fs, s = wav.read('wave.wav')
该信号具有 44100 Hz 采样频率，我想使用 scipy.signal.resample(s,s.size/5.525)但是第二个元素不能浮点，那么，我们如何使用这个函数来重新映射语音信号呢？
我们如何在python中使用scipy.signal.resample将语音信号从44100下采样到8000 Hz？
好吧，另一个解决方案，这个真正的 scipy。正是所要求的。
这是 scipy.signal.resample(( 的文档字符串：
"""
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.
"""
Hz，这意味着一秒钟包含 44100 个样本。
然后，只需计算 8000 Hz 需要多少个样本，并将该数字用作 scipy.signal.resample(( 的第二个参数。
您可以使用内森·怀特黑德(Nathan Whitehead(在重采样函数中使用的方法，我在其他答案中复制了该方法(带有缩放(，
或穿越时间，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)
这是我
从Nathan Whitehead编写的SWMixer模块中挑选的：
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，如果你正在使用scipy，这意味着你也有numpy。如果 scipy 不是"必须#，请使用这个，它可以完美地工作。
如果要
对某个文件夹中的所有 WAV 文件重新采样，请执行此代码：
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)