如何将 base64 格式的音频文件转换为 .wav 文件而不将它们存储在 Python 的当前目录中？

Question

我想知道是否有更好的方法可以将 base64 格式的音频文件转换为 .wav 文件而不将它们存储在当前目录中。

问题是我从用户上传的 POST 请求中获取了 base64 格式的音频文件我为预处理和转录文件而创建的函数，这些函数使用 .wav 文件的 wave 模块。由于我创建 .wav 文件只是为了转录它们，所以我不需要存储它们，最后我用 os.unlink 函数删除了它们。

import fastapi_server.preprocessing_f as pr

app = FastAPI()

class AudioBase64(BaseModel):
    audio_name: str = Field(..., min_length=1, example="my-audio")
    data_base64: str = Field(..., min_length=1)

@app.post(
    path="/upload-base64-audios/",
    status_code=status.HTTP_200_OK
)
async def upload_base64_audios(audios: list[AudioBase64] = Body(...)):

    model: str = "~/models"
    dir_name = os.path.expanduser(model)
    output_graph, scorer = pr.resolve_models(dir_name)
    model_retval: List[str] = pr.load_model(output_graph, scorer)

    all_names: list[str] = []
    all_datas: list[str] = []
    all_decode: list[str] = []
    aggresive = 1
    transcriptions: list[str] = []
    new_data: list[str] = []
    final_data: list[str] = []
    header: list[str] = ["audio_name", "transcriptions"]

    for i in range(len(audios)):
        name = audios[i].audio_name
        data = audios[i].data_base64
        decode = base64.b64decode(data)
        all_names.append(name)
        all_datas.append(data)
        all_decode.append(decode)

        filename = "%s.wav" % name
        with open(filename, "wb") as f:
            f.write(decode)

        cwd = os.getcwd()
        files = glob.glob(cwd + "/" + name + ".wav")

        segments, sample_rate, audio_length = pr.vad_segment_generator(
            files[0], aggresive
        )
        for k, segment in enumerate(segments):
            audio = np.frombuffer(segment, dtype=np.int16)
            output = pr.stt(model_retval[0], audio)
            transcript = output[0]
        transcriptions.append(transcript)
        new_data = [all_names[i], transcriptions[i]]
        final_data.append(new_data)

    dir_files = glob.glob(cwd + "/*.wav")
    for file in dir_files:
        os.unlink(file)

    new_df = pd.DataFrame(final_data, columns=header)
    stream = io.StringIO()
    new_df.to_csv(stream, index=False)
    response: Response = StreamingResponse(
        iter([stream.getvalue()]), media_type="text/csv"
    )
    response.headers["Content-Disposition"] = "attachment; filename=my-file.csv"
    return response

Answer 1

作为，尝试中间写入一个 io.BytesIO，但写入后，调用 .seek(0) 到 return 蒸汽位置到开始，而不是调用 getbuffer()
（写入后流位置会在最后，准备更多数据）

with io.BytesIO() as buffer:
    buffer.write(decode)
    buffer.seek(0)  # rewind stream
    ...
    segments, sample_rate, audio_length = pr.vad_segment_generator(
        buffer, aggresive)

如何将 base64 格式的音频文件转换为 .wav 文件而不将它们存储在 Python 的当前目录中？

How to convert base64-format audio files into .wav files without storage them on current directory in Python?

python

base64

wav

fastapi