[中] 固定选眼路径的邻域过滤为 O(n²)
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Assessment
- Difficulty
- 2/5
- Estimated time
- 1-3 hours
- Newbie friendliness
- 75/100
- Issue type
- Refactor
- Clarity
- Clearly specified
- Activity status
- Active
- Tech stack
- numpy, python
- Domain
- performance
Research direction
Open windows/birdlock/core/analysis.py around lines 471-473, inside build_track(), where actual comes from np.flatnonzero(...) and the per-index np.any(np.abs(actual - i) <= 8) check runs. Replace that check with the suggested np.searchsorted(actual, i+9) - np.searchsorted(actual, i-8) > 0. Done means the two forms return identical results on the same arrays and a 3000-frame rebuild is measurably faster; no test file is named, so verify by re-running a sequence through build_track() after a manual eye correction.
Written by the indexing model from the issue text.
Description
位置：windows/birdlock/core/analysis.py:471-473
现象：actual 已由 np.flatnonzero(...) 得到有序数组，随后对每个 i 执行 np.any(np.abs(actual - i) <= 8)。n=3000 时约 900 万次运算，而 build_track() 在每次手动修正/选眼后都会重建。
复现：在 3000 帧序列上反复点选/修正眼睛，观察每次重建耗时。
建议：用 np.searchsorted(actual, i+9) - np.searchsorted(actual, i-8) > 0。
来源：2026-11-15 全仓只读审计，PR #7 未包含此项。 未在 Windows 实机复现，复现路径由代码审计推得。
- Dominant language
- Swift
- Stars
- 2
- Forks
- 2
- Avg merge
- 1h 11m
- Merged PRs (30d)
- 2
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