mean(), anyNA(), and members of "Summary" group generic should "untranspose"
还没有人认领这个 Issue。
评估
- 难度
- 5/5
- 预计耗时
- 一周以上
- 新手友好度
- 32/100
- Issue 类型
- 功能
- 描述清晰度
- 基本清楚
- 活跃度
- 停滞
- 技术栈
- r
- 领域
- data, performance
调研方向
首先定位 DelayedArray 对象的 mean()、anyNA() 和 "Summary" 组泛型的实现,然后检查 DelayedAperm 和 simplify()。跟踪 issue 中描述的延迟操作主干,并确定如何在计算前应用反向 aperm()。当简化缩短树时,这些操作能够保持结果,同时避免不必要的延迟转置,即视为完成。
由索引模型根据 Issue 内容生成。
描述
Delayed transposition (t() or aperm()) significantly slows down block processing of a DelayedMatrix or DelayedArray object. However the result of block-processed operations like mean(), anyNA(), and members of the "Summary" group generic does not change if the input is transposed. So these operations should be smart enough to "untranspose" their input in order to be faster.
The exact algorithm for "untransposing" could be:
- Go up the tree of delayed ops in
xuntil a DelayedAperm op is found. Only climb the trunk of the tree i.e. start fromx@seedand go up only if there is exactly 1 "next seed", that is, if the current seed is a DelayedUnaryOp object. Stop on the first DelayedAperm op (i.e. the most recently applied DelayedAperm op), or when the next seed is no longer a DelayedUnaryOp object. - If no DelayedAperm op was found then there is nothing to do.
- If a DelayedAperm op is found, do
y <- aperm(x, ....)where the exactaperm()transformation is the reverse of this DelayedAperm op. The tree trunk inyshould be either shorter than the tree trunk inx(if the 2 DelayedAperm ops could be simplified) or longer (if they couldn't). If it's shorter then replacexwithybefore computingmean(),anyNA(), etc...
The advantage of this algo is that it doesn't need to know anything about the delayed ops found between the root of the tree and the first DelayedAperm op found on the trunk. It just relies on simplify().
- 主要语言
- R
- 星标
- 29
- 派生
- 12
- PR 合并指标
- 30 天内没有已合并 PR
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- Fork 仓库,在一个分支上完成修改。
- 提交 Pull Request,并在描述里引用这个 Issue 编号。
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