在 Python 中混合/叠加 wav 音频文件

Question

我一直在寻找一种仅使用波形库将两个 WAV 音频文件叠加/混合在一起的解决方案。

我找到了以下解决方案：Mixing two audio files together with python

其中一个答案提供了以下代码：

import wave

w1 = wave.open("/path/to/wav/1")
w2 = wave.open("/path/to/wav/2")

#get samples formatted as a string.
samples1 = w1.readframes(w1.getnframes())
samples2 = w2.readframes(w2.getnframes())

#takes every 2 bytes and groups them together as 1 sample. ("123456" -> ["12", "34", "56"])
samples1 = [samples1[i:i+2] for i in xrange(0, len(samples1), 2)]
samples2 = [samples2[i:i+2] for i in xrange(0, len(samples2), 2)]

#convert samples from strings to ints
def bin_to_int(bin):
    as_int = 0
    for char in bin[::-1]: #iterate over each char in reverse (because little-endian)
        #get the integer value of char and assign to the lowest byte of as_int, shifting the rest up
        as_int <<= 8
        as_int += ord(char) 
    return as_int

samples1 = [bin_to_int(s) for s in samples1] #['\x04\x08'] -> [0x0804]
samples2 = [bin_to_int(s) for s in samples2]

#average the samples:
samples_avg = [(s1+s2)/2 for (s1, s2) in zip(samples1, samples2)]

代码是用 Python 2 编写的，而 ord() 在 Python 3 中被贬值了，所以代码看起来像这样，去掉了 ord() 并在 samples_avg 处加倍 // 以避免创建浮点数

import wave

w1 = wave.open("/path/to/wav/1")
w2 = wave.open("/path/to/wav/2")

#get samples formatted as a string.
samples1 = w1.readframes(w1.getnframes())
samples2 = w2.readframes(w2.getnframes())

#takes every 2 bytes and groups them together as 1 sample. ("123456" -> ["12", "34", "56"])
samples1 = [samples1[i:i+2] for i in range(0, len(samples1), 2)]
samples2 = [samples2[i:i+2] for i in range(0, len(samples2), 2)]

#convert samples from strings to ints
def bin_to_int(bin):
    as_int = 0
    for char in bin[::-1]: #iterate over each char in reverse (because little-endian)
        #get the integer value of char and assign to the lowest byte of as_int, shifting the rest up
        as_int <<= 8
        as_int += char
    return as_int

samples1 = [bin_to_int(s) for s in samples1] #['\x04\x08'] -> [0x0804]
samples2 = [bin_to_int(s) for s in samples2]

#average the samples:
samples_avg = [(s1+s2)//2 for (s1, s2) in zip(samples1, samples2)]

代码只是部分代码。缺少的是将 samples_avg 恢复为二进制字符串。这就是我遇到麻烦的地方。我已经尝试使用以下代码将以下代码用于 bin()、chr()

samples_avg = [ chr(s) for s in samples_avg]

samples_avg = [ bin(s) + "'" for s in samples_avg]

我已经尝试过一百万种其他解决方案，但都不好意思发布，但都失败了。

谁能帮忙完成这段代码？我认为这将是一个非常有用的代码，因为它只依赖于 wave 库并且可以在虚拟环境中使用。

我对 Python 比较陌生，对音频处理完全陌生，所以对于任何愚蠢的问题和错误，我深表歉意。

只是为了澄清我所说的混合/溢出的意思。如果我有两个长度为 4 秒的音频文件，我想将它们混合在一起形成一个长度为 4 秒的音频文件，同时播放两个音频文件。

Answer 1

因此，经过一些尝试，@Ponkadoodle 提供了一个错误和帮助，我得到了它的工作。它适用于我在同一台计算机上使用 quicktime 和在线 wav 转换器完成的两个录音。如果我使用来自互联网的 wav 文件，最后的样本听起来真的很乱，我不知道这是不是由于频率等原因。

这是最终代码

import wave
import array


w1 = wave.open("path/to/file/audiofile1.wav")
w2 = wave.open("path/to/file/audiofile2.wav")

#get samples formatted as a string.
samples1 = w1.readframes(w1.getnframes())
samples2 = w2.readframes(w2.getnframes())


#takes every 2 bytes and groups them together as 1 sample. ("123456" -> ["12", "34", "56"])
samples1 = [samples1[i:i+2] for i in range(0, len(samples1), 2)]
samples2 = [samples2[i:i+2] for i in range(0, len(samples2), 2)]

#convert samples from strings to ints
def bin_to_int(bin):
    as_int = 0
    for char in bin[::-1]: #iterate over each char in reverse (because little-endian)
        #get the integer value of char and assign to the lowest byte of as_int, shifting the rest up
        as_int <<= 8
        as_int += char
    return as_int

samples1 = [bin_to_int(s) for s in samples1] #['\x04\x08'] -> [0x0804]
samples2 = [bin_to_int(s) for s in samples2]

#average the samples:
samples_avg = [(s1+s2) for (s1, s2) in zip(samples1, samples2)]

samples_array = array.array('i')
samples_array.fromlist(samples_avg)

wave_out = wave.open ("out.wav", "wb")
wave_out.setnchannels(1)
wave_out.setsampwidth(2)
wave_out.setframerate(w1.getframerate()*4) 
wave_out.writeframes(samples_array)

我仍然对 setframerate() 有疑问。我将它乘以 4 并且它起作用了，这可能取决于您原始录制的频率/帧速率等。

wave_out.setframerate(w1.getframerate()*4)