问题描述
我有一个信号不是等距采样的；进行进一步处理.我以为scipy.signal.resample可以做到，但我不了解它的行为.
I have a signal that is not sampled equidistant; for further processing it needs to be. I thought that scipy.signal.resample would do it, but I do not understand its behavior.
信号在y中，对应的时间在x中.预计将在yy进行重新采样，所有相应的时间在xx中.有人知道我做错了什么或如何实现我所需要的吗?
The signal is in y, corresponding time in x.The resampled is expected in yy, with all corresponding time in xx. Does anyone know what I do wrong or how to achieve what I need?
此代码不起作用:xx不是时间:
This code does not work: xx is not time:
import numpy as np
from scipy import signal
import matplotlib.pyplot as plt
x = np.array([0,1,2,3,4,5,6,6.5,7,7.5,8,8.5,9])
y = np.cos(-x**2/4.0)
num=50
z=signal.resample(y, num, x, axis=0, window=None)
yy=z[0]
xx=z[1]
plt.plot(x,y)
plt.plot(xx,yy)
plt.show()
推荐答案
即使您提供x坐标(与t参数相对应)，也会 resample 假定采样是统一的.
Even when you give the x coordinates (which corresponds to the t argument), resample assumes that the sampling is uniform.
请考虑在 scipy.interpolate 中使用单变量插值器之一.
Consider using one of the univariate interpolators in scipy.interpolate.
例如，此脚本:
import numpy as np
from scipy import interpolate
import matplotlib.pyplot as plt
x = np.array([0,1,2,3,4,5,6,6.5,7,7.5,8,8.5,9])
y = np.cos(-x**2/4.0)
f = interpolate.interp1d(x, y)
num = 50
xx = np.linspace(x[0], x[-1], num)
yy = f(xx)
plt.plot(x,y, 'bo-')
plt.plot(xx,yy, 'g.-')
plt.show()
生成此图:
检查 interp1d 用于控制插值的选项，并查看其他插值类.
Check the docstring of interp1d for options to control the interpolation, and also check out the other interpolation classes.
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