Long-only backtest with costs

mediumTradingacceptance —~30 min

Implement backtest(closes, signals, cost_bps) — a long-only backtest with transaction costs. Start with equity 1.0. On each bar t >= 1, if signals[t - 1] == 1 the strategy was long, so multiply equity by closes[t] / closes[t - 1]. Whenever the position changes between signals[t - 1] and signals[t] (in or out), charge cost_bps / 10_000 of equity. Return the final equity.

The same signals that look brilliant at zero cost usually die at ten basis points — which is why the /learn backtest chapter has a cost slider. Return the equity rounded to 4 decimal places.

example 1

input

closes=[100.0, 102.0, 101.0, 105.0, 104.0], signals=[1, 1, 0, 1, 0], cost_bps=0.0

output

1.0004

example 2

input

closes=[100.0, 102.0, 101.0, 105.0, 104.0], signals=[1, 1, 0, 1, 0], cost_bps=10.0

output

0.9974

constraints

2 ≤ len(closes) == len(signals) ≤ 10_000
signals ∈ {0, 1}
0 ≤ cost_bps ≤ 100

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

●solution.py· utf-8 · spaces:4

def backtest(closes: list, signals: list, cost_bps: float) -> float:
    """Final equity of a long-only backtest with costs, rounded to 4 dp."""
    eq = 1.0
    for t in range(1, len(closes)):
        if signals[t - 1] == 1:
            eq *= closes[t] / closes[t - 1]
        # TODO: charge cost_bps / 10_000 of equity whenever the position changes
    return round(eq, 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to backtest()

closes=[100.0, 102.0, 101.0, 105.0, 104.0], signals=[1, 1, 0, 1, 0], cost_bps=0.0

expected

1.0004

the judge calls backtest() 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