用于拆分和合并mp3文件的Python库
有很多库可以用来处理mp3标签,但我只需要两个功能:一个是把mp3文件分成两部分,另一个是把五个mp3合并在一起。
你能推荐一些吗?谢谢!
5 个回答
5
这是我用Python尝试在不重新编码的情况下分割MP3文件的方法。并不是所有类型的MP3文件都能支持这个方法,如果你有好的建议或改进意见,我非常欢迎。这个脚本是固定设置在55秒处进行分割,但代码展示了基本的原理。
from __future__ import print_function
import struct
import sys
#MP3 frames are not independent because of the byte reservoir. This script does not account for
#that in determining where to do the split.
def SplitMp3(fi, splitSec, out):
#Constants for MP3
bitrates = {0x0: "free", 0x1: 32, 0x2: 40, 0x3: 48, 0x4: 56, 0x5: 64, 0x6: 80, 0x7: 96, 0x8: 112,
0x9: 128, 0xa: 160, 0xb: 192, 0xc: 224, 0xd: 256, 0xe: 320, 0xf: "bad"}
freqrates = {0x0: 44100, 0x1: 48000, 0x2: 32000, 0x3: "reserved"}
countMpegFrames = 0
frameDuration = 0.026
unrecognizedBytes = 0
splitFrame = int(round(splitSec / frameDuration))
while True:
startPos = fi.tell()
#Check for 3 byte headers
id3Start = fi.read(3)
if len(id3Start) == 3:
if id3Start == b'TAG':
print ("Found ID3 v1/1.1 header")
fi.seek(startPos + 256)
continue
if id3Start == b'ID3':
#Possibly a ID3v2 header
majorVer, minorVer, flags, encSize = struct.unpack(">BBBI", fi.read(7))
if majorVer != 0xFF and minorVer != 0xFF:
encSize1 = (encSize & 0x7f000000) >> 24
encSize2 = (encSize & 0x7f0000) >> 16
encSize3 = (encSize & 0x7f00) >> 8
encSize4 = (encSize & 0x7f)
if encSize1 < 0x80 and encSize2 < 0x80 and encSize3 < 0x80 and encSize4 < 0x80:
size = ((encSize & 0x7f000000) >> 3) + ((encSize & 0x7f0000) >> 2) + ((encSize & 0x7f00) >> 1) + (encSize & 0x7f)
unsync = (flags >> 7) & 0x1
extendedHeader = (flags >> 6) & 0x1
experimental = (flags >> 5) & 0x1
print ("Found ID3v2 header")
print ("version", majorVer, minorVer, unsync, extendedHeader, experimental)
print ("size", size)
#TODO extendedHeader not supported yet
fi.seek(startPos + 10 + size)
continue
#Check for 4 byte headers
fi.seek(startPos)
headerRaw = fi.read(4)
if len(headerRaw) == 4:
headerWord = struct.unpack(">I", headerRaw)[0]
#Check for MPEG-1 audio frame
if headerWord & 0xfff00000 == 0xfff00000:
print ("Possible MPEG-1 audio header", hex(headerWord))
countMpegFrames += 1
ver = (headerWord & 0xf0000) >> 16
bitrateEnc = (headerWord & 0xf000) >> 12
freqEnc = (headerWord & 0xf00) >> 8
mode = (headerWord & 0xf0) >> 4
cpy = (headerWord & 0xf)
if ver & 0xe == 0xa and freqEnc != 0xf:
print ("Probably an MP3 frame")
bitrate = bitrates[bitrateEnc]
freq = freqrates[freqEnc >> 2]
padding = ((freqEnc >> 1) & 0x1) == 1
print ("bitrate", bitrate, "kbps")
print ("freq", freq, "Hz")
print ("padding", padding)
frameLen = int((144 * bitrate * 1000 / freq ) + padding)
#Copy frame to output
fi.seek(startPos)
frameData = fi.read(frameLen)
if countMpegFrames >= splitFrame:
out.write(frameData)
fi.seek(startPos + frameLen)
continue
else:
raise RuntimeError("Unsupported format:", hex(ver), "header:", hex(headerWord))
#If no header can be detected, move on to the next byte
fi.seek(startPos)
nextByteRaw = fi.read(1)
if len(nextByteRaw) == 0:
break #End of file
unrecognizedBytes += 1
print ("unrecognizedBytes", unrecognizedBytes)
print ("countMpegFrames", countMpegFrames)
print ("duration", countMpegFrames * frameDuration, "sec")
if __name__=="__main__":
fi = open(sys.argv[1], "rb")
out = open("test.mp3", "wb")
SplitMp3(fi, 55.0, out)
out.close()
合并MP3文件的过程也差不多,就是从两个不同的MP3文件中提取和添加音频帧。
12
你可以去看看维基百科上的MP3文件结构。在Python中使用二进制读取模式来编辑MP3文件。你可以用s = open(file_name, 'rb').read()
这行代码把整个文件读进来,这样就会得到一个字符串对象,里面包含了文件的原始字节(比如说\xeb\xfe\x80
)。接下来,你可以在这个字符串里搜索和编辑,使用方括号来根据索引访问字节,比如s[n]
。最后,你只需要把你想要的MP3帧以二进制方式写入新的文件,并把ID3头信息附加到你想要的帧集合中,这样就能组成每个文件。
149
我写了一个库(pydub),就是为了处理这种情况:
from pydub import AudioSegment
sound = AudioSegment.from_mp3("/path/to/file.mp3")
# len() and slicing are in milliseconds
halfway_point = len(sound) / 2
second_half = sound[halfway_point:]
# Concatenation is just adding
second_half_3_times = second_half + second_half + second_half
# writing mp3 files is a one liner
second_half_3_times.export("/path/to/new/file.mp3", format="mp3")
添加静音间隔
如果你想在声音的不同部分之间添加一些静音:
two_sec_silence = AudioSegment.silent(duration=2000)
sound_with_gap = sound[:1000] + two_sec_silence + sound[1000:]