tuple unpacking is slower than tuple(list comprehension)
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還沒有人認領這個 Issue。
interpreter-core
performance
type-feature
- 主要語言
- Python
- 星號
- 77.2k
- 分支
- 36k
- PR 合併指標
- PR 指標待擷取
描述
Feature or enhancement
Proposal:
In []: %timeit (*(x**2 for x in range(1000)),) # (A)
47.2 μs ± 1.18 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
In []: %timeit tuple(x**2 for x in range(1000)) # most idiomatic
45.4 μs ± 5.57 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
In []: %timeit (*[x**2 for x in range(1000)],) # (B)
36.5 μs ± 77.8 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
In []: %timeit tuple([x**2 for x in range(1000)]) # fastest
33.8 μs ± 710 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
Currently (A) and (B) are slower than the last one, even though it doesn't need to be so. There doesn't seem to be any bottleneck such as global namespace lookup.
Also, we want people to write the most idiomatic code, so it would be preferable to make the most idiomatic version to be as fast as the fastest version. Possibly with something like
if tuple is builtins.tuple:
MAGIC
else:
code as usual...
Has this already been discussed elsewhere?
This is a minor feature, which does not need previous discussion elsewhere
Links to previous discussion of this feature:
No response
Linked PRs
- gh-149960
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研究方向
首先重現 issue 中的四個基準表達式,並比較 tuple unpacking 與 tuple 建構。查看已連結的 PR gh-149960,了解已在進行的工作;當慣用的 generator 形式不再落後於最快的形式,且不破壞所述的 fallback 行為時,即表示完成。
由索引模型根據 Issue 內容生成。
評估
- 技術堆疊
- python
- 領域
- compilers, performance
- Issue 類型
- 功能
- 難度
- 5/5
- 預估耗時
- 一週以上
- 活躍度
- 停滯
- 描述清晰度
- 基本清楚
- 新手友好度
- 25/100