如何将可格式化的字符串传递给 python 函数？答案

【问题标题】：How do I pass a format-able string to a python function?如何将可格式化的字符串传递给 python 函数？
【发布时间】：2021-06-10 23:21:08
【问题描述】：

我正在使用一个将其结果保存在几个声音文件中的软件库。我想将它们以更好的名称保存在默认目录以外的其他目录中。软件为此提供了可选参数filename_format。文档说以这种方式使用它：

separator.separate_to_file("/path/to/audio.mp3",  "/path/to/output",  filename_format="{filename}_{instrument}.{codec}")

文档说，“您可以在格式化字符串中的大括号之间使用关键字文件名（输入文件名）、仪器（仪器名称）、文件夹名（输入文件所在文件夹的名称）和编解码器。 "

这是我尝试调用该函数的方式：

separator.separate_to_file(audioDescriptor, destination, filename_format='{}_{}.{}'.format(basename, "accompaniment", "wav"))

这会引发错误：

spleeter.SpleeterError: Separated source path conflict : /Users/.../separation-component/music/audio_OUTPUT/Yellow_accompaniment.wav,please check your filename format

这里是我调用的函数供参考。

    def save_to_file(
    self,
    sources: Dict,
    audio_descriptor: AudioDescriptor,
    destination: str,
    filename_format: str = "{filename}/{instrument}.{codec}",
    codec: Codec = Codec.WAV,
    audio_adapter: Optional[AudioAdapter] = None,
    bitrate: str = "128k",
    synchronous: bool = True,
) -> None:
    """
    Export dictionary of sources to files.

    Parameters:
        sources (Dict):
            Dictionary of sources to be exported. The keys are the name
            of the instruments, and the values are `N x 2` numpy arrays
            containing the corresponding intrument waveform, as
            returned by the separate method
        audio_descriptor (AudioDescriptor):
            Describe song to separate, used by audio adapter to
            retrieve and load audio data, in case of file based audio
            adapter, such descriptor would be a file path.
        destination (str):
            Target directory to write output to.
        filename_format (str):
            (Optional) Filename format.
        codec (Codec):
            (Optional) Export codec.
        audio_adapter (Optional[AudioAdapter]):
            (Optional) Audio adapter to use for I/O.
        bitrate (str):
            (Optional) Export bitrate.
        synchronous (bool):
            (Optional) True is should by synchronous.
    """
    if audio_adapter is None:
        audio_adapter = AudioAdapter.default()
    foldername = basename(dirname(audio_descriptor))
    filename = splitext(basename(audio_descriptor))[0]
    generated = []
    for instrument, data in sources.items():
        path = join(
            destination,
            filename_format.format(
                filename=filename,
                instrument=instrument,
                foldername=foldername,
                codec=codec,
            ),
        )
        directory = os.path.dirname(path)
        if not os.path.exists(directory):
            os.makedirs(directory)
        if path in generated:
            raise SpleeterError(
                (
                    f"Separated source path conflict : {path},"
                    "please check your filename format"
                )
            )
        generated.append(path)
        if self._pool:
            task = self._pool.apply_async(
                audio_adapter.save, (path, data, self._sample_rate, codec, bitrate)
            )
            self._tasks.append(task)
        else:
            audio_adapter.save(path, data, self._sample_rate, codec, bitrate)
    if synchronous and self._pool:
        self.join()

【问题讨论】：

标签： python spleeter

【解决方案1】：

看看它在哪里调用

filename_format.format(
                filename=filename,
                instrument=instrument,
                foldername=foldername,
                codec=codec,
            ),

它期望filename_format arg 只包含字符串（在您的情况下为"{filename}_{instrument}.{codec}" - 它为您调用format。它使用命名参数，因此您需要将名称放在大括号之间匹配format 调用中调用的名称。

【讨论】：