Problème
Decision trees split data into groups that are as pure as possible. Gini impurity measures how mixed a group's labels are:
Gini = 1 - Σ pₖ²
where pₖ is the fraction of labels equal to class k. A group with a single class has impurity 0; an even mix of two classes has impurity 0.5.
Implement gini_impurity(labels).
Input
labels: The class labels in the group.
Output
Return the Gini impurity as a decimal number.
Contraintes
1 ≤ length ≤ 10,000
Each label is 1 to 20 characters
Exemples
Exemple 1
Entrée :
labels = ["yes", "yes", "no", "no"]
Sortie attendue :
expected = 0.5Entrée :
std::vector<std::string> labels = {"yes", "yes", "no", "no"};
Sortie attendue :
double expected = 0.5;Entrée :
string[] labels = new string[] {"yes", "yes", "no", "no"};
Sortie attendue :
double expected = 0.5;Explication
Half are yes and half are no: 1 - (0.5² + 0.5²) = 1 - 0.5 = 0.5, the most mixed a two-class group can be.
Exemple 2
Entrée :
labels = ["cat", "cat", "cat"]
Sortie attendue :
expected = 0.0Entrée :
std::vector<std::string> labels = {"cat", "cat", "cat"};
Sortie attendue :
double expected = 0.0;Entrée :
string[] labels = new string[] {"cat", "cat", "cat"};
Sortie attendue :
double expected = 0.0;Explication
Only one class: 1 - 1² = 0. The group is pure.
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.