fs, s = wav.read('wave.wav')
此信号具有44100 Hz的采样频率，我想将此信号采样到8Khz
scipy.signal.resample(s,s.size/5.525)但第二个元素不能浮动，那么，我们如何使用此函数重新映射语音信号呢？
我们如何使用scipy.signal.resample在python中将语音信号从44100下采样到8000 Hz？
答案 0 :(得分：3)
那么，另一个解决方案，这个与scipy for real。正是要求的。
这是scipy.signal.resample（）的 doc 字符串：
"""
Resample `x` to `num` samples using Fourier method along the given axis.
The resampled signal starts at the same value as `x` but is sampled
with a spacing of ``len(x) / num * (spacing of x)``. Because a
Fourier method is used, the signal is assumed to be periodic.
Parameters
----------
x : array_like
The data to be resampled.
num : int
The number of samples in the resampled signal.
t : array_like, optional
If `t` is given, it is assumed to be the sample positions
associated with the signal data in `x`.
axis : int, optional
The axis of `x` that is resampled. Default is 0.
window : array_like, callable, string, float, or tuple, optional
Specifies the window applied to the signal in the Fourier
domain. See below for details.
Returns
-------
resampled_x or (resampled_x, resampled_t)
Either the resampled array, or, if `t` was given, a tuple
containing the resampled array and the corresponding resampled
positions.
Notes
-----
The argument `window` controls a Fourier-domain window that tapers
the Fourier spectrum before zero-padding to alleviate ringing in
the resampled values for sampled signals you didn't intend to be
interpreted as band-limited.
If `window` is a function, then it is called with a vector of inputs
indicating the frequency bins (i.e. fftfreq(x.shape[axis]) ).
If `window` is an array of the same length as `x.shape[axis]` it is
assumed to be the window to be applied directly in the Fourier
domain (with dc and low-frequency first).
For any other type of `window`, the function `scipy.signal.get_window`
is called to generate the window.
The first sample of the returned vector is the same as the first
sample of the input vector. The spacing between samples is changed
from dx to:
dx * len(x) / num
If `t` is not None, then it represents the old sample positions,
and the new sample positions will be returned as well as the new
samples.
"""
正如您所知，8000 Hz意味着您的信号的一秒钟包含8000个样本，而对于44100 Hz，这意味着一秒钟包含44100个样本。
然后，只计算8000 Hz需要多少个样本，并使用该数字作为scipy.signal.resample（）的第二个参数。
你可以使用Nathan Whitehead在重新采样函数中使用的方法，我在其他答案中进行了修改（使用缩放），
或经历时间，即
secs = len(X)/44100.0 # Number of seconds in signal X
samps = secs*8000 # Number of samples to downsample
Y = scipy.signal.resample(X, samps)
答案 1 :(得分：0)
这是我从Nathan Whitehead编写的SWMixer模块中选择的：
import numpy
def resample(smp, scale=1.0):
"""Resample a sound to be a different length
Sample must be mono. May take some time for longer sounds
sampled at 44100 Hz.
Keyword arguments:
scale - scale factor for length of sound (2.0 means double length)
"""
# f*ing cool, numpy can do this with one command
# calculate new length of sample
n = round(len(smp) * scale)
# use linear interpolation
# endpoint keyword means than linspace doesn't go all the way to 1.0
# If it did, there are some off-by-one errors
# e.g. scale=2.0, [1,2,3] should go to [1,1.5,2,2.5,3,3]
# but with endpoint=True, we get [1,1.4,1.8,2.2,2.6,3]
# Both are OK, but since resampling will often involve
# exact ratios (i.e. for 44100 to 22050 or vice versa)
# using endpoint=False gets less noise in the resampled sound
return numpy.interp(
numpy.linspace(0.0, 1.0, n, endpoint=False), # where to interpret
numpy.linspace(0.0, 1.0, len(smp), endpoint=False), # known positions
smp, # known data points
)
所以，如果你使用scipy，这意味着你也有numpy。如果scipy不是＆＃34;必须＃，使用它，它可以很好地工作。
答案 2 :(得分：0)
如果要重新采样某个文件夹中的所有wav文件，请使用以下代码：
from tqdm.notebook import tqdm
from scipy.io import wavfile
from scipy.signal import resample
def scip_rs(file):
try:
sr, wv = wavfile.read(file)
if sr==16000:
pass
else:
sec = len(wv)/sr
nsmp = int(sec*16000)+1
rwv = resample(wv, nsmp)
wavfile.write(file, 16000, rwv/(2**15-1))
except Exception as e:
print('Error in file : ', file)
for file in tqdm(np.array(df.filepath.tolist())):
scip_rs(file)