Z-Score Standardization

Standardization rescales a feature so it has mean 0 and standard deviation 1, which many models (and gradient descent) work best with:

z = (x - mean) / std

Use the population standard deviation: std = √(Σ(x - mean)² / n). If std is 0 (all values equal), return 0 for every value.

Implement z_scores(values).

Input

values: The values of one feature.

Output

Return the z-score of each value, in the same order.

Constraints

1 ≤ length ≤ 10,000

-1,000,000 ≤ each value ≤ 1,000,000

Examples

Input:
values = [2.0, 4.0, 4.0, 4.0, 5.0, 5.0, 7.0, 9.0]

Expected output:
expected = [-1.5, -0.5, -0.5, -0.5, 0.0, 0.0, 1.0, 2.0]

Input:
std::vector<double> values = {2.0, 4.0, 4.0, 4.0, 5.0, 5.0, 7.0, 9.0};

Expected output:
std::vector<double> expected = {-1.5, -0.5, -0.5, -0.5, 0.0, 0.0, 1.0, 2.0};

Input:
double[] values = new double[] {2.0, 4.0, 4.0, 4.0, 5.0, 5.0, 7.0, 9.0};

Expected output:
double[] expected = new double[] {-1.5, -0.5, -0.5, -0.5, 0.0, 0.0, 1.0, 2.0};

Input:
values = [3.0, 3.0]

Expected output:
expected = [0.0, 0.0]

Input:
std::vector<double> values = {3.0, 3.0};

Expected output:
std::vector<double> expected = {0.0, 0.0};

Input:
double[] values = new double[] {3.0, 3.0};

Expected output:
double[] expected = new double[] {0.0, 0.0};

References

This challenge is adapted for CodeUtility. Follow the statement and constraints above.

• scikit-learn, Preprocessing data
• Goodfellow, Bengio and Courville, Deep Learning