A Fixed Budget of Tries — easy Machine Learning in Code problem | Incognition

A Fixed Budget of Tries

10 pts · 10 coins

Machine Learning in CodeEasy

A Fixed Budget of Tries

A full grid of tree settings would take too long. You decide how many tries you can afford and let a random search spend them.

Task: write random_tree_search(n_iter, seed).

Load the data with load_breast_cancer(return_X_y=True, as_frame=True).
Split it with train_test_split(X, y, test_size=0.25, random_state=0, stratify=y).
Run RandomizedSearchCV(DecisionTreeClassifier(random_state=0), space, n_iter=n_iter, cv=5, random_state=seed) on the training rows, where space is

setting	values
`max_depth`	1, 2, 3, 4, 5, 6, 8, 10
`min_samples_leaf`	1, 2, 4, 8, 16, 32

Return a dict:

"tried": how many combinations the search tried, as a whole number,
"best": its best_params_, exactly as scikit-learn gives it,
"cv": its best cross-validated score, rounded to 3 decimal places.

Examples

n_iter = 8

seed = 0

returns {'tried': 8, 'best': {'min_samples_leaf': 4, 'max_depth': 5}, 'cv': 0.927}

n_iter = 4

seed = 1

returns {'tried': 4, 'best': {'min_samples_leaf': 4, 'max_depth': 8}, 'cv': 0.927}

Hints

solution.py

1from sklearn.datasets import load_breast_cancer
2from sklearn.model_selection import RandomizedSearchCV, train_test_split
3from sklearn.tree import DecisionTreeClassifier
4
5
6def random_tree_search(n_iter, seed):
7    pass
8