Sharpe ratio, annualised

easyQuantacceptance —~15 min

Implement sharpe(returns, rf, periods) — the annualised Sharpe ratio of a list of per-period simple returns: the mean excess return over its sample standard deviation, scaled by sqrt(periods). rf is the ANNUAL risk-free rate; convert it to a per-period rate by dividing by periods before subtracting.

It is the number every backtest is judged on, and the one most often quoted wrong. Use the sample standard deviation (statistics.stdev, n − 1 in the denominator). Return the ratio rounded to 4 decimal places.

example 1

input

returns=[0.01, -0.005, 0.012, 0.003, -0.002, 0.008], rf=0.0, periods=252

output

10.0698

example 2

input

returns=[0.01, -0.005, 0.012, 0.003, -0.002, 0.008], rf=0.05, periods=252

output

9.6087

note

Same returns, 5% annual risk-free: the ratio drops because the excess return does.

constraints

2 ≤ len(returns) ≤ 10_000
0 ≤ rf ≤ 0.1
periods ∈ {12, 52, 252}
returns are not all equal

teacher · 3 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 math import sqrt
from statistics import mean, stdev


def sharpe(returns: list, rf: float, periods: int) -> float:
    """Annualised Sharpe ratio, rounded to 4 dp. rf is annual."""
    # TODO: subtract the per-period risk-free rate (rf / periods) from each return
    excess = list(returns)
    return round(mean(excess) / stdev(excess) * sqrt(periods), 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to sharpe()

returns=[0.01, -0.005, 0.012, 0.003, -0.002, 0.008], rf=0.0, periods=252

expected

10.0698

the judge calls sharpe() 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