Compare candidates on common evidence

Understand the idea

Comparisons need the same population, outcome definition, split and folds when the aim is a paired predictive comparison. Nominate before the final test. Discovery and PCA use different evidence and must not be ranked by supervised accuracy.

Use matching evaluation designs and practical constraints.Scroll the diagram horizontally if needed.

What the syntax does

fold_errors['tree'] - fold_errors['line']: Subtracts errors on matching folds.
A negative value favours the tree on that fold.

Outputs you create

answer: The requested result, with the values, labels and row order described by this task.

LINE24 · first 8 rows
`distance`	`duration`
1	5.30501
1.47826	11.2361
1.95652	11.8488
2.43478	14.809
2.91304	10.7589
3.3913	19.4972
3.86957	17.89
4.34783	15.531

Your task

Across five matching folds, line RMSE is [4.0, 5.0, 4.5, 5.5, 4.0] and tree RMSE is [3.5, 4.8, 5.0, 4.9, 3.8].
Store tree minus line RMSE for each fold in answer.

answer

Hints

Think: Compare aligned fold-score columns before summarising candidate performance.

Tools: Use fold_errors['tree'] - fold_errors['line']. Read the visible syntax meanings before editing.

Approach: Across five matching folds, line RMSE is [4.0, 5.0, 4.5, 5.5, 4.0] and tree RMSE is [3.5, 4.8, 5.0, 4.9, 3.8]. Store tree minus line RMSE for each fold in answer. Keep the supplied row order and inspect the named output after running.

Explained solution

fold_errors = pd.DataFrame({'line': [4.0, 5.0, 4.5, 5.5, 4.0], 'tree': [3.5, 4.8, 5.0, 4.9, 3.8]})
answer = fold_errors['tree'] - fold_errors['line']

The tree has smaller RMSE in four of these illustrative folds but larger error in one. Pairing the evidence preserves the common evaluation population. The mean difference alone is not a universal ranking or proof of statistical significance.