I am developing a machine learning model to detect bones from a skeleton image.
I am using pytorch, and the model i am using is the hourglass model.
When i use binary_cross_entropy_with_logits i can see the loss decrease, but when i try to test the model, i notice that:
The output is never greater than zero.
The output is just incorrect (the bones are not detected).
This is how i am calling binary_cross_entropy_with_logits
loss = F.binary_cross_entropy_with_logits(ouputs[i], Y, weight=Mask, reduction='sum') / Mask.sum()
This is how i am doing testing
ouput = model(X)
ouput_sig = torch.sigmoid(ouput)
plot_voxel2d(ouput_soft1)
The exact same model, inputs, targets work if i use the mse loss like this:
loss = torch.sum(((ouputs[i] - Y) ** 2) * Mask) / torch.sum(Mask)
I made sure the target is between 0 and 1.
Will appreciate your help.
The following code block is proposed by G. Hinton in his course: http://www.cs.toronto.edu/~tijmen/csc321/slides/lecture_slides_lec6.pdf
optimizer = optim.RMSprop(net.parameters(), lr=0.005, weight_decay=1e-8)
if net.n_classes > 1:
criterion = nn.CrossEntropyLoss()
else:
criterion = nn.BCEWithLogitsLoss()
Then you will need to use sigmoid (Torch Functional: F.sigmoid) in a similar manner as the code example below:
for isample, sample in enumerate(ds):
mask_torch = net2(sample['image'][None, :, :, :].type(torch.cuda.FloatTensor))
mask = (F.sigmoid(mask_torch.type(torch.cuda.FloatTensor)) > 0.4925099).type(torch.FloatTensor)
print(mask)
for ichan in range(3):
ax[isample, ichan].imshow(sample['image'][ichan].cpu())
ax[isample, 3].imshow(sample['mask'][0].cpu())
ax[isample, 4].imshow(mask[0, 0].cpu().detach().numpy())
Place the sigmoid at the end after al the layers. It will look something like this:
For sigmoid, It will look something like this:
def forward(self, x):
#print(x.shape)
x = self.layer_1(x)
x = self.layer_2(x)
x = self.layer_3(x)
logits = F.sigmoid(self.outc(x))
return logits
I've ran into this multiple times, each time reason was labels are not between 0 and 1
Related
i'm trying to define the loss function of a two-class classification problem. However, the target label is not hard label 0,1, but a float number between 0~1.
torch.nn.CrossEntropy in Pytorch do not support soft label so i'm trying to write a cross entropy function by my self.
My function looks like this
def cross_entropy(self, pred, target):
loss = -torch.mean(torch.sum(target.flatten() * torch.log(pred.flatten())))
return loss
def step(self, batch: Any):
x, y = batch
logits = self.forward(x)
loss = self.criterion(logits, y)
preds = logits
# torch.argmax(logits, dim=1)
return loss, preds, y
however it does not work at all.
Can anyone give me a suggestion is there any mistake in my loss function?
It seems like BCELoss and the robust version BCEWithLogitsLoss are working with fuzzy targets "out of the box". They do not expect target to be binary" any number between zero and one is fine.
Please read the doc.
So i am new to deep learning and started learning PyTorch. I created a classifier model with following structure.
class model(nn.Module):
def __init__(self):
super(model, self).__init__()
resnet = models.resnet34(pretrained=True)
layers = list(resnet.children())[:8]
self.features1 = nn.Sequential(*layers[:6])
self.features2 = nn.Sequential(*layers[6:])
self.classifier = nn.Sequential(nn.BatchNorm1d(512), nn.Linear(512, 3))
def forward(self, x):
x = self.features1(x)
x = self.features2(x)
x = F.relu(x)
x = nn.AdaptiveAvgPool2d((1,1))(x)
x = x.view(x.shape[0], -1)
return self.classifier(x)
So basically I wanted to classify among three things {0,1,2}. While evaluating, I passed the image it returned a Tensor with three values like below
(tensor([[-0.1526, 1.3511, -1.0384]], device='cuda:0', grad_fn=<AddmmBackward>)
So my question is what are these three numbers? Are they probability ?
P.S. Please pardon me If I asked something too silly.
The final layer nn.Linear (fully connected layer) of self.classifier of your model produces values, that we can call a scores, for example, it may be: [10.3, -3.5, -12.0], the same you can see in your example as well: [-0.1526, 1.3511, -1.0384] which are not normalized and cannot be interpreted as probabilities.
As you can see it's just a kind of "raw unscaled" network output, in other words these values are not normalized, and it's hard to use them or interpret the results, that's why the common practice is converting them to normalized probability distribution by using softmax after the final layer, as #skinny_func has already described. After that you will get the probabilities in the range of 0 and 1, which is more intuitive representation.
So after training what you would want to do is to apply softmax to the output tensor to extract the probability of each class, then you choose the maximal value (highest probability).
in your case:
prob = torch.nn.functional.softmax(model(x), dim=1)
_, pred_class = torch.max(prob, dim=1)
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])
Sorry for a nub's question:
Having the NN that is trained in fit_generator mode, say something like:
Lambda(...)
or
Dense(...)
and the custom loss function, what are input tensors?
Am I correct expecting (batch size, previous layer's output) in case of a Lambda layer?
Is it going to be the same (batch size, data) in case of a custom loss function that looks like:
triplet_loss(y_true, y_pred)
Are y_true, y_pred in format (batch,previous layer's output) and (batch, true 'expected' data we fed to NN)?
I would probaly duplicate the dense layers. Instead of having 2 layers with 128 units, have 4 layers with 64 units. The result is the same, but you will be able to perform the cross products better.
from keras.models import Model
#create dense layers and store their output tensors, they use the output of models 1 and to as input
d1 = Dense(64, ....)(Model_1.output)
d2 = Dense(64, ....)(Model_1.output)
d3 = Dense(64, ....)(Model_2.output)
d4 = Dense(64, ....)(Model_2.output)
cross1 = Lambda(myFunc, output_shape=....)([d1,d4])
cross2 = Lambda(myFunc, output_shape=....)([d2,d3])
#I don't really know what kind of "merge" you want, so I used concatenate, there are
Add, Multiply and others....
output = Concatenate()([cross1,cross2])
#use the "axis" attribute of the concatenate layer to define better which axis will
be doubled due to the concatenation
model = Model([Model_1.input,Model_2.input], output)
Now, for the lambda function:
import keras.backend as K
def myFunc(x):
return x[0] * x[1]
custom loss function, what are input tensors?
It depends on how you define your model outputs.
For example, let's define a simple model that returns the input unchanged.
model = Sequential([Lambda(lambda x: x, input_shape=(1,))])
Let's use dummy input X and label Y
x = [[0]]
x = np.array(x)
y = [[4]]
y = np.array(y)
If our custom loss function looks like this
def mce(y_true, y_pred):
print(y_true.shape)
print(y_pred.shape)
return K.mean(K.pow(K.abs(y_true - y_pred), 3))
model.compile('sgd', mce)
and then we can see the shape of y_true and y_pred will be
y_true: (?, ?)
y_pred: (?, 1)
However, for triplet loss the input for the loss function also can be received like this-
ALPHA = 0.2
def triplet_loss(x):
anchor, positive, negative = x
pos_dist = tf.reduce_sum(tf.square(tf.subtract(anchor, positive)), 1)
neg_dist = tf.reduce_sum(tf.square(tf.subtract(anchor, negative)), 1)
basic_loss = tf.add(tf.subtract(pos_dist, neg_dist), ALPHA)
loss = tf.reduce_mean(tf.maximum(basic_loss, 0.0), 0)
return loss
# Source: https://github.com/davidsandberg/facenet/blob/master/src/facenet.py
def build_model(input_shape):
# Standardizing the input shape order
K.set_image_dim_ordering('th')
positive_example = Input(shape=input_shape)
negative_example = Input(shape=input_shape)
anchor_example = Input(shape=input_shape)
# Create Common network to share the weights along different examples (+/-/Anchor)
embedding_network = faceRecoModel(input_shape)
positive_embedding = embedding_network(positive_example)
negative_embedding = embedding_network(negative_example)
anchor_embedding = embedding_network(anchor_example)
loss = merge([anchor_embedding, positive_embedding, negative_embedding],
mode=triplet_loss, output_shape=(1,))
model = Model(inputs=[anchor_example, positive_example, negative_example],
outputs=loss)
model.compile(loss='mean_absolute_error', optimizer=Adam())
return model
Given a simple 2 layer neural network, the traditional idea is to compute the gradient w.r.t. the weights/model parameters. For an experiment, I want to compute the gradient of the error w.r.t the input. Are there existing Pytorch methods that can allow me to do this?
More concretely, consider the following neural network:
import torch.nn as nn
import torch.nn.functional as F
class NeuralNet(nn.Module):
def __init__(self, n_features, n_hidden, n_classes, dropout):
super(NeuralNet, self).__init__()
self.fc1 = nn.Linear(n_features, n_hidden)
self.sigmoid = nn.Sigmoid()
self.fc2 = nn.Linear(n_hidden, n_classes)
self.dropout = dropout
def forward(self, x):
x = self.sigmoid(self.fc1(x))
x = F.dropout(x, self.dropout, training=self.training)
x = self.fc2(x)
return F.log_softmax(x, dim=1)
I instantiate the model and an optimizer for the weights as follows:
import torch.optim as optim
model = NeuralNet(n_features=args.n_features,
n_hidden=args.n_hidden,
n_classes=args.n_classes,
dropout=args.dropout)
optimizer_w = optim.SGD(model.parameters(), lr=0.001)
While training, I update the weights as usual. Now, given that I have values for the weights, I should be able to use them to compute the gradient w.r.t. the input. I am unable to figure out how.
def train(epoch):
t = time.time()
model.train()
optimizer.zero_grad()
output = model(features)
loss_train = F.nll_loss(output[idx_train], labels[idx_train])
acc_train = accuracy(output[idx_train], labels[idx_train])
loss_train.backward()
optimizer_w.step()
# grad_features = loss_train.backward() w.r.t to features
# features -= 0.001 * grad_features
for epoch in range(args.epochs):
train(epoch)
It is possible, just set input.requires_grad = True for each input batch you're feeding in, and then after loss.backward() you should see that input.grad holds the expected gradient. In other words, if your input to the model (which you call features in your code) is some M x N x ... tensor, features.grad will be a tensor of the same shape, where each element of grad holds the gradient with respect to the corresponding element of features. In my comments below, I use i as a generalized index - if your parameters has for instance 3 dimensions, replace it with features.grad[i, j, k], etc.
Regarding the error you're getting: PyTorch operations build a tree representing the mathematical operation they are describing, which is then used for differentiation. For instance c = a + b will create a tree where a and b are leaf nodes and c is not a leaf (since it results from other expressions). Your model is the expression, and its inputs as well as parameters are the leaves, whereas all intermediate and final outputs are not leaves. You can think of leaves as "constants" or "parameters" and of all other variables as of functions of those. This message tells you that you can only set requires_grad of leaf variables.
Your problem is that at the first iteration, features is random (or however else you initialize) and is therefore a valid leaf. After your first iteration, features is no longer a leaf, since it becomes an expression calculated based on the previous ones. In pseudocode, you have
f_1 = initial_value # valid leaf
f_2 = f_1 + your_grad_stuff # not a leaf: f_2 is a function of f_1
to deal with that you need to use detach, which breaks the links in the tree, and makes the autograd treat a tensor as if it was constant, no matter how it was created. In particular, no gradient calculations will be backpropagated through detach. So you need something like
features = features.detach() - 0.01 * features.grad
Note: perhaps you need to sprinkle a couple more detaches here and there, which is hard to say without seeing your whole code and knowing the exact purpose.