在 Python 中打开一个波形文件:未知格式:49. 出了什么问题?

问题描述

我尝试使用 wave 模块打开一个波形文件，但无论我尝试什么，都会遇到相同的错误.出现错误的行如下:

I try to open a wave file with the wave module, but I keep getting the same error whatever I try. The line with the error is the following:

wav = wave.open(f)

这是错误信息:

Traceback (most recent call last):
  File "annotate.py", line 47, in <module>
    play(file)
  File "annotate.py", line 33, in play
    wav = wave.open(f)
  File "C:Program Files (x86)Pythonlibwave.py", line 498, in open
    return Wave_read(f)
  File "C:Program Files (x86)Pythonlibwave.py", line 163, in __init__
    self.initfp(f)
  File "C:Program Files (x86)Pythonlibwave.py", line 143, in initfp
    self._read_fmt_chunk(chunk)
  File "C:Program Files (x86)Pythonlibwave.py", line 269, in _read_fmt_chunk
    raise Error('unknown format: %r' % (wFormatTag,))
wave.Error: unknown format: 49

String f 是 .WAV 文件的路径，在我的任何媒体播放器中播放时都可以使用.

String f is a path to a .WAV file and it works when played in any of my media players.

我当然已经导入了 wave 模块.我尝试将 f 作为相对路径和绝对路径.我尝试用wav"替换WAV".

I have of course imported the wave module. I tried f both as a relative and an absolute path. I tried replacing "WAV" by "wav".

错误是由什么引起的?

推荐答案

Python 的 wave 模块适用于特定类型的 WAV:PCM (WAVE_FORMAT_PCM: 0x0001).

Python's wave module works with a specific type of WAV: PCM (WAVE_FORMAT_PCM: 0x0001).

在您的情况下，您使用的是 WAVE_FORMAT_GSM610 [0x0031 = hex(49)] 类型的 WAV.

In your case, you're using a WAV of type WAVE_FORMAT_GSM610 [0x0031 = hex(49)].

您可以使用 Audacity 之类的程序或一些库来转换编解码器以更改 WAV 文件的类型.

You can use a program like Audacity or some lib for converting codecs to change the type of the WAV file.

您可以在此处查看 WAV 类型列表:https://www.videolan.org/developers/vlc/doc/doxygen/html/vlc__codecs_8h.html

You can see a list of WAV types here: https://www.videolan.org/developers/vlc/doc/doxygen/html/vlc__codecs_8h.html

Python的wave模块源码:https://github.com/python/cpython/blob/master/Lib/wave.py

Python's wave module source code: https://github.com/python/cpython/blob/master/Lib/wave.py

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