SMA crossover signals

easyTradingacceptance —~15 min

Implement sma_cross_signals(closes, fast, slow) — for every bar where both simple moving averages exist (from index slow - 1 onward), emit 1 if the fast SMA is strictly above the slow, -1 if strictly below, and 0 when they are exactly equal.

It is the oldest trend signal there is and the workbench's first bot. Return the list of signals.

example 1

input

closes=[10.0, 11.0, 12.0, 13.0, 12.0, 11.0], fast=2, slow=3

output

[1, 1, 1, -1]

example 2

input

closes=[10.0, 12.0, 11.0, 11.0, 11.0, 13.0], fast=2, slow=3

output

[1, -1, 0, 1]

note

Three flat closes make the two averages equal — a 0, not a buy.

constraints

1 ≤ fast < slow ≤ len(closes) ≤ 10_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 sma_cross_signals(closes: list, fast: int, slow: int) -> list:
    """1 fast above slow, -1 below, 0 equal — one per bar from slow - 1."""
    def sma(t, period):
        return sum(closes[t - period + 1:t + 1]) / period

    out = []
    for t in range(slow - 1, len(closes)):
        f = sma(t, fast)
        s = sma(t, slow)
        # TODO: a tie is not a cross — equal averages should be 0, not 1
        out.append(1 if f >= s else -1)
    return out

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to sma_cross_signals()

closes=[10.0, 11.0, 12.0, 13.0, 12.0, 11.0], fast=2, slow=3

expected

[1, 1, 1, -1]

the judge calls sma_cross_signals() 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