Out-of-fold stacking features

hardMLacceptance —~40 min

Implement oof_predictions(y, k) — out-of-fold predictions from the simplest level-1 model there is: the mean. Cut the n observations into k contiguous folds (remainder to the last); for each fold, predict every one of its observations with the mean of y over all the OTHER folds. Return the n predictions in order.

A stacked meta-learner trains on level-1 predictions; if those predictions saw the answer, the stack is a leak wearing a lab coat. This is the ai quants stacking lever with the model reduced to its skeleton. Return values rounded to 4 decimal places.

example 1

input

y=[1.0, 2.0, 1.0, 2.0, 1.0, 2.0], k=3

output

[1.5, 1.5, 1.5, 1.5, 1.5, 1.5]

example 2

input

y=[1.0, 2.0, 3.0, 4.0, 5.0, 6.0], k=3

output

[4.5, 4.5, 3.5, 3.5, 2.5, 2.5]

note

Fold means differ from the full mean here, so the leak shows.

constraints

2 ≤ k ≤ n ≤ 5_000

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

▶ open in the workbench

run the same math on a real tape

●solution.py· utf-8 · spaces:4

def oof_predictions(y: list, k: int) -> list:
    """Out-of-fold mean predictions, one per observation, 4 dp."""
    n = len(y)
    size = n // k
    out = [0.0] * n
    for f in range(k):
        start = f * size
        end = n if f == k - 1 else start + size
        # TODO: exclude the held-out fold — this mean includes the rows it is predicting
        m = sum(y) / n
        for i in range(start, end):
            out[i] = round(m, 4)
    return out

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to oof_predictions()

y=[1.0, 2.0, 1.0, 2.0, 1.0, 2.0], k=3

expected

[1.5, 1.5, 1.5, 1.5, 1.5, 1.5]

the judge calls oof_predictions() with exactly these inputs · outputs are compared to 6 decimal places

● standing bypython 3.13 in your browser · pyodide 0.28.3 · judged on the server to 6 dp⌘↵ run · ⌘⇧↵ submit

paper only · educational · not advice