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:
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
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:
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.
Related
I've got the following simple script that plots a graph:
import matplotlib.pyplot as plt
import numpy as np
T = np.array([6, 7, 8, 9, 10, 11, 12])
power = np.array([1.53E+03, 5.92E+02, 2.04E+02, 7.24E+01, 2.72E+01, 1.10E+01, 4.70E+00])
plt.plot(T,power)
plt.show()
As it is now, the line goes straight from point to point which looks ok, but could be better in my opinion. What I want is to smooth the line between the points. In Gnuplot I would have plotted with smooth cplines.
Is there an easy way to do this in PyPlot? I've found some tutorials, but they all seem rather complex.
You could use scipy.interpolate.spline to smooth out your data yourself:
from scipy.interpolate import spline
# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300)
power_smooth = spline(T, power, xnew)
plt.plot(xnew,power_smooth)
plt.show()
spline is deprecated in scipy 0.19.0, use BSpline class instead.
Switching from spline to BSpline isn't a straightforward copy/paste and requires a little tweaking:
from scipy.interpolate import make_interp_spline, BSpline
# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300)
spl = make_interp_spline(T, power, k=3) # type: BSpline
power_smooth = spl(xnew)
plt.plot(xnew, power_smooth)
plt.show()
Before:
After:
For this example spline works well, but if the function is not smooth inherently and you want to have smoothed version you can also try:
from scipy.ndimage.filters import gaussian_filter1d
ysmoothed = gaussian_filter1d(y, sigma=2)
plt.plot(x, ysmoothed)
plt.show()
if you increase sigma you can get a more smoothed function.
Proceed with caution with this one. It modifies the original values and may not be what you want.
See the scipy.interpolate documentation for some examples.
The following example demonstrates its use, for linear and cubic spline interpolation:
import matplotlib.pyplot as plt
import numpy as np
from scipy.interpolate import interp1d
# Define x, y, and xnew to resample at.
x = np.linspace(0, 10, num=11, endpoint=True)
y = np.cos(-x**2/9.0)
xnew = np.linspace(0, 10, num=41, endpoint=True)
# Define interpolators.
f_linear = interp1d(x, y)
f_cubic = interp1d(x, y, kind='cubic')
# Plot.
plt.plot(x, y, 'o', label='data')
plt.plot(xnew, f_linear(xnew), '-', label='linear')
plt.plot(xnew, f_cubic(xnew), '--', label='cubic')
plt.legend(loc='best')
plt.show()
Slightly modified for increased readability.
One of the easiest implementations I found was to use that Exponential Moving Average the Tensorboard uses:
def smooth(scalars: List[float], weight: float) -> List[float]: # Weight between 0 and 1
last = scalars[0] # First value in the plot (first timestep)
smoothed = list()
for point in scalars:
smoothed_val = last * weight + (1 - weight) * point # Calculate smoothed value
smoothed.append(smoothed_val) # Save it
last = smoothed_val # Anchor the last smoothed value
return smoothed
ax.plot(x_labels, smooth(train_data, .9), x_labels, train_data)
I presume you mean curve-fitting and not anti-aliasing from the context of your question. PyPlot doesn't have any built-in support for this, but you can easily implement some basic curve-fitting yourself, like the code seen here, or if you're using GuiQwt it has a curve fitting module. (You could probably also steal the code from SciPy to do this as well).
Here is a simple solution for dates:
from scipy.interpolate import make_interp_spline
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as dates
from datetime import datetime
data = {
datetime(2016, 9, 26, 0, 0): 26060, datetime(2016, 9, 27, 0, 0): 23243,
datetime(2016, 9, 28, 0, 0): 22534, datetime(2016, 9, 29, 0, 0): 22841,
datetime(2016, 9, 30, 0, 0): 22441, datetime(2016, 10, 1, 0, 0): 23248
}
#create data
date_np = np.array(list(data.keys()))
value_np = np.array(list(data.values()))
date_num = dates.date2num(date_np)
# smooth
date_num_smooth = np.linspace(date_num.min(), date_num.max(), 100)
spl = make_interp_spline(date_num, value_np, k=3)
value_np_smooth = spl(date_num_smooth)
# print
plt.plot(date_np, value_np)
plt.plot(dates.num2date(date_num_smooth), value_np_smooth)
plt.show()
It's worth your time looking at seaborn for plotting smoothed lines.
The seaborn lmplot function will plot data and regression model fits.
The following illustrates both polynomial and lowess fits:
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
T = np.array([6, 7, 8, 9, 10, 11, 12])
power = np.array([1.53E+03, 5.92E+02, 2.04E+02, 7.24E+01, 2.72E+01, 1.10E+01, 4.70E+00])
df = pd.DataFrame(data = {'T': T, 'power': power})
sns.lmplot(x='T', y='power', data=df, ci=None, order=4, truncate=False)
sns.lmplot(x='T', y='power', data=df, ci=None, lowess=True, truncate=False)
The order = 4 polynomial fit is overfitting this toy dataset. I don't show it here but order = 2 and order = 3 gave worse results.
The lowess = True fit is underfitting this tiny dataset but may give better results on larger datasets.
Check the seaborn regression tutorial for more examples.
Another way to go, which slightly modifies the function depending on the parameters you use:
from statsmodels.nonparametric.smoothers_lowess import lowess
def smoothing(x, y):
lowess_frac = 0.15 # size of data (%) for estimation =~ smoothing window
lowess_it = 0
x_smooth = x
y_smooth = lowess(y, x, is_sorted=False, frac=lowess_frac, it=lowess_it, return_sorted=False)
return x_smooth, y_smooth
That was better suited than other answers for my specific application case.
I've got the following simple script that plots a graph:
import matplotlib.pyplot as plt
import numpy as np
T = np.array([6, 7, 8, 9, 10, 11, 12])
power = np.array([1.53E+03, 5.92E+02, 2.04E+02, 7.24E+01, 2.72E+01, 1.10E+01, 4.70E+00])
plt.plot(T,power)
plt.show()
As it is now, the line goes straight from point to point which looks ok, but could be better in my opinion. What I want is to smooth the line between the points. In Gnuplot I would have plotted with smooth cplines.
Is there an easy way to do this in PyPlot? I've found some tutorials, but they all seem rather complex.
You could use scipy.interpolate.spline to smooth out your data yourself:
from scipy.interpolate import spline
# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300)
power_smooth = spline(T, power, xnew)
plt.plot(xnew,power_smooth)
plt.show()
spline is deprecated in scipy 0.19.0, use BSpline class instead.
Switching from spline to BSpline isn't a straightforward copy/paste and requires a little tweaking:
from scipy.interpolate import make_interp_spline, BSpline
# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300)
spl = make_interp_spline(T, power, k=3) # type: BSpline
power_smooth = spl(xnew)
plt.plot(xnew, power_smooth)
plt.show()
Before:
After:
For this example spline works well, but if the function is not smooth inherently and you want to have smoothed version you can also try:
from scipy.ndimage.filters import gaussian_filter1d
ysmoothed = gaussian_filter1d(y, sigma=2)
plt.plot(x, ysmoothed)
plt.show()
if you increase sigma you can get a more smoothed function.
Proceed with caution with this one. It modifies the original values and may not be what you want.
See the scipy.interpolate documentation for some examples.
The following example demonstrates its use, for linear and cubic spline interpolation:
import matplotlib.pyplot as plt
import numpy as np
from scipy.interpolate import interp1d
# Define x, y, and xnew to resample at.
x = np.linspace(0, 10, num=11, endpoint=True)
y = np.cos(-x**2/9.0)
xnew = np.linspace(0, 10, num=41, endpoint=True)
# Define interpolators.
f_linear = interp1d(x, y)
f_cubic = interp1d(x, y, kind='cubic')
# Plot.
plt.plot(x, y, 'o', label='data')
plt.plot(xnew, f_linear(xnew), '-', label='linear')
plt.plot(xnew, f_cubic(xnew), '--', label='cubic')
plt.legend(loc='best')
plt.show()
Slightly modified for increased readability.
One of the easiest implementations I found was to use that Exponential Moving Average the Tensorboard uses:
def smooth(scalars: List[float], weight: float) -> List[float]: # Weight between 0 and 1
last = scalars[0] # First value in the plot (first timestep)
smoothed = list()
for point in scalars:
smoothed_val = last * weight + (1 - weight) * point # Calculate smoothed value
smoothed.append(smoothed_val) # Save it
last = smoothed_val # Anchor the last smoothed value
return smoothed
ax.plot(x_labels, smooth(train_data, .9), x_labels, train_data)
I presume you mean curve-fitting and not anti-aliasing from the context of your question. PyPlot doesn't have any built-in support for this, but you can easily implement some basic curve-fitting yourself, like the code seen here, or if you're using GuiQwt it has a curve fitting module. (You could probably also steal the code from SciPy to do this as well).
Here is a simple solution for dates:
from scipy.interpolate import make_interp_spline
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.dates as dates
from datetime import datetime
data = {
datetime(2016, 9, 26, 0, 0): 26060, datetime(2016, 9, 27, 0, 0): 23243,
datetime(2016, 9, 28, 0, 0): 22534, datetime(2016, 9, 29, 0, 0): 22841,
datetime(2016, 9, 30, 0, 0): 22441, datetime(2016, 10, 1, 0, 0): 23248
}
#create data
date_np = np.array(list(data.keys()))
value_np = np.array(list(data.values()))
date_num = dates.date2num(date_np)
# smooth
date_num_smooth = np.linspace(date_num.min(), date_num.max(), 100)
spl = make_interp_spline(date_num, value_np, k=3)
value_np_smooth = spl(date_num_smooth)
# print
plt.plot(date_np, value_np)
plt.plot(dates.num2date(date_num_smooth), value_np_smooth)
plt.show()
It's worth your time looking at seaborn for plotting smoothed lines.
The seaborn lmplot function will plot data and regression model fits.
The following illustrates both polynomial and lowess fits:
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
T = np.array([6, 7, 8, 9, 10, 11, 12])
power = np.array([1.53E+03, 5.92E+02, 2.04E+02, 7.24E+01, 2.72E+01, 1.10E+01, 4.70E+00])
df = pd.DataFrame(data = {'T': T, 'power': power})
sns.lmplot(x='T', y='power', data=df, ci=None, order=4, truncate=False)
sns.lmplot(x='T', y='power', data=df, ci=None, lowess=True, truncate=False)
The order = 4 polynomial fit is overfitting this toy dataset. I don't show it here but order = 2 and order = 3 gave worse results.
The lowess = True fit is underfitting this tiny dataset but may give better results on larger datasets.
Check the seaborn regression tutorial for more examples.
Another way to go, which slightly modifies the function depending on the parameters you use:
from statsmodels.nonparametric.smoothers_lowess import lowess
def smoothing(x, y):
lowess_frac = 0.15 # size of data (%) for estimation =~ smoothing window
lowess_it = 0
x_smooth = x
y_smooth = lowess(y, x, is_sorted=False, frac=lowess_frac, it=lowess_it, return_sorted=False)
return x_smooth, y_smooth
That was better suited than other answers for my specific application case.
I've plotted a 3-d mesh in Matlab by below little m-file:
[x,n] = meshgrid(0:0.1:20, 1:1:100);
mu = 0;
sigma = sqrt(2)./n;
f = normcdf(x,mu,sigma);
mesh(x,n,f);
I am going to acquire the same result by utilization of Python and its corresponding modules, by below code snippet:
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt
sigma = 1
def integrand(x, n):
return (n/(2*sigma*np.sqrt(np.pi)))*np.exp(-(n**2*x**2)/(4*sigma**2))
tt = np.linspace(0, 20, 2000)
nn = np.linspace(1, 100, 100)
T = np.zeros([len(tt), len(nn)])
for i,t in enumerate(tt):
for j,n in enumerate(nn):
T[i, j], _ = quad(integrand, -np.inf, t, args=(n,))
x, y = np.mgrid[0:20:0.01, 1:101:1]
plt.pcolormesh(x, y, T)
plt.show()
But the output of the Python is is considerably different with the Matlab one, and as a matter of fact is unacceptable.
I am afraid of wrong utilization of the functions just like linespace, enumerate or mgrid...
Does somebody have any idea about?!...
PS. Unfortunately, I couldn't insert the output plots within this thread...!
Best
..............................
Edit: I changed the linespace and mgrid intervals and replaced plot_surface method... The output is 3d now with the suitable accuracy and smoothness...
From what I see the equivalent solution would be:
import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
x, n = np.mgrid[0:20:0.01, 1:100:1]
mu = 0
sigma = np.sqrt(2)/n
f = norm.cdf(x, mu, sigma)
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_surface(x, n, f, rstride=x.shape[0]//20, cstride=x.shape[1]//20, alpha=0.3)
plt.show()
Unfortunately 3D plotting with matplotlib is not as straight forward as with matlab.
Here is the plot from this code:
Your Matlab code generate 201 points through x:
[x,n] = meshgrid(0:0.1:20, 1:1:100);
While your Python code generate only 20 points:
tt = np.linspace(0, 19, 20)
Maybe it's causing accuracy problems?
Try this code:
tt = np.linspace(0, 20, 201)
The seminal points to resolve the problem was:
1- Necessity of the equivalence regarding the provided dimensions of the linespace and mgrid functions...
2- Utilization of a mesh with more density to make a bee line into a high degree of smoothness...
3- Application of a 3d plotter function, like plot_surf...
The current code is totally valid...
I use Matplotlib with Python 2.7.6 and my code
import matplotlib.pyplot as plt
import numpy as np
from scipy.interpolate import spline
x = np.array([1,2,3])
y = np.array([1,5,15])
x_smooth = np.linspace(x.min(), x.max(), 100)
y_smooth = spline(x, y, x_smooth)
plt.plot(x_smooth, y_smooth)
plt.show()
When I run it show image
How to get the length of Spline ? Help me
import math
distance = 0
for count in range(1,len(y_smooth)):
distance += math.sqrt(math.pow(x_smooth[count]-x_smooth[count-1],2) + math.pow(y_smooth[count]-y_smooth[count-1],2))
in cartesian geometry the distance between two points is calculated as
p1(x,y), p2(a,b)
[p1p2] = sqrt((a-x)^2 + (b-y)^2)
I'm sure there is a more elegant way to do this, probably in a one-liner
It appears as if matplotlib.tri.Triangulation uses a buggy and possibly incorrect implementation of Delaunay triangulation that is due to be replaced by qHull.
I'm trying to plot a trisurf using mpl_toolkits.mplot3d.plot_trisurf() and running into a bunch of exceptions that are unhelpful (IndexErrors and KeyErrors mostly, with no indication of what exactly went wrong).
Since scipy.spatial.Delaunay already uses qHull, I was wondering if there was a way to build a matplotlib.tri.Triangulation object for use with mpl_toolkits.mplot3d.plot_trisurf() using scipy's implementation of Delaunay triangulation.
I've tried passing the delaunay.points directly to matplotlib.tri.Triangulate via the triangles parameter, but this results in a ValueError: triangles min element is out of bounds.
http://docs.scipy.org/doc/scipy-0.13.0/reference/generated/scipy.spatial.Delaunay.html
http://matplotlib.org/dev/api/tri_api.html
So you need to pass both the points and the triangles from qhull to the Triangulation constructor:
import numpy as np
import scipy.spatial
import matplotlib
import math
import matplotlib.tri as mtri
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
# First create the x and y coordinates of the points.
n_angles = 20
n_radii = 10
min_radius = 0.15
radii = np.linspace(min_radius, 0.95, n_radii)
angles = np.linspace(0, 2*math.pi, n_angles, endpoint=False)
angles = np.repeat(angles[..., np.newaxis], n_radii, axis=1)
angles[:, 1::2] += math.pi/n_angles
x = (radii*np.cos(angles)).flatten()
y = (radii*np.sin(angles)).flatten()
# Create the Delaunay tessalation using scipy.spatial
pts = np.vstack([x, y]).T
tess = scipy.spatial.Delaunay(pts)
# Create the matplotlib Triangulation object
x = tess.points[:, 0]
y = tess.points[:, 1]
tri = tess.vertices # or tess.simplices depending on scipy version
triang = mtri.Triangulation(x=pts[:, 0], y=pts[:, 1], triangles=tri)
# Plotting
z = x*x + y*y
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_trisurf(triang, z)
plt.show()
output (with matplotlib current master):
#Marco was curious to know how to run this for a 2d array. I hope this would be useful. The list of vertices according to coordinates should be made an array and can be tessellated using mtri.Triangulation.
Sample code below:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.tri as mtri
verts = np.array([[0.6,0.8],[0.2,0.9],[0.1,-0.5],[0.2,-2]])
triang = mtri.Triangulation(verts[:,0], verts[:,1])
plt.triplot(triang, marker="o")
plt.show()`enter code here`