[英]Failed to open file file.wav as a WAV due to: file does not start with RIFF id
嘗試在 python 中打開 RIFF 文件(據我所知它是一種 WAV)時出現此錯誤。
Failed to open file file.wav as a WAV due to: file does not start with RIFF id
當我用各種工具檢查它這使我相信這是一個真正的WAV / RIFF文件。
$ file file.wav
file.wav: MBWF/RF64 audio, stereo 96000 Hz
$ file -i file.wav
file.wav: audio/x-wav; charset=binary
$ mediainfo file.wav
General
Complete name : file.wav
Format : Wave
Format profile : RF64
File size : 4.10 GiB
Duration : 2h 7mn
Overall bit rate mode : Constant
Overall bit rate : 4 608 Kbps
Audio
Format : PCM
Format settings, Endianness : Little
Format settings, Sign : Signed
Codec ID : 1
Duration : 2h 7mn
Bit rate mode : Constant
Bit rate : 4 608 Kbps
Channel(s) : 2 channels
Sampling rate : 96.0 KHz
Bit depth : 24 bits
Stream size : 4.10 GiB (100%)
您擁有的是64 位 RIFF 。 wave
不支持 64 位 RIFF 文件。
如果您的音頻沒問題,並且您可以使用 librosa 或 scipy.io 讀取文件,我們可以簡單地讀取文件,將其寫回臨時 wav 文件,然后再次使用 wave 包讀取它。
例子。 下面,我們得到 RIFF id 錯誤。
>>> import wave
>>> wave.open('./SA1.WAV')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/home/pytorch/anaconda3/lib/python3.6/wave.py", line 499, in open
return Wave_read(f)
File "/home/pytorch/anaconda3/lib/python3.6/wave.py", line 163, in __init__
self.initfp(f)
File "/home/pytorch/anaconda3/lib/python3.6/wave.py", line 130, in initfp
raise Error('file does not start with RIFF id')
wave.Error: file does not start with RIFF id
我們用 librosa 讀入 numpy,用 soundfile 寫回。
import librosa
import soundfile as sf
>>> x,_ = librosa.load('./SA1.WAV', sr=16000)
>>> sf.write('tmp.wav', x, 16000)
>>> wave.open('tmp.wav','r')
<wave.Wave_read object at 0x7fbcb4c8cf28>
類似於@kakrafoon 的答案,但使用soundfile
進行讀寫(以防您關心限制外部依賴項的數量):
import soundfile
import wave
file_path = "your_file.wav"
# Read and rewrite the file with soundfile
data, samplerate = soundfile.read(file_path)
soundfile.write(file_path, data, samplerate)
# Now try to open the file with wave
with wave.open(file_path) as file:
print('File opened!')
我有一個單詞,我將文件的后綴重命名為“mp3”並將其轉換為“wav”,然后我就可以閱讀它了。
subprocess.call(['ffmpeg', '-i', 'XXX.mp3', 'XXX.wav'])
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.