Smooth curved line between 3 points in plot

Question

I have 3 data points on the x axis and 3 on the y axis:

x = [1,3,5]
y=[0,5,0]

I would like a curved line that starts at (1,0), goes to the highest point at (3,5) and then finishes at (5,0)

I think I need to use interpolation, but unsure how. If I use spline from scipy like this:

import bokeh.plotting as bk
from scipy.interpolate import spline
p = bk.figure()
xvals=np.linspace(1, 5, 10)
y_smooth = spline(x,y,xvals)
p.line(xvals, y_smooth)

bk.show(p)

I get the highest point before (3,5) and it looks unbalanced:

Doing exactly what you do but plotting it with matplotlib (can't get bokeh to work because p is undefined) get's me a perfectly symmetric graph. — Nathan, Feb 19 '18 at 23:10
@Nathan hmm, thanks. so it might be related to bokeh. I updated with the value of p, but yeah it might be just bokeh — Claudiu Creanga, Feb 19 '18 at 23:26
@Nathan: That's odd, because that is **not** what I see. I get an asymmetrical graph, just as the OP. This has to do with how `spline`works, and not the bokeh/matplotlib. — JohanL, Feb 20 '18 at 05:46
@JohanL I did it again and now I also get an assymetric plot... Sorry about that — Nathan, Feb 21 '18 at 16:11

JohanL · Accepted Answer · 2018-02-20T11:26:08.067

The issue is due to that spline with no extra argument is of order 3. That means that you do not have points/equations enough to get a spline curve (which manifests itself as a warning of an ill-conditioned matrix). You need to apply a spline of lower order, such as a cubic spline, which is of order 2:

import bokeh.plotting as bk
from scipy.interpolate import spline
p = bk.figure()
xvals=np.linspace(1, 5, 10)
y_smooth = spline(x,y,xvals, order=2) # This fixes your immediate problem
p.line(xvals, y_smooth)

bk.show(p)

In addition, spline is deprecated in SciPy, so you should preferably not use it, even if it is possible. A better solution is to use the CubicSpline class:

import bokeh.plotting as bk
from scipy.interpolate import CubicSpline
p = bk.figure()
xvals=np.linspace(1, 5, 10)
spl = CubicSpline(x, y) # First generate spline function
y_smooth = spl(xvals) # then evalute for your interpolated points
p.line(xvals, y_smooth)

bk.show(p)

Just to show the difference (using pyplot):

As can be seen, the CubicSpline is identical to the spline of order=2

score 4 · Answer 2 · answered Feb 20 '18 at 01:53

4

use pchip_interpolate():

import numpy as np
from scipy import interpolate

x = [1,3,5]
y=[0,5,0]

x2 = np.linspace(x[0], x[-1], 100)
y2 = interpolate.pchip_interpolate(x, y, x2)
pl.plot(x2, y2)
pl.plot(x, y, "o")

the result:

answered Feb 20 '18 at 01:53

HYRY

94,853
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for my actual purpose the pchip provides the best results – Aroc Apr 17 '23 at 15:28

score 2 · Answer 3 · answered Feb 19 '18 at 23:56

You can use quadratic interpolation. This is possible by making use of scipy.interpolate.interp1d.

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



x = [1, 3, 5]
y = [0, 5, 0]

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

x_interpol = np.linspace(1, 5, 1000)
y_interpol = f(x_interpol)

plt.plot(x_interpol, y_interpol)
plt.show()

Check the documentation for more details.

Smooth curved line between 3 points in plot

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