomRegularization layer is not doing what I thought it would do, therefore it does not do the same things as in the description.
It seems like the main issue is the use of binary cross-entropy to minimize the difference between encoder and decoder. The internal representation in the network is not going to be a single class probability like the output might be if you were classifying MNIST digits. I was able to get your network to output some reasonable-looking reconstructions with these simple changes:
Using objectives.mean_squared_error instead of objectives.binary_crossentropy in the CustomRegularization class
Changing number of epochs to 5
Changing learning rate to .01
Changes 2 and 3 were simply made to speed up the testing. Change 1 is the key here. Cross entropy is designed for problems where there is a binary "ground truth" variable and an estimate of that variable. However, you do not have a binary truth value in the middle of your network, only at the output layer. Thus a cross entropy loss function in the middle of the network doesn't make much sense (at least to me) -- it will be trying to measure entropy for a variable that isn't binary. Mean squared error, on the other hand, is a bit more generic and should work for this case since you are simply minimizing the difference between two real values. In essence, the middle of the network is performing regression (difference between activations in two continuous values, i.e. layers), not classification, so it needs a loss function that is appropriate for regression.
I also want to suggest that there may be a better approach to accomplish what you want. If you really want the encoder and decoder to be exactly the same, you can share weights between them. Then they will be identical, not just highly similar, and your model will have fewer parameters to train. There is a decent explanation of shared (tied) weights autoencoders with Keras here if you're curious.
Reading your code it does seem like it is doing what you want in your illustration, but I am not really sure how to verify that.
I am trying to set up an image classifier using Pytorch. My sample images have 4 channels and are 28x28 pixels in size. I am trying to use the built-in torchvision.models.inception_v3() as my model. Whenever I try to run my code, I get this error:
RuntimeError: Calculated padded input size per channel: (1 x 1).
Kernel size: (3 x 3). Kernel size can't greater than actual input size
at
/opt/conda/conda-bld/pytorch_1524584710464/work/aten/src/THNN/generic/SpatialConvolutionMM.c:48
I can't find how to change the padded input size per channel or quite figure out what the error means. I figure that I must modify the padded input size per channel since I can't edit the Kernel size in the pre-made model.
I have tried padding, but it didn't help.
Here is a shortened part of my code that throws the error when I call train():
import torch
import torchvision as tv
import torch.optim as optim
from torch import nn
from torch.utils.data import DataLoader
model = tv.models.inception_v3()
criterion = nn.CrossEntropyLoss()
optimizer = optim.Adam(model.parameters(), lr=0.0001, weight_decay=0)
lr_scheduler = optim.lr_scheduler.StepLR(optimizer, step_size=4, gamma=0.9)
trn_dataset = tv.datasets.ImageFolder(
"D:/tests/classification_test_data/trn",
transform=tv.transforms.Compose([tv.transforms.RandomRotation((0,275)), tv.transforms.RandomHorizontalFlip(),
tv.transforms.ToTensor()]))
trn_dataloader = DataLoader(trn_dataset, batch_size=32, num_workers=4, shuffle=True)
for epoch in range(0, 10):
train(trn_dataloader, model, criterion, optimizer, lr_scheduler, 6, 32)
print("End of training")
def train(train_loader, model, criterion, optimizer, scheduler, num_classes, batch_size):
model.train()
scheduler.step()
for index, data in enumerate(train_loader):
inputs, labels = data
optimizer.zero_grad()
outputs = model(inputs)
outputs_flatten = flatten_outputs(outputs, num_classes)
loss = criterion(outputs_flatten, labels)
loss.backward()
optimizer.step()
def flatten_outputs(predictions, number_of_classes):
logits_permuted = predictions.permute(0, 2, 3, 1)
logits_permuted_cont = logits_permuted.contiguous()
outputs_flatten = logits_permuted_cont.view(-1, number_of_classes)
return outputs_flatten
It could be due the following. Pytorch documentation for Inception_v3 model notes that the model expects input of shape Nx3x299x299. This is because the architecture contains a fully connected layer which fixed shape.
Important: In contrast to the other models the inception_v3 expects tensors with a size of N x 3 x 299 x 299, so ensure your images are sized accordingly.
https://pytorch.org/docs/stable/torchvision/models.html#inception-v3
May be this is a late post, but i tried to sort out this with a simple technique.
In got this kind of error, i was using custom conv2d module, and somehow i missed sending the padding to my nn.conv2d.
I found out this error by,
In my conv2d implementation, i printed out the shape of the output variable, and found the exact bug in my code.
model = VGG_BNN_ReLU('VGG11',10)
import torch
x = torch.randn(1,3,32,32)
model.forward(x)
Hope this helps.Happy learning
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))