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普遍从源头获得档案,这种档案既可以是焦炭,也可以是双倍?
原标题:Universally getting files from a source-Path that can be either a zip or a folder?

这是一个trick小的问题。

鉴于以下法典,你将如何采用通用的取用文件方式?

from pathlib import Path
import re
import zipfile as zip

def check_if_zip(path: Path) -> bool:
    return path.is_file() and (
        re.search(r".zip$", path.name, re.IGNORECASE) is not None
    )

sources: list[Path] = [
    Path("Foo"),  # Directory
    Path("Bar.zip"),  # Zip
]

for source in sources:
    source_is_zip = check_if_zip(source)

    # Is there a way to universally wrap each source
    # as some kind of "streamed I/O object" or class
    # that reads live from either the folder or zip
    # file? I need a storage-medium agnostic way
    # of fetching files from the "source".
    #
    # It needs to use a fast byte-stream or similar,
    # NOT reading the entire source into memory! :)
    # And the stream needs to be seekable so that
    # it works as input for other libraries, for
    # decoding files contained within the folders/zips.

也许不能在沙尔这样做?

我希望有人会发现这个问题令人感兴趣!

问题回答

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