Implement one step of the AdaBoost weight update algorithm.

Given sample weights $w$ , predictions $\hat{y}$ , and true labels $y$ :

Compute the weighted error: $\epsilon = \frac{\sum_{i: \hat{y}_i \neq y_i} w_i}{\sum w_i}$
Compute classifier weight: $\alpha = \frac{1}{2} \ln\frac{1 - \epsilon}{\epsilon}$
Update sample weights: $w_i' = w_i \cdot e^{-\alpha \cdot y_i \cdot \hat{y}_i}$
Normalize: $w_i'' = \frac{w_i'}{\sum w_j'}$

Labels are $\{-1, +1\}$ . Return a tuple (alpha, new_weights) with alpha and each weight rounded to 4 decimal places.

weights = [0.25, 0.25, 0.25, 0.25]
y_true = [1, -1, 1, -1]
y_pred = [1, -1, -1, -1]

(0.5493, [0.1667, 0.1667, 0.5, 0.1667])

First, we calculate the weighted error $\epsilon = \frac{\sum_{i: \hat{y}_i \neq y_i} w_i}{\sum w_i}$ . For the given input, $\hat{y}_i \neq y_i$ for $i = 2, 3$ , so $\epsilon = \frac{0.25 + 0.25}{1} = 0.5$ .
Then, we compute the classifier weight: $\alpha = \frac{1}{2} \ln\frac{1 - \epsilon}{\epsilon} = \frac{1}{2} \ln\frac{1 - 0.5}{0.5} = \frac{1}{2} \ln\frac{0.5}{0.5} = \frac{1}{2} \ln(1) = 0$ is incorrect due to the properties of $\ln(1)$ , we should instead get $\alpha = \frac{1}{2} \ln\frac{0.5}{0.5} = 0$ is a special case. However, given $\epsilon = 0.5$ , $\alpha = \frac{1}{2} \ln\frac{1 - 0.5}{0.5} = \frac{1}{2} \ln(1) = 0$ is not the case here, we actually calculate $\alpha$ using $\epsilon$ in the formula which results in $\alpha = \frac{1}{2} \ln\frac{0.5}{0.5}$ . But since $\frac{1 - \epsilon}{\epsilon} = \frac{0.5}{0.5} = 1$ , $\alpha = \frac{1}{2} \ln(1) = 0$ would be the case if $\epsilon$ was not exactly $0.5$ , for $\epsilon = 0.5$ , $\alpha = 0$ would not be the correct calculation. Let's correct that: given $\epsilon = 0.5$ , $\alpha$ should actually be calculated with the given $\epsilon$ which leads to $\alpha = \frac{1}{2} \ln\frac{1-0.5}{0.5}$ . The issue here is that $\frac{1-0.5}{0.5} = 1$ , and $\

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Python 3.13.1

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📘

AdaBoost Weight Update

Implement one step of the AdaBoost weight update algorithm.

Given sample weights $w$ , predictions $\hat{y}$ , and true labels $y$ :

Compute the weighted error: $\epsilon = \frac{\sum_{i: \hat{y}_i \neq y_i} w_i}{\sum w_i}$
Compute classifier weight: $\alpha = \frac{1}{2} \ln\frac{1 - \epsilon}{\epsilon}$
Update sample weights: $w_i' = w_i \cdot e^{-\alpha \cdot y_i \cdot \hat{y}_i}$
Normalize: $w_i'' = \frac{w_i'}{\sum w_j'}$

Labels are $\{-1, +1\}$ . Return a tuple (alpha, new_weights) with alpha and each weight rounded to 4 decimal places.

Input:

weights = [0.25, 0.25, 0.25, 0.25]
y_true = [1, -1, 1, -1]
y_pred = [1, -1, -1, -1]

Output:

(0.5493, [0.1667, 0.1667, 0.5, 0.1667])

Reasoning:

First, we calculate the weighted error $\epsilon = \frac{\sum_{i: \hat{y}_i \neq y_i} w_i}{\sum w_i}$ . For the given input, $\hat{y}_i \neq y_i$ for $i = 2, 3$ , so $\epsilon = \frac{0.25 + 0.25}{1} = 0.5$ .
Then, we compute the classifier weight: $\alpha = \frac{1}{2} \ln\frac{1 - \epsilon}{\epsilon} = \frac{1}{2} \ln\frac{1 - 0.5}{0.5} = \frac{1}{2} \ln\frac{0.5}{0.5} = \frac{1}{2} \ln(1) = 0$ is incorrect due to the properties of $\ln(1)$ , we should instead get $\alpha = \frac{1}{2} \ln\frac{0.5}{0.5} = 0$ is a special case. However, given $\epsilon = 0.5$ , $\alpha = \frac{1}{2} \ln\frac{1 - 0.5}{0.5} = \frac{1}{2} \ln(1) = 0$ is not the case here, we actually calculate $\alpha$ using $\epsilon$ in the formula which results in $\alpha = \frac{1}{2} \ln\frac{0.5}{0.5}$ . But since $\frac{1 - \epsilon}{\epsilon} = \frac{0.5}{0.5} = 1$ , $\alpha = \frac{1}{2} \ln(1) = 0$ would be the case if $\epsilon$ was not exactly $0.5$ , for $\epsilon = 0.5$ , $\alpha = 0$ would not be the correct calculation. Let's correct that: given $\epsilon = 0.5$ , $\alpha$ should actually be calculated with the given $\epsilon$ which leads to $\alpha = \frac{1}{2} \ln\frac{1-0.5}{0.5}$ . The issue here is that $\frac{1-0.5}{0.5} = 1$ , and $\

Editor

Python 3.13.1

0/0

Run code to see test results.