import numpy as np
%matplotlib widget
import matplotlib.pyplot as plt
from lab_utils_common import plot_data, sigmoid, dlc
plt.style.use('./deeplearning.mplstyle')
Dataset¶
Let's start with the same dataset as was used in the decision boundary lab.
X_train = np.array([[0.5, 1.5], [1,1], [1.5, 0.5], [3, 0.5], [2, 2], [1, 2.5]]) #(m,n)
y_train = np.array([0, 0, 0, 1, 1, 1]) #(m,)
We will use a helper function to plot this data. The data points with label $y=1$ are shown as red crosses, while the data points with label $y=0$ are shown as blue circles.
fig,ax = plt.subplots(1,1,figsize=(4,4))
plot_data(X_train, y_train, ax)
# Set both axes to be from 0-4
ax.axis([0, 4, 0, 3.5])
ax.set_ylabel('$x_1$', fontsize=12)
ax.set_xlabel('$x_0$', fontsize=12)
plt.show()
Canvas(toolbar=Toolbar(toolitems=[('Home', 'Reset original view', 'home', 'home'), ('Back', 'Back to previous …
Cost function¶
In a previous lab, you developed the logistic loss function. Recall, loss is defined to apply to one example. Here you combine the losses to form the cost, which includes all the examples.
Recall that for logistic regression, the cost function is of the form
$$ J(\mathbf{w},b) = \frac{1}{m} \sum_{i=0}^{m-1} \left[ loss(f_{\mathbf{w},b}(\mathbf{x}^{(i)}), y^{(i)}) \right] \tag{1}$$
where
$loss(f_{\mathbf{w},b}(\mathbf{x}^{(i)}), y^{(i)})$ is the cost for a single data point, which is:
$$loss(f_{\mathbf{w},b}(\mathbf{x}^{(i)}), y^{(i)}) = -y^{(i)} \log\left(f_{\mathbf{w},b}\left( \mathbf{x}^{(i)} \right) \right) - \left( 1 - y^{(i)}\right) \log \left( 1 - f_{\mathbf{w},b}\left( \mathbf{x}^{(i)} \right) \right) \tag{2}$$
where m is the number of training examples in the data set and: $$ \begin{align} f_{\mathbf{w},b}(\mathbf{x^{(i)}}) &= g(z^{(i)})\tag{3} \\ z^{(i)} &= \mathbf{w} \cdot \mathbf{x}^{(i)}+ b\tag{4} \\ g(z^{(i)}) &= \frac{1}{1+e^{-z^{(i)}}}\tag{5} \end{align} $$
Code Description¶
The algorithm for compute_cost_logistic loops over all the examples calculating the loss for each example and accumulating the total.
Note that the variables X and y are not scalar values but matrices of shape ($m, n$) and ($𝑚$,) respectively, where $𝑛$ is the number of features and $𝑚$ is the number of training examples.
def compute_cost_logistic(X, y, w, b):
"""
Computes cost
Args:
X (ndarray (m,n)): Data, m examples with n features
y (ndarray (m,)) : target values
w (ndarray (n,)) : model parameters
b (scalar) : model parameter
Returns:
cost (scalar): cost
"""
m = X.shape[0]
cost = 0.0
for i in range(m):
z_i = np.dot(X[i],w) + b
f_wb_i = sigmoid(z_i)
cost += -y[i]*np.log(f_wb_i) - (1-y[i])*np.log(1-f_wb_i)
cost = cost / m
return cost
Check the implementation of the cost function using the cell below.
w_tmp = np.array([1,1])
b_tmp = -3
print(compute_cost_logistic(X_train, y_train, w_tmp, b_tmp))
0.36686678640551745
Expected output: 0.3668667864055175
Example¶
Now, let's see what the cost function output is for a different value of $w$.
In a previous lab, you plotted the decision boundary for $b = -3, w_0 = 1, w_1 = 1$. That is, you had
b = -3, w = np.array([1,1]).Let's say you want to see if $b = -4, w_0 = 1, w_1 = 1$, or
b = -4, w = np.array([1,1])provides a better model.
Let's first plot the decision boundary for these two different $b$ values to see which one fits the data better.
- For $b = -3, w_0 = 1, w_1 = 1$, we'll plot $-3 + x_0+x_1 = 0$ (shown in blue)
- For $b = -4, w_0 = 1, w_1 = 1$, we'll plot $-4 + x_0+x_1 = 0$ (shown in magenta)
import matplotlib.pyplot as plt
# Choose values between 0 and 6
x0 = np.arange(0,6)
# Plot the two decision boundaries
x1 = 3 - x0
x1_other = 4 - x0
fig,ax = plt.subplots(1, 1, figsize=(4,4))
# Plot the decision boundary
ax.plot(x0,x1, c=dlc["dlblue"], label="$b$=-3")
ax.plot(x0,x1_other, c=dlc["dlmagenta"], label="$b$=-4")
ax.axis([0, 4, 0, 4])
# Plot the original data
plot_data(X_train,y_train,ax)
ax.axis([0, 4, 0, 4])
ax.set_ylabel('$x_1$', fontsize=12)
ax.set_xlabel('$x_0$', fontsize=12)
plt.legend(loc="upper right")
plt.title("Decision Boundary")
plt.show()
Canvas(toolbar=Toolbar(toolitems=[('Home', 'Reset original view', 'home', 'home'), ('Back', 'Back to previous …
You can see from this plot that b = -4, w = np.array([1,1]) is a worse model for the training data. Let's see if the cost function implementation reflects this.
w_array1 = np.array([1,1])
b_1 = -3
w_array2 = np.array([1,1])
b_2 = -4
print("Cost for b = -3 : ", compute_cost_logistic(X_train, y_train, w_array1, b_1))
print("Cost for b = -4 : ", compute_cost_logistic(X_train, y_train, w_array2, b_2))
Cost for b = -3 : 0.36686678640551745 Cost for b = -4 : 0.5036808636748461
Expected output
Cost for b = -3 : 0.3668667864055175
Cost for b = -4 : 0.5036808636748461
You can see the cost function behaves as expected and the cost for b = -4, w = np.array([1,1]) is indeed higher than the cost for b = -3, w = np.array([1,1])
Congratulations!¶
In this lab you examined and utilized the cost function for logistic regression.