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Curve fitting with cubic spline

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:

Generating non-random normally distributed values between two points

I've stumbled across this code in an answer to a question and I'd like to automate the process of getting the distribution to fit neatly between two bounds.
import numpy as np
from scipy import stats
bounds = [0, 100]
n = np.mean(bounds)
# your distribution:
distribution = stats.norm(loc=n, scale=20)
# percentile point, the range for the inverse cumulative distribution function:
bounds_for_range = distribution.cdf(bounds)
# Linspace for the inverse cdf:
pp = np.linspace(*bounds_for_range, num=1000)
x = distribution.ppf(pp)
# And just to check that it makes sense you can try:
from matplotlib import pyplot as plt
plt.hist(x)
plt.show()
Let's say I have the values [720, 965], or any other bounds, that I would like to fit my distribution across. Is there a way to soft-code the adjustment of scale in stats.norm to fit this distribution across my bounds without any unreasonable gaps? Or are there any functions that have this type of functionality?
A scale of ~20 works well for the example code, but I have to adjust it to ~50 for the example of [720, 965]

I am not sure, but truncated normal distribution should be exactly what you are looking for.
from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

Simple plots of eigenvectors for sklearn.decomposition.PCA

I'm trying to understand how Principal Component Analysis works and I am testing it on the sklearn.datasets.load_iris dataset. I understand how each step works (e.g. standardize the data, covariance, eigendecomposition, sort for highest eigenvalue, transform original data to new axis using K selected dimensions).
The next step is to visualize where these eigenvectors are being projected into on the dataset (on the PC1 vs. PC2 plot, right?).
Can someone explain how to plot [PC1, PC2, PC3] eigenvectors on a 3D plot of the reduced dimension dataset?
Also, am I plotting this 2D version correctly? I'm not sure why my first eigenvector has a shorter length. Should I multiply by the eigenvalue?
Here is some of the research I have done to accomplish this:
The PCA method that I'm following is from :
https://plot.ly/ipython-notebooks/principal-component-analysis/#Shortcut---PCA-in-scikit-learn (although I don't want to use plotly. I want to stick with pandas, numpy, sklearn, matplotlib, scipy, and seaborn)
I've been following this tutorial for plotting eigenvectors it seems pretty simple: Basic example for PCA with matplotlib but I can't seem to replicate the results with my data.
I found this but it seems overly complicated for what I'm trying to do and I don't want to have to create a FancyArrowPatch: plotting the eigenvector of covariance matrix using matplotlib and np.linalg
I've tried to make my code as straightforward as possible to follow the other tutorials:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.datasets import load_iris
from sklearn.preprocessing import StandardScaler
from sklearn import decomposition
import seaborn as sns; sns.set_style("whitegrid", {'axes.grid' : False})
%matplotlib inline
np.random.seed(0)
# Iris dataset
DF_data = pd.DataFrame(load_iris().data,
index = ["iris_%d" % i for i in range(load_iris().data.shape[0])],
columns = load_iris().feature_names)
Se_targets = pd.Series(load_iris().target,
index = ["iris_%d" % i for i in range(load_iris().data.shape[0])],
name = "Species")
# Scaling mean = 0, var = 1
DF_standard = pd.DataFrame(StandardScaler().fit_transform(DF_data),
index = DF_data.index,
columns = DF_data.columns)
# Sklearn for Principal Componenet Analysis
# Dims
m = DF_standard.shape[1]
K = 2
# PCA (How I tend to set it up)
M_PCA = decomposition.PCA(n_components=m)
DF_PCA = pd.DataFrame(M_PCA.fit_transform(DF_standard),
columns=["PC%d" % k for k in range(1,m + 1)]).iloc[:,:K]
# Plot the eigenvectors
#https://stackoverflow.com/questions/18299523/basic-example-for-pca-with-matplotlib
# This is where stuff gets weird...
data = DF_standard
mu = data.mean(axis=0)
eigenvectors, eigenvalues = M_PCA.components_, M_PCA.explained_variance_ #eigenvectors, eigenvalues, V = np.linalg.svd(data.T, full_matrices=False)
projected_data = DF_PCA #np.dot(data, eigenvectors)
sigma = projected_data.std(axis=0).mean()
fig, ax = plt.subplots(figsize=(10,10))
ax.scatter(projected_data["PC1"], projected_data["PC2"])
for axis, color in zip(eigenvectors[:K], ["red","green"]):
# start, end = mu, mu + sigma * axis ### leads to "ValueError: too many values to unpack (expected 2)"
# So I tried this but I don't think it's correct
start, end = (mu)[:K], (mu + sigma * axis)[:K]
ax.annotate('', xy=end,xytext=start, arrowprops=dict(facecolor=color, width=1.0))
ax.set_aspect('equal')
plt.show()

I think you are asking the wrong question. The eigenvectors ARE the the principal components (PC1, PC2, etc.). So plotting the eigenvectors in the [PC1, PC2, PC3] 3D plot is simply plotting the three orthogonal axes of that plot.
You probably want to visualize how the eigenvectors look in your original coordinate system. This is what is discussed in your second link: Basic example for PCA with matplotlib.

Intuitive interpolation between unevenly spaced points

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 to make this matplotlib plot less noisy?

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).