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

Question

I would like to know if there is a better way to convert base64-format audio files into .wav files without storage them on current directory.

The thing is that I get base64-format audio files from user uploading whith a POST request with FastAPI, then I decode them and convert them into .wav files because I need to pass the files over some functions that I created in order to preprocess and transcript the files and those functions use the wave module for .wav files. Due to I created .wav files for nothing more than transcripting them, so I don't need to store them and I finally delete them with os.unlink function.

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

You could store the data in memory in a [`io.Bytesio`](https://docs.python.org/3/library/io.html#io.BytesIO) instance. — martineau, Feb 26 '22 at 22:29
@martineau, I tried what you suggested me like this: `buffer =io.BytesIO(decode) with io.BytesIO() as f: f.write(buffer.getvalue())` and I send buffer.getbuffer() to my function like this: `segments, sample_rate, audio_length = pr.vad_segment_generator( buffer.getbuffer(), aggresive)` , but I got the error: `AttributeError: 'memoryview' object has no attribute 'read'` — 0x55b1E06FF, Feb 26 '22 at 23:59
@martineau, as I described before I need decoded data turns into a .wav file because then I read this using Python' wave module — 0x55b1E06FF, Feb 27 '22 at 00:03
A `io.BytesIO` is like an open file and can be `read()` like one. Your code appears to be expecting a file name, but they don't have one because they are not real files. Perhaps a ramdisk would be more suitable: [How can I create a ramdisk in Python?](https://stackoverflow.com/questions/4351048/how-can-i-create-a-ramdisk-in-python) — martineau, Feb 27 '22 at 00:32

ti7 · Accepted Answer · 2022-02-28T03:48:39.527

2

As suggested by @martineau, try writing intermediately to an io.BytesIO, but after writing, call .seek(0) to return the steam position to the start, rather than calling getbuffer()
(after writing, the stream position will be at the end, ready for more data)

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

edited Feb 28 '22 at 03:48

answered Feb 27 '22 at 06:13

ti7

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I tried this but I got a `with io.BytesIO("wb") as buffer: TypeError: a bytes-like object is required, not 'str'`. – 0x55b1E06FF Feb 28 '22 at 00:24
1

@nath_ no need to pass `"wb"` to `BytesIO` which is what it's complaining about! the stream is already readable, writable, and seekable, and the argument is just for pre-allocating (which is rarely necessary) https://docs.python.org/3/library/io.html#io.BytesIO – ti7 Feb 28 '22 at 03:47
@nath_ oh, sorry, I was traveling and I see _I put that there_ fixed now – ti7 Feb 28 '22 at 03:48
1

thanks!, it finally works. Just the function call must be one level inside (indentation thing) – 0x55b1E06FF Feb 28 '22 at 05:36

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

1 Answers1