Question

我编写了一个简单的Python程序来读取wave文件，更改后将其存储为新文件。

import codecs, wave

#convert a number to its two's complemented value (For positive it is equal itself)
def convert_to_twos(value, wid_len=16):
    if value < 0 :
        value = value + (1 << wid_len)
    return value

#receive the value of a two's complemented number.
def twos_back_value(value, wid_len=16):
    if value & (1 << wid_len -1):
        value = value - (1 << wid_len)
    return value

#opening files
input_file = wave.open(r"<address of input wave file>", 'r')
output_file = wave.open(r"<an address for output wave file>", 'w')

#Get input file parameters and set them to the output file after modifing the channel number.
out_params = [None, None, None, None, None, None]
in_params = input_file.getparams()
out_params[0] = 1 # I want to have a mono type wave file in output. so I set the channels = 1
out_params[1] = in_params[1] #Frame Width
out_params[2] = in_params[2] #Sample Rate
out_params[3] = in_params[3] #Number of Frames
out_params[4] = in_params[4] #Type
out_params[5] = in_params[5] #Compressed or not
output_file.setparams(out_params)

#reading frames from first file and storing in the second file
for frame in range(out_params[2]):
    value = int(codecs.getencoder('hex')(input_file.readframes(1))[0][:4], 16) #converting first two bytes of each frame (let assume each channel has two bytes frame length) to int (from byte string).
    t_back_value = twos_back_value( value ,out_params[1]*8)
    new_value = int(t_back_value * 1)
    new_twos = convert_to_twos(new_value, out_params[1]*8)
    to_write = new_twos.to_bytes((new_twos.bit_length() + 7) // 8, 'big')
    output_file.writeframes(to_write)


#closing files
input_file.close()
output_file.close()

问题是，当我运行上述程序并播放输出文件时，我只能听到噪音而没有别的！（虽然我希望只在一种频道模式下使用相同的文件！）

更新

我有些奇怪。根据文档，函数readframes(n) 读取并返回最多n帧音频，作为字节串。所以我希望这个函数只返回十六进制值。但实际上我可以看到一些奇怪的非十六进制值：

read_frame = input_file.readframes(1)
print (read_frame)
print (codecs.getencoder('hex')(read_frame)[0])
print ("")

上面的代码，在for循环中返回：

b'\xe3\x00\xc7\xf5'
b'e300c7f5'

b'D\xe8\xa1\xfd'
b'44e8a1fd'

b'\xde\x08\xb2\x1c'
b'de08b21c'

b'\x17\xea\x10\xe9'
b'17ea10e9'

b'{\xf7\xbc\xf5'
b'7bf7bcf5'

b'*\xf6K\x08'
b'2af64b08'

如您所见，read_frame中有一些非十六进制值！（*，}，D，......例如）。这些是什么？

Answer 1

您看到的值是每帧的四个字节，即第一个通道的两个字节和第二个通道的两个字节。对于单声道WAV，您只能看到两个字节。

以下方法可让您走正确的道路。您需要使用Python的struct库将二进制帧值转换为有符号整数。然后，您可以根据需要操作它们。对于我的例子，我简单地乘以2/3：

 <tr class="gradeA {{ cycle(['even', 'odd'], loop.index0) }}">

注意，像这样一次处理一个帧的波形文件会非常慢。可以通过减少对<tr class="gradeA {{ cycle(['even', 'odd'], loop.index) }}">的调用次数加快速度。

import wave import codecs import struct #opening files input_file = wave.open(r"sample.wav", 'rb') output_file = wave.open(r"sample_out.wav", 'wb') #Get input file parameters and set them to the output file after modifing the channel number. in_params = list(input_file.getparams()) out_params = in_params[:] out_params[0] = 1 output_file.setparams(out_params) nchannels, sampwidth, framerate, nframes, comptype, compname = in_params format = '<{}h'.format(nchannels) #reading frames from first file and storing in the second file for index in range(nframes): frame = input_file.readframes(1) data = struct.unpack(format, frame) value = data[0] # first (left) channel only value = (value * 2) // 3 # apply a simple function to each value output_file.writeframes(struct.pack('<h', value)) #closing files input_file.close() output_file.close()包含解压缩二进制值所需的格式。对于双通道WAV文件，这将包含4个字节。然后writeframes将format配置为format，这意味着使用<hh将产生两个字段，每个字段包含每个频道的整数表示。因此，四个字节成为两个整数的列表，每个通道一个。

使用Python的WAV文件修饰符

1 个答案: