为什么scipy和librosa在读取wav文件方面有所不同？

Question

So I'm trying to get the samples from a wave file and I noticed that it's a different value depending on whether I use scipy or librosa.

sampleFloats, fs = librosa.load('hi.wav', sr=48000)
print('{0:.15f}'.format(sampleFloats[len(sampleFloats)-1]))

from scipy.io.wavfile import read as wavread
# from python_speech_features import mfcc

[samplerate, x] = wavread('hi.wav') # x is a numpy array of integer, representing the samples 

# scale to -1.0 -- 1.0
if x.dtype == 'int16':
    nb_bits = 16 # -> 16-bit wav files
elif x.dtype == 'int32':
    nb_bits = 32 # -> 32-bit wav files
max_nb_bit = float(2 ** (nb_bits - 1))
samples = x / (max_nb_bit + 1.0) # samples is a numpy array of float representing the samples 

print(samples[len(samples)-1])

The print statements read:

0.001251220703125
0.001274064182641886

The sample rate for the file is 48000.

Why might they be different? Is librosa using a different normalization?

Answer 1

It's a type mismatch. It is often useful to print not only the value, but also its type. In this case, because of the way the normalisation is done, type of samples values is float64 , while librosa returns float32 .

This answer can help to figure out how to normalise (also, as pointed above, it is indeed max_nb_bit - 1 , not + )