在python中，如何在绘图中用平滑线连接点？

Question

我试图用样条曲线绘制点+平滑线。但这条线“超出”了一些点，例如在下面的代码中，超过了0.85点。

import numpy as np 
import matplotlib.pyplot as plt
from scipy.interpolate import spline

x=np.array([0.1, 0.3, 0.5, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7, 1.9, 2])
y=np.array([0.57,0.85,0.66,0.84,0.59,0.55,0.61,0.76,0.54,0.55,0.48])

x_new = np.linspace(x.min(), x.max(),500)
y_smooth = spline(x, y, x_new)

plt.plot (x_new,y_smooth)
plt.scatter (x, y)

我该怎么修？

Answer 1

您可以尝试在scipy.interpolate中使用interp1d：

import numpy as np 
import matplotlib.pyplot as plt
from scipy.interpolate import interp1d

x=np.array([0.1, 0.3, 0.5, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7, 1.9, 2])
y=np.array([0.57,0.85,0.66,0.84,0.59,0.55,0.61,0.76,0.54,0.55,0.48])

x_new = np.linspace(x.min(), x.max(),500)

f = interp1d(x, y, kind='quadratic')
y_smooth=f(x_new)

plt.plot (x_new,y_smooth)
plt.scatter (x, y)

结果是：

文档中提供了kind参数的其他一些选项：

kind : str or int, optional Specifies the kind of interpolation as a string (‘linear’, ‘nearest’, ‘zero’, ‘slinear’, ‘quadratic’, ‘cubic’ where ‘zero’, ‘slinear’, ‘quadratic’ and ‘cubic’ refer to a spline interpolation of zeroth, first, second or third order) or as an integer specifying the order of the spline interpolator to use. Default is ‘linear’.