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

Zheng picture Zheng · Nov 18, 2017 · Viewed 13.5k times · Source

I am trying to plot points + smooth line using spline. But the line "overshoots" some points, e.g in following codes, over the point 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)

how do I fix it?

Answer

Joshua R. picture Joshua R. · Nov 18, 2017

You might try the using interp1d in scipy.interpolate:

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)

which yields:

enter image description here

some other options for the kind parameter are in the docs:

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’.