I am trying to interpolate a cumulated distribution of e.g. i) number of people to ii) number of owned cars, showing that e.g. the top 20% of people own much more than 20% of all cars - off course 100% of people own 100% of cars. Also I know that there are e.g. 100mn people and 200mn cars.
Now coming to my code:
#import libraries (more than required here)
import pandas as pd
from scipy import interpolate
from scipy.interpolate import interp1d
from sympy import symbols, solve, Eq
import matplotlib.pyplot as plt
from matplotlib import pyplot as plt
%matplotlib inline
import plotly.express as px
from scipy import interpolate
curve=pd.read_excel('inputs.xlsx',sheet_name='inputdata')
Input data: Curveplot (cumulated people (x) on the left // cumulated cars (y) on the right)
#Input data in list form (I am not sure how to interpolate from a list for the moment)
cumulatedpeople = [0, 0.453086, 0.772334, 0.950475, 0.978981, 0.999876, 0.999990, 1]
cumulatedcars= [0, 0.016356, 0.126713, 0.410482, 0.554976, 0.950073, 0.984913, 1]
x, y = points[:,0], points[:,1]
interpolation = interp1d(x, y, kind = 'cubic')
number_of_people_mn= 100000000
oneperson = 1 / number_of_people_mn
dataset = pd.DataFrame(range(number_of_people_mn + 1))
dataset.columns = ["nr_of_one_person"]
dataset.drop(dataset.index[:1], inplace=True)
#calculating the position of every single person on the cumulated x-axis (between 0 and 1)
dataset["cumulatedpeople"] = dataset["nr_of_one_person"] / number_of_people_mn
#finding the "cumulatedcars" to the "cumulatedpeople" via interpolation (between 0 and 1)
dataset["cumulatedcars"] = interpolation(dataset["cumulatedpeople"])
plt.plot(dataset["cumulatedpeople"], dataset["cumulatedcars"])
plt.legend(['Cubic interpolation'], loc = 'best')
plt.xlabel('Cumulated people')
plt.ylabel('Cumulated cars')
plt.title("People-to-car cumulated curve")
plt.show()
However when looking at the actual plot, I get the following result which is false: Cubic interpolation
In fact, the curve should look almost like the one from a linear interpolation with the exact same input data - however this is not accurate enough for my purpose: Linear interpolation
Is there any relevant step I am missing out or what would be the best way to get an accurate interpolation from the inputs that almost looks like the one from a linear interpolation?
Short answer: your code is doing the right thing, but the data is unsuitable for cubic interpolation.
Let me explain. Here is your code that I simplified for clarity
from scipy.interpolate import interp1d
from matplotlib import pyplot as plt
cumulatedpeople = [0, 0.453086, 0.772334, 0.950475, 0.978981, 0.999876, 0.999990, 1]
cumulatedcars= [0, 0.016356, 0.126713, 0.410482, 0.554976, 0.950073, 0.984913, 1]
interpolation = interp1d(cumulatedpeople, cumulatedcars, kind = 'cubic')
number_of_people_mn= 100#000000
cumppl = np.arange(number_of_people_mn + 1)/number_of_people_mn
cumcars = interpolation(cumppl)
plt.plot(cumppl, cumcars)
plt.plot(cumulatedpeople, cumulatedcars,'o')
plt.show()
note the last couple of lines -- I am plotting, on the same graph, both the interpolated results and the input date. Here is the result
orange dots are the original data, blue line is cubic interpolation. The interpolator passes through all the points so technically is doing the right thing
Clearly it is not doing what you would want
The reason for such strange behavior is mostly at the right end where you have a few x-points that are very close together -- the interpolator produces massive wiggles trying to fit very closely spaced points.
If I remove two right-most points from the interpolator:
interpolation = interp1d(cumulatedpeople[:-2], cumulatedcars[:-2], kind = 'cubic')
it looks a bit more reasonable:
But still one would argue linear interpolation is better. The wiggles on the left end now because the gaps between initial x-poonts are too large
The moral here is that cubic interpolation should really be used only if gaps between x points are roughly the same
Your best bet here, I think, is to use something like curve_fit
a related discussion can be found here
specifically monotone interpolation as explained here yields good results on your data. Copying the relevant bits here, you would replace the interpolator with
from scipy.interpolate import pchip
interpolation = pchip(cumulatedpeople, cumulatedcars)
and get a decent-looking fit:
Related
I am trying to to calculate and plot the numerical derivative (dy/dx) from two lists x and y. I am using the scipy.interpolate.UnivariateSpline and scipy.interpolate.UnivariateSpline.derivative to compute the slope. The plot of y vs x seems to be C1 continuous and I was expecting the slope dy/dx to be smooth as well when plotted against x. But then what is causing the little bump in the plot here? Also any suggestion on how I can massage the code to make it C1 continuous?
import numpy as np
from matplotlib import pyplot as plt
from scipy.interpolate import UnivariateSpline
x=[20.14141131550861, 20.29161104293003, 20.458574567775457, 20.653802880772922, 20.910446090013004, 21.404599384233677, 21.427939384233678, 21.451279384233676, 21.474619384233677, 21.497959384233678, 21.52129938423368, 21.52130038423368, 21.54463938423368, 21.56797938423368, 21.59131938423368, 21.61465938423368, 21.63799938423368, 22.132152678454354, 22.388795887694435, 22.5840242006919]
y=[-1.6629252348586834, -1.7625046339166028, -1.875358801338162, -2.01040013818419, -2.193327440415778, -2.5538174545988306, -2.571799827167608, -2.5896274995868005, -2.607298426787476, -2.624811539182082, -2.642165776735291, -2.642165776735291, -2.659360089028171, -2.6763934353217587, -2.693264784620056, -2.7099731157324367, -2.7265165368570314, -3.0965791078676754, -3.290845721407758, -3.440799238587583]
spl1 = UnivariateSpline(x,y,s=0)
dydx = spl1.derivative(n=1)
T = dydx(x)
plt.plot(x,y,'-x')
plt.plot(x,T,'-')
plt.show()
The given data points look like they define a nice C1-smooth curve, but they do not. Plotting the slopes (difference of y over difference of x) shows this:
plt.plot(np.diff(y)/np.diff(x))
There are some duplicate values of y in the array, which look like they don't belong, also some near-duplicate (but not duplicate) values of x.
The easiest way to fix the spline is to allow a tiny bit of smoothing:
spl1 = UnivariateSpline(x, y, s=1e-5)
makes the derivative what you expected:
Removing the "bad apple" also helps, though not as much.
spl1 = UnivariateSpline(x[:10] + x[11:], y[:10] + y[11:], s=0)
I have the following graph that I want to digitize to a high-quality publication grade figure using Python and Matplotlib:
I used a digitizer program to grab a few samples from one of the 3 data sets:
x_data = np.array([
1,
1.2371,
1.6809,
2.89151,
5.13304,
9.23238,
])
y_data = np.array([
0.0688824,
0.0490012,
0.0332843,
0.0235889,
0.0222304,
0.0245952,
])
I have already tried 3 different methods of fitting a curve through these data points. The first method being to draw a spline through the points using scipy.interpolate import spline
This results in (with the actual data points drawn as blue markers):
This is obvisously no good.
My second attempt was to draw a curve fit using a series of different order polinimials using scipy.optimize import curve_fit. Even up to a fourth order polynomial the answer is useless (the lower order ones were even more useless):
Finally, I used scipy.interpolate import interp1d to try and interpolate between the data points. Linear interpolation obviously yields expected results but the line are straight and the whole purpose of this exercise is to get a nice smooth curve:
If I then use cubic interpolation I get a rubish result, however quadratic interpolation yields a slightly better result:
But it's not quite there yet, and I don't think interp1d can do higher order interpolation.
Is there anyone out there who has a good method of doing this? Maybe I would be better off trying to do it in IPE or something?
Thank you!
A standard cubic spline is not very good at reasonable looking interpolations between data points that are very unevenly spaced. Fortunately, there are plenty of other interpolation algorithms and Scipy provides a number of them. Here are a few applied to your data:
import numpy as np
from scipy.interpolate import spline, UnivariateSpline, Akima1DInterpolator, PchipInterpolator
import matplotlib.pyplot as plt
x_data = np.array([1, 1.2371, 1.6809, 2.89151, 5.13304, 9.23238])
y_data = np.array([0.0688824, 0.0490012, 0.0332843, 0.0235889, 0.0222304, 0.0245952])
x_data_smooth = np.linspace(min(x_data), max(x_data), 1000)
fig, ax = plt.subplots(1,1)
spl = UnivariateSpline(x_data, y_data, s=0, k=2)
y_data_smooth = spl(x_data_smooth)
ax.plot(x_data_smooth, y_data_smooth, 'b')
bi = Akima1DInterpolator(x_data, y_data)
y_data_smooth = bi(x_data_smooth)
ax.plot(x_data_smooth, y_data_smooth, 'g')
bi = PchipInterpolator(x_data, y_data)
y_data_smooth = bi(x_data_smooth)
ax.plot(x_data_smooth, y_data_smooth, 'k')
ax.plot(x_data_smooth, y_data_smooth)
ax.scatter(x_data, y_data)
plt.show()
I suggest looking through these, and also a few others, and finding one that matches what you think looks right. Also, though, you may want to sample a few more points. For example, I think the PCHIP algorithm wants to keep the fit monotonic between data points, so digitizing your minimum point would be useful (and probably a good idea regardless of the algorithm you use).
How can I plot the following noisy data with a smooth, continuous line without considering each individual value? I would like to only show the behavior in a nicer way, without caring about noisy and extreme values. This is the code I am using:
import numpy
import sys
import matplotlib.pyplot as plt
from scipy.interpolate import spline
dataset = numpy.genfromtxt(fname='data', delimiter=",")
dic = {}
for d in dataset:
dic[d[0]] = d[1]
plt.plot(range(len(dic)), dic.values(),linestyle='-', linewidth=2)
plt.savefig('plot.png')
plt.show()
In a previous answer, I was introduced to the Savitzky Golay filter, a particular type of low-pass filter, well adapted for data smoothing. How "smooth" you want your resulting curve to be is a matter of preference, and this can be adjusted by both the window-size and the order of the interpolating polynomial. Using the cookbook example for sg_filter:
import numpy as np
import sg_filter
import matplotlib.pyplot as plt
# Generate some sample data similar to your post
X = np.arange(1,1000,1)
Y = np.log(X**3) + 10*np.random.random(X.shape)
Y2 = sg_filter.savitzky_golay(Y, 101, 3)
plt.plot(X,Y,linestyle='-', linewidth=2,alpha=.5)
plt.plot(X,Y2,color='r')
plt.show()
There is more than one way to do it!
Here I show how to reduce noise using a variety of techniques:
Moving average
LOWESS regression
Low pass filter
Interpolation
Sticking with #Hooked example data for consistency:
import numpy as np
import matplotlib.pyplot as plt
X = np.arange(1, 1000, 1)
Y = np.log(X ** 3) + 10 * np.random.random(X.shape)
plt.plot(X, Y, alpha = .5)
plt.show()
Moving average
Sometimes all you need is a moving average.
For example, using pandas with a window size of 100:
import pandas as pd
df = pd.DataFrame(Y, X)
df_mva = df.rolling(100).mean() # moving average with a window size of 100
df_mva.plot(legend = False);
You will probably have to try several window sizes with your data. Note that the first 100 values of df_mva will be NaN but these can be removed with the dropna method.
Usage details for the pandas rolling function.
LOWESS regression
I've used LOWESS (Locally Weighted Scatterplot Smoothing) successfully to remove noise from repeated measures datasets. More information on local regression methods, including LOWESS and LOESS, here. It's a simple method with only one parameter to tune which in my experience gives good results.
Here is how to apply the LOWESS technique using the statsmodels implementation:
import statsmodels.api as sm
y_lowess = sm.nonparametric.lowess(Y, X, frac = 0.3) # 30 % lowess smoothing
plt.plot(y_lowess[:, 0], y_lowess[:, 1]) # some noise removed
plt.show()
It may be necessary to vary the frac parameter, which is the fraction of the data used when estimating each y value. Increase the frac value to increase the amount of smoothing. The frac value must be between 0 and 1.
Further details on statsmodels lowess usage.
Low pass filter
Scipy provides a set of low pass filters which may be appropriate.
After application of the lfiter:
from scipy.signal import lfilter
n = 50 # larger n gives smoother curves
b = [1.0 / n] * n # numerator coefficients
a = 1 # denominator coefficient
y_lf = lfilter(b, a, Y)
plt.plot(X, y_lf)
plt.show()
Check scipy lfilter documentation for implementation details regarding how numerator and denominator coefficients are used in the difference equations.
There are other filters in the scipy.signal package.
Interpolation
Finally, here is an example of radial basis function interpolation:
from scipy.interpolate import Rbf
rbf = Rbf(X, Y, function = 'multiquadric', smooth = 500)
y_rbf = rbf(X)
plt.plot(X, y_rbf)
plt.show()
Smoother approximation can be achieved by increasing the smooth parameter. Alternative function parameters to consider include 'cubic' and 'thin_plate'. When considering the function value, I usually try 'thin_plate' first followed by 'cubic'; however both 'thin_plate' and 'cubic' seemed to struggle with the noise in this dataset.
Check other Rbf options in the scipy docs. Scipy provides other univariate and multivariate interpolation techniques (see this tutorial).
I have some data over a 2D range that I am interested in analyzing. These data were originally in lists x,y, and z where z[i] was the value for the point located at (x[i],y[i]). I then interpolated this data onto a regular grid using
x=np.array(x)
y=np.array(y)
z=np.array(z)
xi=np.linspace(minx,maxx,100)
yi=np.linspace(miny,maxy,100)
zi=griddata(x,y,z,xi,yi)
I then plotted the xi,yi,zi data using
plt.contour(xi,yi,zi)
plt.pcolormesh(xi,yi,zi,cmap=plt.get_cmap('PRGn'),norm=plt.Normalize(-10,10),vmin=-10,vmax=10)
This produced this plot:
In this plot you can see the S-like curve where the values are equal to zero (aside: the data doesn't vary as rapidly as shown in the colorbar -- that's simply a result of me normalizing the data to -10-10 when it actually extends far beyond that range; I did this to make the zero-valued region show up better -- maybe there's a better way of doing this too...).
The scattered dots are simply the points at which I have original data (yes, in this case my data was already on a regular grid). What I'm curious about is whether there is a good way for me to extract the values for which the curve is zero and obtain x,y pairs that, if plotted as a line, would trace that zero-region in the colormesh. I could interpolate to a really fine grid and then just brute force search for the values which are closest to zero. But is there a more automatic way of doing this, or a more automatic way of plotting this "zero-line"?
And a secondary question: I am using griddata correctly, right? I have these simple 1D arrays although elsewhere people use various meshgrids, loading texts, etc., before calling griddata.
Here is a full example:
import numpy as np
import matplotlib.pyplot as plt
y, x = np.ogrid[-1.5:1.5:200j, -1.5:1.5:200j]
f = (x**2 + y**2)**4 - (x**2 - y**2)**2
plt.figure(figsize=(9,4))
plt.subplot(121)
extent = [np.min(x), np.max(x), np.min(y), np.max(y)]
cs = plt.contour(f, extent=extent, levels=[0.1],
colors=["b", "r"], linestyles=["solid", "dashed"], linewidths=[2, 2])
plt.subplot(122)
# get the points on the lines
for c in cs.collections:
data = c.get_paths()[0].vertices
plt.plot(data[:,0], data[:,1],
color=c.get_color()[0], linewidth=c.get_linewidth()[0])
plt.show()
here is the output:
I need to (numerically) calculate the first and second derivative of a function for which I've attempted to use both splrep and UnivariateSpline to create splines for the purpose of interpolation the function to take the derivatives.
However, it seems that there's an inherent problem in the spline representation itself for functions who's magnitude is order 10^-1 or lower and are (rapidly) oscillating.
As an example, consider the following code to create a spline representation of the sine function over the interval (0,6*pi) (so the function oscillates three times only):
import scipy
from scipy import interpolate
import numpy
from numpy import linspace
import math
from math import sin
k = linspace(0, 6.*pi, num=10000) #interval (0,6*pi) in 10'000 steps
y=[]
A = 1.e0 # Amplitude of sine function
for i in range(len(k)):
y.append(A*sin(k[i]))
tck =interpolate.UnivariateSpline(x, y, w=None, bbox=[None, None], k=5, s=2)
M=tck(k)
Below are the results for M for A = 1.e0 and A = 1.e-2
http://i.imgur.com/uEIxq.png Amplitude = 1
http://i.imgur.com/zFfK0.png Amplitude = 1/100
Clearly the interpolated function created by the splines is totally incorrect! The 2nd graph does not even oscillate the correct frequency.
Does anyone have any insight into this problem? Or know of another way to create splines within numpy/scipy?
Cheers,
Rory
I'm guessing that your problem is due to aliasing.
What is x in your example?
If the x values that you're interpolating at are less closely spaced than your original points, you'll inherently lose frequency information. This is completely independent from any type of interpolation. It's inherent in downsampling.
Nevermind the above bit about aliasing. It doesn't apply in this case (though I still have no idea what x is in your example...
I just realized that you're evaluating your points at the original input points when you're using a non-zero smoothing factor (s).
By definition, smoothing won't fit the data exactly. Try putting s=0 in instead.
As a quick example:
import matplotlib.pyplot as plt
import numpy as np
from scipy import interpolate
x = np.linspace(0, 6.*np.pi, num=100) #interval (0,6*pi) in 10'000 steps
A = 1.e-4 # Amplitude of sine function
y = A*np.sin(x)
fig, axes = plt.subplots(nrows=2)
for ax, s, title in zip(axes, [2, 0], ['With', 'Without']):
yinterp = interpolate.UnivariateSpline(x, y, s=s)(x)
ax.plot(x, yinterp, label='Interpolated')
ax.plot(x, y, 'bo',label='Original')
ax.legend()
ax.set_title(title + ' Smoothing')
plt.show()
The reason that you're only clearly seeing the effects of smoothing with a low amplitude is due to the way the smoothing factor is defined. See the documentation for scipy.interpolate.UnivariateSpline for more details.
Even with a higher amplitude, the interpolated data won't match the original data if you use smoothing.
For example, if we just change the amplitude (A) to 1.0 in the code example above, we'll still see the effects of smoothing...
The problem is in choosing suitable values for the s parameter. Its values depend on the scaling of the data.
Reading the documentation carefully, one can deduce that the parameter should be chosen around s = len(y) * np.var(y), i.e. # of data points * variance. Taking for example s = 0.05 * len(y) * np.var(y) gives a smoothing spline that does not depend on the scaling of the data or the number of data points.
EDIT: sensible values for s depend of course also on the noise level in the data. The docs seem to recommend choosing s in the range (m - sqrt(2*m)) * std**2 <= s <= (m + sqrt(2*m)) * std**2 where std is the standard deviation associated with the "noise" you want to smooth over.