Sum of squares

When the data contains the observations $(x_1,y_1),(x_2,y_2),\dots,(x_n,y_n)$ and the line $y=ax+b$ is fitted to the data, the error can be computed with the sum of squares formula $$\sum_{i=1}^{n}(y_i-(ax_i+b))^2.$$ For example, when the data is $(1,1),(3,2),(5,3)$ and the line is $y=x-1$ (i.e., $a=1$ and $b=-1$), the error is $$(1-(1-1))^2+(2-(3-1))^2+(3-(5-1))^2=2.$$

In a file squaresum.py, implement the class DataAnalyzer with the methods

• add_point(x, y): add an observation to the data
• calculate_error(a, b): return the sum of squares error for the given line parameters

The time complexity of both methods should be $O(1)$.

class DataAnalyzer:
    def __init__(self):
        # TODO

    def add_point(self, x, y):
        # TODO

    def calculate_error(self, a, b):
        # TODO

if __name__ == "__main__":
    analyzer = DataAnalyzer()

    analyzer.add_point(1, 1)
    analyzer.add_point(3, 2)
    analyzer.add_point(5, 3)
    print(analyzer.calculate_error(1, 0)) # 5
    print(analyzer.calculate_error(1, -1)) # 2
    print(analyzer.calculate_error(3, 2)) # 293

    analyzer.add_point(4, 2)
    print(analyzer.calculate_error(1, 0)) # 9
    print(analyzer.calculate_error(1, -1)) # 3
    print(analyzer.calculate_error(3, 2)) # 437

You can test the efficiency of your solution with the following code. In this case too the code should finish almost immediately.

    analyzer = DataAnalyzer()
    total = 0
    for i in range(10**5):
        analyzer.add_point(i, i % 100)
        total += analyzer.calculate_error(i % 97, i % 101)
    print(total) # 25745448974503313754828