問題
Entropy measures the uncertainty of a set of labels: H(S) = -Σ pₖ × log₂(pₖ), where pₖ is the fraction of labels of class k. Information gain is how much entropy drops when you split the set by a feature:
IG = H(labels) - Σ (|Sᵥ| / |S|) × H(Sᵥ)
where Sᵥ contains the labels of the samples whose feature equals v. Decision trees pick the feature with the highest gain.
Implement information_gain(feature, labels).
Input
feature: The feature value of each sample.
labels: The class label of each sample (same length).
Output
Return the information gain in bits, as a decimal number.
制約
1 ≤ length ≤ 10,000
Each value and label is 1 to 20 characters
例
入力：
labels = ["no", "no", "yes", "yes"]
feature = ["sunny", "sunny", "rain", "rain"]
期待される出力：
expected = 1.0入力：
std::vector<std::string> labels = {"no", "no", "yes", "yes"};
std::vector<std::string> feature = {"sunny", "sunny", "rain", "rain"};
期待される出力：
double expected = 1.0;入力：
string[] labels = new string[] {"no", "no", "yes", "yes"};
string[] feature = new string[] {"sunny", "sunny", "rain", "rain"};
期待される出力：
double expected = 1.0;解説
The labels are half yes and half no, so H = 1 bit. Splitting by weather gives two pure groups (H = 0 each), so the gain is 1 - 0 = 1: the feature tells you everything.
入力：
labels = ["yes", "yes", "no", "no"]
feature = ["a", "b", "a", "b"]
期待される出力：
expected = 0.0入力：
std::vector<std::string> labels = {"yes", "yes", "no", "no"};
std::vector<std::string> feature = {"a", "b", "a", "b"};
期待される出力：
double expected = 0.0;入力：
string[] labels = new string[] {"yes", "yes", "no", "no"};
string[] feature = new string[] {"a", "b", "a", "b"};
期待される出力：
double expected = 0.0;解説
Each group a and b still has one yes and one no (H = 1), so the remainder is 0.5 × 1 + 0.5 × 1 = 1 and the gain is 1 - 1 = 0: the feature is useless.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。