python – 我们如何使用scipy.signal.resample对44100到8000 Hz信号的语音信号进行下采样？
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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？
我们如何使用scipy.signal.resample在python中对44100到8000 Hz的语音信号进行下采样？
解决方法:
好吧,那么,另一个解决方案,这个与scipy真实.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.
"""
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需要多少样本,并将该数字用作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),
或经过时间,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)
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