Rolling z-score

easyQuantacceptance —~15 min

Implement rolling_z(values, window) — the z-score of the LAST value relative to the trailing window values (inclusive of the last): (last - mean) / stdev using the sample standard deviation of exactly those window values.

A z-score of −2 means today is two standard deviations below its recent self — the raw material of every mean-reversion bot in the workbench. Return the score rounded to 4 decimal places.

example 1

input

values=[100.0, 101.0, 99.0, 102.0, 98.0, 97.0], window=6

output

-1.3363

example 2

input

values=[100.0, 101.0, 99.0, 102.0, 98.0, 97.0], window=3

output

-0.7559

constraints

2 ≤ window ≤ len(values) ≤ 10_000
the window values are not all equal

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

from statistics import mean, stdev


def rolling_z(values: list, window: int) -> float:
    """z-score of the last value against the trailing window, 4 dp."""
    # TODO: only the trailing `window` values count, not the whole series
    w = values
    return round((w[-1] - mean(w)) / stdev(w), 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to rolling_z()

values=[100.0, 101.0, 99.0, 102.0, 98.0, 97.0], window=6

expected

-1.3363

the judge calls rolling_z() 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