What type of file is the "sound fragment" parameter for audioop?

Question

The Python audioop documentation states that most of the available functions require "sound fragments."

The audioop module contains some useful operations on sound fragments. It operates on sound fragments consisting of signed integer samples 8, 16 or 32 bits wide, stored in Python strings.

What exactly is a sound fragment and how can I turn an existing .wav file into one?

Thanks.

Scoreboard

Answer 1

You can do so by using the wave module

The open() method opens the file and readframes(n) returns (maximum) n frames of audio as a string of bytes, just what audioop wants.

For example, let's say you need to use the avg() method from audioop. This is how you could do it:

import wave
import audioop


wav = wave.open("piano2.wav")
print(audioop.avg(wav.readframes(wav.getnframes()), wav.getsampwidth()))

Outputs:

-2

Also, you may be interested in the rewind() method from the wave module. It puts the reading position back to the beginning of the wav file.

If you need to read through your wav file twice you can write this:

wav = wave.open("piano2.wav")
print(audioop.avg(wav.readframes(wav.getnframes()), wav.getsampwidth()))

# if you don't call rewind, next readframes() call 
# will return nothing and audioop will fail

wav.rewind() 
print(audioop.max(wav.readframes(wav.getnframes()), wav.getsampwidth()))

Or alternatively you can cache the string:

wav = wave.open("piano2.wav")
string_wav = wav.readframes(wav.getnframes())
print(audioop.avg(string_wav, wav.getsampwidth()))
# wav.rewind()
print(audioop.max(string_wav, wav.getsampwidth()))

Ansermet

Answer 2

You may want to look into the wave module. You probably want to open a file in read mode and use readframes to get the sample you need for audiooop.

Ostentation

Answer 3

To answer what exactly a fragment is, it's a bytes object, which is just a string of bytes. I believe that for 8-bit audio files, there would be one byte for each frame for 8-bit audio, two bytes per frame for 16-bit audio, and four bytes for 32-bit audio.

Cleromancy

Answer 4

A sound fragment represents a sequence of signed integer sound samples encoded in a bytes-like object. audioop supports representations of 1, 2, 3 or 4 bytes per sample.

A single sample can be converted with struct.pack (let's use 0, 1, 2, -1, 42 as examples):

from struct import pack

for sample in [0, 1, 2, -1, 42]:
   print(f'sample value {sample}, 1 byte/sample:', pack('b', sample))
   print(f'             {sample}, 2 byte/sample:', pack('h', sample))
   print(f'             {sample}, 4 byte/sample:', pack('i', sample))

This prints:

sample value 0, 1 byte/sample: b'\x00'
             0, 2 byte/sample: b'\x00\x00'
             0, 4 byte/sample: b'\x00\x00\x00\x00'
sample value 1, 1 byte/sample: b'\x01'
             1, 2 byte/sample: b'\x01\x00'
             1, 4 byte/sample: b'\x01\x00\x00\x00'
sample value 2, 1 byte/sample: b'\x02'
             2, 2 byte/sample: b'\x02\x00'
             2, 4 byte/sample: b'\x02\x00\x00\x00'
sample value -1, 1 byte/sample: b'\xff'
             -1, 2 byte/sample: b'\xff\xff'
             -1, 4 byte/sample: b'\xff\xff\xff\xff'
sample value 42, 1 byte/sample: b'*'
             42, 2 byte/sample: b'*\x00'
             42, 4 byte/sample: b'*\x00\x00\x00'

Let's assume we want to convert some sound samples we have in Python (signed) integer representation to a sound fragment using 2 byte per sample (allowing input sample values between -32768 to 32767; that is -2**15 to 2**15-1) like used in an audio CD:

import audioop
import array

samples = [0, 1000, 32767, 1, -1, -32768, -1000]  # 7 samples of "music"
fragment = array.array('h', samples).tobytes()
print(f'Fragment {fragment} of length {len(fragment)}')

# convert back with audioop function
print([audioop.getsample(fragment, 2, i) for i in range(len(fragment) // 2)])

This prints:

Fragment b'\x00\x00\xe8\x03\xff\x7f\x01\x00\xff\xff\x00\x80\x18\xfc' of length 14
[0, 1000, 32767, 1, -1, -32768, -1000]

As last example, write a 3 second stereo sine wave as .wav file and read it again:

import audioop
import wave
from array import array
from math import sin, pi

bytes_per_sample = 2
duration = 3.  # seconds
sample_rate = 16000.  # Hz
frequency = 440.  # Hz
max_amplitude = 2**(bytes_per_sample * 8 - 1) - 1
amp = max_amplitude * 0.8
time = [i / sample_rate for i in range(int(sample_rate * duration))]
samples = [int(round(amp * sin(2 * pi * frequency * t))) for t in time]
fragment_mono = array('h', samples).tobytes()
fragment_stereo = audioop.tostereo(fragment_mono, bytes_per_sample, 1, 1)
with wave.open('sine_440hz_stereo.wav', 'wb') as wav:
    wav.setnchannels(2)  # stereo
    wav.setsampwidth(bytes_per_sample)
    wav.setframerate(sample_rate)
    wav.writeframes(fragment_stereo)

# read wave file again
with wave.open('sine_440hz_stereo.wav', 'rb') as wav:
    fragment = wav.readframes(wav.getnframes())

# test whether written fragment and read fragment are same
assert fragment == fragment_stereo

Begotten

Answer 5

import audioop
import wave
read = wave.open("C:/Users/Pratik/sampy/cat.wav")
string_wav = read.readframes(read.getnframes())
a = audioop.lin2alaw(string_wav,read.getsampwidth())
print(a)
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how to convert wav file  to alaw

Baucis