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mean(), anyNA(), and members of "Summary" group generic should "untranspose"

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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 x until a DelayedAperm op is found. Only climb the trunk of the tree i.e. start from x@seed and 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 exact aperm() transformation is the reverse of this DelayedAperm op. The tree trunk in y should be either shorter than the tree trunk in x (if the 2 DelayedAperm ops could be simplified) or longer (if they couldn't). If it's shorter then replace x with y before computing mean(), 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
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29
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12
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30 天内没有已合并 PR

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  4. 提交 Pull Request,并在描述里引用这个 Issue 编号。

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