Intermittent Demand MASE

Problem

Compute MASE for intermittent demand using the mean absolute period-to-period change of the training series as the scale: denominator = mean(|train[t] - train[t-1]|). Return the mean absolute forecast error divided by that denominator.

Return 0.0 when the denominator is zero, train is shorter than two points, or actual is empty.

Hint

The period-to-period scale is robust to long runs of zero demand.

Test cases

case 1

in: [[1,0,2],[1.5,0.5,1.5],[0,0,2,0,2]]

expected: 0.3333333333333333

case 2

in: [[0,0],[1,1],[0,0,0]]

expected: 0

case 3

in: [[5],[5],[1,2]]

expected: 0

case 4

in: [[2,4],[3,3],[1,2,3,4]]

expected: 1

Starter codePython

def intermittent_mase(actual, forecast, train):
    # Your code here
    pass

4 test cases · run your code against them in the editor