如何在函数之间传递已编辑的WAV而不在两者之间保存WAV？

Question

我有2个人的wav对话（客户和技术支持），我有3个独立的功能，可提取1个声音，缩短10秒钟并将其转换为嵌入。

def get_customer_voice(file):

    print('getting customer voice only')
    wav = wf.read(file)
    ch = wav[1].shape[1]#customer voice always in 1st track
    sr = wav[0]
    c1 = wav[1][:,1]
    #print('c0 %i'%c0.size)

    if ch==1:
        exit()
    vad = VoiceActivityDetection()
    vad.process(c1)
    voice_samples = vad.get_voice_samples()
    #this is trouble - how to pass it without saving anywhere as wav?
    wf.write('%s_customer.wav'%file,sr,voice_samples)

下面的功能比上面的功能减少10秒的wav文件。

import sys
from pydub import AudioSegment

def get_customer_voice_10_seconds(file):
    voice = AudioSegment.from_wav(file)
    new_voice = voice[0:10000]
    file = str(file) + '_10seconds.wav'
    new_voice.export(file, format='wav')


if __name__ == '__main__':
    if len(sys.argv) < 2:
        print('give wav file to process!')
    else:
        print(sys.argv)
        get_customer_voice_10_seconds(sys.argv[1])

如何将其以wav或其他格式传递而不将其保存到某个目录？它将在rest api中使用，我不知道它将在哪里保存该wav，因此最好以某种方式传递它。

Answer 1

我想通了-下面的函数可以正常工作，而无需保存，缓冲等。它接收一个wav文件并对其进行编辑，然后直接发送给get math embedding函数：

def get_customer_voice_and_cutting_10_seconds_embedding(file):

    print('getting customer voice only')
    wav = read(file)
    ch = wav[1].shape[1]
    sr = wav[0]

    c1 = wav[1][:,1]

    vad = VoiceActivityDetection()
    vad.process(c1)
    voice_samples = vad.get_voice_samples()
    audio_segment = AudioSegment(voice_samples.tobytes(), frame_rate=sr,sample_width=voice_samples.dtype.itemsize, channels=1)
    audio_segment = audio_segment[0:10000]
    file = str(file) + '_10seconds.wav'

    return get_embedding(file)

关键是音频段中的tobytes（），它将它们全部重新组合到1个轨道中