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<!DOCTYPE html>
<html>
<head>
<title>Classification in Deep Learning [Marc Lelarge]</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/>
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<textarea id="source">
class: center, middle, title-slide
count: false
# Module 3:
## Loss functions for classification in Deep Learning
<br/><br/>
.bold[Marc Lelarge]
---
# Overview of the course:
1- .grey[Course overview: machine learning pipeline]
2- .grey[PyTorch tensors and automatic differentiation]
3- Classification with deep learning
* .red[Loss] and .red[Optimizer] - in PyTorch: [`torch.optim`](https://pytorch.org/docs/stable/optim.html) and [Loss functions](https://pytorch.org/docs/stable/nn.html#loss-functions) in `torch.nn`
---
# Deep Learning pipeline
## Dataset and Dataloader + Model + .red[Loss and Optimizer] = .grey[Training]
.center[
<img src="images/lesson1/ml4.png" style="width: 1000px;" />
]
---
# Supervised learning basics
## Linear regression
## Gradient descent algorithms
## Logistic regression
## Classification and softmax regression
---
## Linear regression
### a probabilistic model
The .bold[dataset] is made of $m$ .bold[training examples] $(x(i), y(i))\_{i\in [m]}$, where $x(i) \in \mathbb{R}^d$ are the .bold[features] and $y(i)\in \mathbb{R}$ are the .bold[target] variables.
--
count: false
Assumption, there exists $\theta\in \mathbb{R}^d$ such that:
$$
y(i) = \theta^T x(i) +\epsilon(i),
$$
with $\epsilon(i)$ i.i.d. Gaussian random variables (mean zero and variance $\sigma^2$).
--
count: false
Computing the .bold[likelihood] function gives:
$$\begin{aligned}
L(\theta) &= \prod\_{i=1}^m p\_\theta(y(i) | x(i))\\\\
& = \prod\_{i=1}^m \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y(i)-\theta^T x(i))^2}{2\sigma^2}\right)
\end{aligned}
$$
---
## Linear regression
Maximizing the .bold[log likelihood]:
$$\begin{aligned}
\ell(\theta) &= \log L(\theta)\\\\
& = -m\log(\sigma\sqrt{2\pi}) -\frac{1}{2\sigma^2}\sum\_{i=1}^m(y(i)-\theta^T x(i))^2
\end{aligned}$$
is the same as minimizing the .bold[cost function]:
$$
J(\theta) = \frac{1}{2}\sum\_{i=1}^m(y(i)-\theta^T x(i))^2,
$$
giving rise to the .bold[ordinary least squares] regression model.
--
count: false
The gradient of the least-squares cost function is:
$$
\frac{\partial}{\partial \theta\_j}J(\theta) = \sum\_{i=0}^m (y(i)-\theta^T x(i))\frac{\partial}{\partial \theta\_j}\left( y(i)-\sum\_{k=0}^d\theta\_k x\_k(i) \right) = \sum\_{i=0}^m (y(i)-\theta^T x(i))x\_j(i)
$$
---
## Gradient descent algorithms
.bold[Batch gradient descent] perfoms the update:
$$
\theta\_j := \theta\_j +\alpha \sum\_{i=0}^m (y(i)-\theta^T x(i))x\_j(i),
$$
where $\alpha$ is the .bold[learning rate].
This method looks at every example in the entire training set on every step.
--
count: false
.bold[Stochastic gradient descent] works very well. The sum above is 'replaced' by a loop over the training examples, so that the update becomes:
for $i=1$ to $m$:
$$
\theta\_j := \theta\_j +\alpha (y(i)-\theta^T x(i))x\_j(i).
$$
---
## Linear regression
Recall that under mild assumptions, the explicit solution for the ordinary least squares can be written explicitely as:
$$
\theta^* = (X^TX)^{-1}X^T Y,
$$
where the linear model is written in matrix form $Y = X\theta +\epsilon$, with $Y=(y(1), \dots y(m))\in \mathbb{R}^m$ and $X = (x(1),\dots x(m))\in \mathbb{R}^{m\times d}$.
Exercise: show the above formula is valid as soon as $\text{rk}(X) = d$.
--
count: false
In other words, (if you are able to invert $X^TX$) __optimization is useless for linear regression!__
Nevertheless, gradient descent algorithms are important optimization algorithms with theoretical guarantees when the cost function is .bold[convex].
We used linear regression mainly as an example to understand PyTorch framework.
---
## PyTorch implementation for batch gradient descent
```py
import torch
model = torch.nn.Sequential(torch.nn.Linear(2, 1))
loss_fn = torch.nn.MSELoss(reduction='sum')
model.train()
optimizer = torch.optim.SGD(model.parameters(), lr=learning_rate)
for epoch in range(n_epochs):
y_pred = model(x)
loss = loss_fn(y_pred, y)
optimizer.zero_grad()
loss.backward()
optimizer.step()
```
code taken from the [notebook](https://github.com/dataflowr/notebooks/blob/master/Module2/02b_linear_reg.ipynb) of Module 2b.
--
count: false
But the important piece of code is (assuming x and y are numpy array):
```py
*dtype = torch.FloatTensor
*x = torch.from_numpy(x).type(dtype)
*y = torch.from_numpy(y).type(dtype).unsqueeze(1)
```
The default type for weights and biases are torch.FloatTensor. So, you'll need to cast your inputs to torch.FloatTensor. Moreover, the shape of the target y needs to be $[\text{batch size},1]$.
---
## Logistic regression
### a probabilistic model
A natural (generalized) linear model for .bold[binary classification]:
$$
\begin{aligned}
p\_\theta(y=1|x) &= \sigma(\theta^T x),\\\\
p\_\theta(y=0|x) &= 1-\sigma(\theta^T x),
\end{aligned}
$$
where $\sigma(z) = \frac{1}{1+e^{-z}}$ is the .bold[sigmoid function] (or .bold[logistic function]).
.center[
<img src="images/module3/Logistic-curve.png" style="width: 400px;" />
]
---
## A pause on notations
We (mainly) use the compact notations used in .bold[information theory]: the function $p(.)$ is defined by its argument so that $p(x)$ is the distribution of the r.v. $X$ and $p(y)$ the distribution of the r.v. $Y$:
$$
\begin{aligned}
p(x) &= P(X=x)\\\\
p(y) &= P(Y=y)\\\\
p(x,y) &= P(X=x, Y=y)\\\\
p(y|x) &= P(Y=y|X=x)\\\\
&= \frac{p(x,y)}{p(x)} = \frac{p(x,y)}{\sum\_y p(x,y)}
\end{aligned}
$$
--
count: false
In particular, we can write our model more compactly as:
$$
p\_\theta(y|x) = \sigma(\theta^T x)^y(1-\sigma(\theta^T x))^{(1-y)}.
$$
---
## Logistic regression
The likelihood function is now:
$$
L(\theta) = \prod\_{i=1}^m \sigma(\theta^T x(i))^{y(i)}\left( 1-\sigma(\theta^T x(i))\right)^{1-y(i)}
$$
--
count: false
So that the log likelihood is:
$$\ell(\theta) = \sum\_{i=1}^m y(i)\log \sigma(\theta^T x(i))+(1-y(i))\log \left( 1-\sigma(\theta^T x(i))\right)
$$
There is no closed form fomula for $\arg\max \ell(\theta)$ so that we need now to use iterative algorithms.
--
count:false
The gradient of the log likelihood is:
$$
\frac{\partial}{\partial \theta\_j}\ell(\theta) = \sum\_{i=1}^m (y(i)-\sigma(\theta^T x(i))x\_j(i),
$$
where we used the fact that $\sigma'(z) = \sigma(z)(1-\sigma(z))$.
---
## Logistic regression
### Binary cross entropy loss
torch.nn.BCELoss computes Binary Cross Entropy between the targets $y=(y(1),\dots, y(m))\in \\{0,1\\}^m$ and the output $z=(z(1),\dots, z(m))\in [0,1]^m$ as follows:
$$\begin{aligned}
\text{loss}(i) &= -[y(i)\log z(i) +(1-y(i))\log(1-z(i))]\\\\
\text{BCELoss}(z,y) &= \frac{1}{m}\sum\_{i=1}^m \text{loss}(i).
\end{aligned}
$$
In summary, we get
$$
\text{BCELoss}(\sigma(X^T \theta),y) = - \frac{1}{m} \ell(\theta),
$$
where the sigmoid function is applied componentwise.
---
## Logistic regression
### Binary cross entropy with logits loss
torch.nn.BCEWithLogitsLoss combines a Sigmoid layer and the BCELoss in one single class.
$$\begin{aligned}
\text{loss}(i) &= -[y(i)\log \sigma(z(i)) +(1-y(i))\log(1-\sigma(z(i)))]\\\\
\text{BCEWithLogitsLoss}(z,y) &= \frac{1}{m}\sum\_{i=1}^m \text{loss}(i).
\end{aligned}
$$
$$
\text{BCEWithLogitsLoss}(z,y) = \text{BCELoss}(\sigma(z),y).
$$
--
count: false
This version is more numerically stable.
Note the default $1/m$ factor, where $m$ is typically the size of the batch. This factor will be directly multiplied with the learning rate. Recall the batch gradient descent update:
$$
\theta\_j := \theta\_j +\alpha \frac{\partial}{\partial \theta\_j}\text{loss}(\theta).
$$
---
## Softmax regression
### a probabilistic model
We now have $c$ classes and for a training example $(x,y)$, the quantity $\theta\_k^T x$ should be related to the probability for the target $y$ to belong to class $k$.
By analogy with the binary case, we assume:
$$
\log p\_\theta(y = k| x) \propto \theta\_k^T x, \quad \text{for all }k=1,\dots, c.
$$
As a consequence, we have with $\theta=(\theta\_1,\dots, \theta\_c)\in\mathbb{R}^{d\times c}$:
$$
p\_\theta(y = k| x) = \frac{e^{\theta\_k^T x}}{\sum\_l e^{\theta\_l^T x}}
$$
and we can write it in vector form:
$$
(p\_\theta(y = k| x))\_{k=1,\dots,c } =\text{softmax}(\theta\_1^T x,\dots,\theta\_c^T x)
$$
---
## Softmax regression
For the logistic regression, we had only one parameter $\theta$ whereas here, for two classes we have two parameters: $\theta_1$ and $\theta_2$.
For 2 classes, we recover the logistic regression:
$$
\begin{aligned}
p\_\theta(y=1|x) &= \frac{e^{\theta\_1^T x}}{e^{\theta\_1^T x}+e^{\theta\_0^T x}}\\\\
&= \frac{1}{1+e^{(\theta\_0-\theta\_1)^T x}}\\\\
&= \sigma\left( (\underbrace{\theta\_1-\theta\_0}_{\theta})^T x\right)
\end{aligned}
$$
---
## Classification and softmax regression
For the .bold[softmax regression], the log-likelihood ca be written as:
$$\begin{aligned}
\ell(\theta) &= \sum\_{i=1}^m \sum\_{k=1}^c 1(y(i) =k) \log \left( \frac{e^{\theta\_k^T x(i)}}{\sum\_l e^{\theta\_l^T x(i)}}\right)\\\\
& = \sum\_{i=1}^m \log \text{softmax}\_{y(i)}(\theta\_1^T x(i),\dots,\theta\_c^T x(i))
\end{aligned}
$$
In PyTorch, if the last layer of your network is a `LogSoftmax()` function, then you can do a softmax regression with the `torch.nn.NLLLoss()`.
--
count: false
Most of the time, we are not interested in softmax regression but in classification, i.e. we need to output a class not a probability vector on the classes.
For classification of the feature $x$, we output the most probable class:
$$
\arg\max \\{ \theta\_1^T x,\dots, \theta\_c^T x\\}
$$
---
## Back to dogs and cats!
In the [dogs and cats example](https://github.com/mlelarge/dataflowr/blob/master/PlutonAI/01_intro_PlutonAI_colab.ipynb), we were interested in classification (and not logistic regression!). Hence we used the following modification of the VGG classifier :
```py
model_vgg.classifier._modules['6'] = nn.Linear(4096, 2)
model_vgg.classifier._modules['7'] = torch.nn.LogSoftmax(dim = 1)
```
so that we get an output of dimension 2 and we then used torch.nn.NLLLoss()
As shown above, this approach extends nicley to more than 2 classes.
---
## Classification in PyTorch
Now you should be able to understand the code of the first lesson!
```py
criterion = nn.NLLLoss()
lr = 0.001
optimizer_vgg = torch.optim.SGD(model_vgg.classifier[6].parameters(),lr = lr)
def train_model(model,dataloader,size,epochs=1,optimizer=None):
model.train()
for epoch in range(epochs):
running_loss = 0.0
running_corrects = 0
for inputs,classes in dataloader:
inputs = inputs.to(device)
classes = classes.to(device)
outputs = model(inputs)
loss = criterion(outputs,classes)
optimizer.zero_grad()
loss.backward()
optimizer.step()
_,preds = torch.max(outputs.data,1)
# statistics
running_loss += loss.data.item()
running_corrects += torch.sum(preds == classes.data)
epoch_loss = running_loss / size
epoch_acc = running_corrects.data.item() / size
print('Loss: {:.4f} Acc: {:.4f}'.format(
epoch_loss, epoch_acc))
```
---
## Why not directly maximizing the accuracy?
Going back to the binary classification task, the binary loss is:
$$\begin{aligned}
\text{Binary loss}(\theta) = \sum\_{j=1}^m 1(y(j) =\not 1(\theta^T x(j)\geq 0)).
\end{aligned}
$$
The binary cross entropy loss was a convex function of $\theta$ (proof left as exercise!) but here, we replace it by a discrete optimization problem.
--
count: false
Note that the loss is 'flat' so that the gradient is zero and the optimizer step will not update weights.
.bold[We use softmax regression as a surrogate for our classification task.]
---
# Classification in deep learning
In practice, we can use previous analysis to treat the last layer of the deep neural network, so that we now have:
$$
\log p\_\theta(y=1| x),\dots, \log p\_\theta(y=c| x) \propto f\_\theta(x)\_1,\dots , f\_\theta(x)\_c,
$$
where the vector $(f\_\theta(x)\_1,\dots, f\_\theta(x)\_c)$ is the output of the last layer on the input $x$ and $\theta$ takes now into account all the weights in the different layers.
In order to solve a classification task, we add a `LogSoftmax()` layer and then train our network with the `torch.nn.NLLLoss()` loss.
--
count: false
# Let see what can go wrong [here](https://github.com/dataflowr/notebooks/blob/master/Module3/03_polynomial_regression.ipynb)
---
class: end-slide, center
count: false
The end.
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