Python 读取 mp3 文件时出现问题

Question

import speech_recognition as srprint(sr.__version__)r = sr.Recognizer()file_audio = sr.AudioFile('damn1.mp3')with file_audio as source:   audio_text = r.record(source)print(type(audio_text))print(r.recognize_google(audio_text))我运行这个程序时遇到问题。我得到的输出如下：Traceback (most recent call last):  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\site-packages\speech_recognition\__init__.py", line 203, in __enter__    self.audio_reader = wave.open(self.filename_or_fileobject, "rb")  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\wave.py", line 510, in open    return Wave_read(f)  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\wave.py", line 164, in __init__    self.initfp(f)  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\wave.py", line 131, in initfp    raise Error('file does not start with RIFF id')wave.Error: file does not start with RIFF id在处理上述异常的过程中，又出现了一个异常：Traceback (most recent call last):  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\site-packages\speech_recognition\__init__.py", line 208, in __enter__    self.audio_reader = aifc.open(self.filename_or_fileobject, "rb")  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\aifc.py", line 917, in open    return Aifc_read(f)  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\aifc.py", line 352, in __init__    self.initfp(file_object)  File "C:\Users\kubar\AppData\Local\Programs\Python\Python38-32\lib\aifc.py", line 316, in initfp    raise Error('file does not start with FORM id')aifc.Error: file does not start with FORM id

Answer 1

MP3 是一种压缩格式。操作音频时切勿使用它，因为大多数处理音频的工具都是在非压缩音频流上进行操作的。因此，即使此类工具接受您的文件，它也可能会首先对其进行转换，这会消耗时间和空间。此外，从事音频工作的专业人士（音乐家、工程师等）从不使用 MP3：避免将其与对您的工作具有一定重要性的音频材料一起使用（即使是存档，因为压缩是不可逆的），总是更喜欢使用非压缩格式作为 WAV 或 AIF 代替（这里库似乎期望 AIF）。