JuniorTheoryCommon

What does this program print?

Python

import numpy as np
from sklearn.metrics import mean_absolute_error, mean_squared_error

y_true = [1, 2, 3, 4]
y_pred = [1, 2, 3, 8]
mae = mean_absolute_error(y_true, y_pred)
rmse = np.sqrt(mean_squared_error(y_true, y_pred))
print(mae, rmse)

1.0 2.0

Expecting 1.0 4.0: mean_squared_error returns MSE, in squared units; the code takes its square root, so RMSE is sqrt(4.0) = 2.0.
Expecting MAE and RMSE to match: squaring weights the single error of 4 heavily, so RMSE is twice MAE here; they are equal only when all errors have the same size.

If rare large errors are costly, would you report mean absolute error (MAE) or root mean squared error (RMSE), and why?
How would both values change if the one error of 4 were split into four errors of 1?

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Back to questions

JuniorTheoryCommon

What does this program print?

Python

import numpy as np
from sklearn.metrics import mean_absolute_error, mean_squared_error

y_true = [1, 2, 3, 4]
y_pred = [1, 2, 3, 8]
mae = mean_absolute_error(y_true, y_pred)
rmse = np.sqrt(mean_squared_error(y_true, y_pred))
print(mae, rmse)

1.0 2.0

Expecting 1.0 4.0: mean_squared_error returns MSE, in squared units; the code takes its square root, so RMSE is sqrt(4.0) = 2.0.
Expecting MAE and RMSE to match: squaring weights the single error of 4 heavily, so RMSE is twice MAE here; they are equal only when all errors have the same size.

If rare large errors are costly, would you report mean absolute error (MAE) or root mean squared error (RMSE), and why?
How would both values change if the one error of 4 were split into four errors of 1?

Back to questions Open topic in roadmap →Answer it yourself — Quiz