Mitchell-JohnsonandClaude Fable 5.1
ec94cb8bff
Co-Authored-By: Claude Fable 5.1 <noreply@anthropic.com>
42 lines
2.1 KiB
TypeScript
42 lines
2.1 KiB
TypeScript
// Which points of a curve does each dot-column of a graph draw?
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//
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// A curve holds up to `cap` points (the whole history window); a box is `cols` columns wide,
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// and a narrow one — the net graph, a phone — has fewer columns than points. Taking "the
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// nearest point" for each column draws one point in three and, worse, a *different* one in
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// three every time the curve advances: a spike is there, gone, there again, and the whole
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// graph reshapes once a second. So:
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//
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// - a column draws the PEAK of all the points that fall in it, so nothing is ever skipped; and
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// - columns are cut from the data, not from the box: point number `a` (counted since the page
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// opened, `advance` being the newest) always belongs to column ⌊a·cols/cap⌋, whatever has
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// arrived since. A finished column therefore never changes — the graph only ever slides
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// left by whole columns, and only its newest column is still moving.
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//
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// With more columns than points (a wide box) each point simply spans several columns.
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// Returns one value per column, oldest first, right-aligned: a young curve has fewer columns
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// than the box and draws in from the right.
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export function columnsOf(values: number[], advance: number, cap: number, cols: number): number[] {
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const n = Math.min(values.length, cap)
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if (n === 0 || cols < 1 || cap < 1) return []
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const at = (a: number) => values[values.length - 1 - (advance - a)] // the point numbered a
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const oldest = advance - n + 1
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const out: number[] = []
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if (cols >= cap) {
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const kLast = Math.ceil(((advance + 1) * cols) / cap) - 1
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for (let k = kLast - cols + 1; k <= kLast; k++) {
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const a = Math.floor((k * cap) / cols)
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if (a >= oldest && a <= advance) out.push(at(a))
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}
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return out
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}
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const kLast = Math.floor((advance * cols) / cap)
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for (let k = kLast - cols + 1; k <= kLast; k++) {
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const a0 = Math.max(oldest, Math.ceil((k * cap) / cols)), a1 = Math.min(advance, Math.ceil(((k + 1) * cap) / cols) - 1)
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if (a1 < a0) continue
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let peak = at(a0)
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for (let a = a0 + 1; a <= a1; a++) peak = Math.max(peak, at(a))
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out.push(peak)
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}
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return out
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}
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