在python中，绘图时如何用平滑的线连接点？

Question

我正在尝试使用样条曲线绘制点 + 平滑线。但是 "overshoots" 行有些点，例如在下面的代码中，超过了点 0.85。

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)

我该如何解决？

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)

产生：

kind 参数的一些其他选项在文档中：

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’.

Answer 2

您也可以尝试 thin plate spline 使用来自 scipy 的径向基函数插值：

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

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)

rbf = Rbf(x, y, function = 'thin_plate', smooth = 0.001)
y_smooth = rbf(x_new)

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

可以通过改变 smooth 参数来获得更好的数据近似值。

要考虑的替代 function 参数值包括 'multiquadric'、'inverse'、'gaussian'、'linear'、'cubic' 和 'quintic'.在考虑 function 值时，我通常先尝试 'thin_plate'，然后再尝试 'cubic'。

检查 scipy.interpolate.Rbf docs 中的其他 Rbf 选项。

在python中，绘图时如何用平滑的线连接点？

in python, how to connect points with smooth line in plotting?

python

matplotlib

spline