我们如何使用 scipy.signal.resample 将语音信号从 44100 下采样到 8000 Hz 信号？

Question

fs, s = wav.read('wave.wav')

该信号的采样频率为 44100 Hz，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，我们如何使用该函数对语音信号进行重新映射？

我们如何scipy.signal.resample在 python 中将语音信号从 44100 下采样到 8000 Hz？

Answer 1

好的，另一个解决方案，这个解决方案是真正的 scipy。正是所求。

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.

"""

您应该知道，8000 Hz 意味着您的信号的一秒包含 8000 个样本，而对于 44100 Hz，则意味着一秒包含 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 Nathan Whitehead 在我在其他答案中复制的重采样函数中使用的方法（带缩放），

或者穿越时空即

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)

Answer 2

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，如果你使用 scipy，那就意味着你也有 numpy。如果 scipy 不是“必须#，请使用它，它可以完美运行。