Fitting empirical distribution to theoretical ones with Scipy (Python)?

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Fitting empirical distribution to theoretical ones with Scipy (Python)?

INTRODUCTION: I have a list of more than 30,000 integer values ranging from 0 to 47, inclusive, e.g.[0,0,0,0,..,1,1,1,1,...,2,2,2,2,...,47,47,47,...] sampled from some continuous distribution. The values in the list are not necessarily in order, but order doesn’t matter for this problem.

PROBLEM: Based on my distribution I would like to calculate p-value (the probability of seeing greater values) for any given value. For example, as you can see p-value for 0 would be approaching 1 and p-value for higher numbers would be tending to 0.

I don’t know if I am right, but to determine probabilities I think I need to fit my data to a theoretical distribution that is the most suitable to describe my data. I assume that some kind of goodness of fit test is needed to determine the best model.

Is there a way to implement such an analysis in Python (Scipy or Numpy)?
Could you present any examples?

Thank you!

Asked By: s_sherly

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Answer #1:

Distribution Fitting with Sum of Square Error (SSE)

This is an update and modification to Saullo’s answer, that uses the full list of the current scipy.stats distributions and returns the distribution with the least SSE between the distribution’s histogram and the data’s histogram.

Example Fitting

Using the El Niño dataset from statsmodels, the distributions are fit and error is determined. The distribution with the least error is returned.

All Distributions

All Fitted Distributions

Best Fit Distribution

Best Fit Distribution

Example Code

%matplotlib inline
import warnings
import numpy as np
import pandas as pd
import scipy.stats as st
import statsmodels as sm
import matplotlib
import matplotlib.pyplot as plt
matplotlib.rcParams['figure.figsize'] = (16.0, 12.0)
matplotlib.style.use('ggplot')
# Create models from data
def best_fit_distribution(data, bins=200, ax=None):
    """Model data by finding best fit distribution to data"""
    # Get histogram of original data
    y, x = np.histogram(data, bins=bins, density=True)
    x = (x + np.roll(x, -1))[:-1] / 2.0
    # Distributions to check
    DISTRIBUTIONS = [
        st.alpha,st.anglit,st.arcsine,st.beta,st.betaprime,st.bradford,st.burr,st.cauchy,st.chi,st.chi2,st.cosine,
        st.dgamma,st.dweibull,st.erlang,st.expon,st.exponnorm,st.exponweib,st.exponpow,st.f,st.fatiguelife,st.fisk,
        st.foldcauchy,st.foldnorm,st.frechet_r,st.frechet_l,st.genlogistic,st.genpareto,st.gennorm,st.genexpon,
        st.genextreme,st.gausshyper,st.gamma,st.gengamma,st.genhalflogistic,st.gilbrat,st.gompertz,st.gumbel_r,
        st.gumbel_l,st.halfcauchy,st.halflogistic,st.halfnorm,st.halfgennorm,st.hypsecant,st.invgamma,st.invgauss,
        st.invweibull,st.johnsonsb,st.johnsonsu,st.ksone,st.kstwobign,st.laplace,st.levy,st.levy_l,st.levy_stable,
        st.logistic,st.loggamma,st.loglaplace,st.lognorm,st.lomax,st.maxwell,st.mielke,st.nakagami,st.ncx2,st.ncf,
        st.nct,st.norm,st.pareto,st.pearson3,st.powerlaw,st.powerlognorm,st.powernorm,st.rdist,st.reciprocal,
        st.rayleigh,st.rice,st.recipinvgauss,st.semicircular,st.t,st.triang,st.truncexpon,st.truncnorm,st.tukeylambda,
        st.uniform,st.vonmises,st.vonmises_line,st.wald,st.weibull_min,st.weibull_max,st.wrapcauchy
    ]
    # Best holders
    best_distribution = st.norm
    best_params = (0.0, 1.0)
    best_sse = np.inf
    # Estimate distribution parameters from data
    for distribution in DISTRIBUTIONS:
        # Try to fit the distribution
        try:
            # Ignore warnings from data that can't be fit
            with warnings.catch_warnings():
                warnings.filterwarnings('ignore')
                # fit dist to data
                params = distribution.fit(data)
                # Separate parts of parameters
                arg = params[:-2]
                loc = params[-2]
                scale = params[-1]
                # Calculate fitted PDF and error with fit in distribution
                pdf = distribution.pdf(x, loc=loc, scale=scale, *arg)
                sse = np.sum(np.power(y - pdf, 2.0))
                # if axis pass in add to plot
                try:
                    if ax:
                        pd.Series(pdf, x).plot(ax=ax)
                    end
                except Exception:
                    pass
                # identify if this distribution is better
                if best_sse > sse > 0:
                    best_distribution = distribution
                    best_params = params
                    best_sse = sse
        except Exception:
            pass
    return (best_distribution.name, best_params)
def make_pdf(dist, params, size=10000):
    """Generate distributions's Probability Distribution Function """
    # Separate parts of parameters
    arg = params[:-2]
    loc = params[-2]
    scale = params[-1]
    # Get sane start and end points of distribution
    start = dist.ppf(0.01, *arg, loc=loc, scale=scale) if arg else dist.ppf(0.01, loc=loc, scale=scale)
    end = dist.ppf(0.99, *arg, loc=loc, scale=scale) if arg else dist.ppf(0.99, loc=loc, scale=scale)
    # Build PDF and turn into pandas Series
    x = np.linspace(start, end, size)
    y = dist.pdf(x, loc=loc, scale=scale, *arg)
    pdf = pd.Series(y, x)
    return pdf
# Load data from statsmodels datasets
data = pd.Series(sm.datasets.elnino.load_pandas().data.set_index('YEAR').values.ravel())
# Plot for comparison
plt.figure(figsize=(12,8))
ax = data.plot(kind='hist', bins=50, normed=True, alpha=0.5, color=plt.rcParams['axes.color_cycle'][1])
# Save plot limits
dataYLim = ax.get_ylim()
# Find best fit distribution
best_fit_name, best_fit_params = best_fit_distribution(data, 200, ax)
best_dist = getattr(st, best_fit_name)
# Update plots
ax.set_ylim(dataYLim)
ax.set_title(u'El Niño sea temp.n All Fitted Distributions')
ax.set_xlabel(u'Temp (°C)')
ax.set_ylabel('Frequency')
# Make PDF with best params 
pdf = make_pdf(best_dist, best_fit_params)
# Display
plt.figure(figsize=(12,8))
ax = pdf.plot(lw=2, label='PDF', legend=True)
data.plot(kind='hist', bins=50, normed=True, alpha=0.5, label='Data', legend=True, ax=ax)
param_names = (best_dist.shapes + ', loc, scale').split(', ') if best_dist.shapes else ['loc', 'scale']
param_str = ', '.join(['{}={:0.2f}'.format(k,v) for k,v in zip(param_names, best_fit_params)])
dist_str = '{}({})'.format(best_fit_name, param_str)
ax.set_title(u'El Niño sea temp. with best fit distribution n' + dist_str)
ax.set_xlabel(u'Temp. (°C)')
ax.set_ylabel('Frequency')
Answered By: tmthydvnprt

Answer #2:

There are 82 implemented distribution functions in SciPy 0.12.0. You can test how some of them fit to your data using their fit() method. Check the code below for more details:

enter image description here

import matplotlib.pyplot as plt
import scipy
import scipy.stats
size = 30000
x = scipy.arange(size)
y = scipy.int_(scipy.round_(scipy.stats.vonmises.rvs(5,size=size)*47))
h = plt.hist(y, bins=range(48))
dist_names = ['gamma', 'beta', 'rayleigh', 'norm', 'pareto']
for dist_name in dist_names:
    dist = getattr(scipy.stats, dist_name)
    param = dist.fit(y)
    pdf_fitted = dist.pdf(x, *param[:-2], loc=param[-2], scale=param[-1]) * size
    plt.plot(pdf_fitted, label=dist_name)
    plt.xlim(0,47)
plt.legend(loc='upper right')
plt.show()

References:

– Fitting distributions, goodness of fit, p-value. Is it possible to do this with Scipy (Python)?

– Distribution fitting with Scipy

And here a list with the names of all distribution functions available in Scipy 0.12.0 (VI):

dist_names = [ 'alpha', 'anglit', 'arcsine', 'beta', 'betaprime', 'bradford', 'burr', 'cauchy', 'chi', 'chi2', 'cosine', 'dgamma', 'dweibull', 'erlang', 'expon', 'exponweib', 'exponpow', 'f', 'fatiguelife', 'fisk', 'foldcauchy', 'foldnorm', 'frechet_r', 'frechet_l', 'genlogistic', 'genpareto', 'genexpon', 'genextreme', 'gausshyper', 'gamma', 'gengamma', 'genhalflogistic', 'gilbrat', 'gompertz', 'gumbel_r', 'gumbel_l', 'halfcauchy', 'halflogistic', 'halfnorm', 'hypsecant', 'invgamma', 'invgauss', 'invweibull', 'johnsonsb', 'johnsonsu', 'ksone', 'kstwobign', 'laplace', 'logistic', 'loggamma', 'loglaplace', 'lognorm', 'lomax', 'maxwell', 'mielke', 'nakagami', 'ncx2', 'ncf', 'nct', 'norm', 'pareto', 'pearson3', 'powerlaw', 'powerlognorm', 'powernorm', 'rdist', 'reciprocal', 'rayleigh', 'rice', 'recipinvgauss', 'semicircular', 't', 'triang', 'truncexpon', 'truncnorm', 'tukeylambda', 'uniform', 'vonmises', 'wald', 'weibull_min', 'weibull_max', 'wrapcauchy']

Answer #3:

fit() method mentioned by @Saullo Castro provides maximum likelihood estimates (MLE). The best distribution for your data is the one give you the highest can be determined by several different ways: such as

1, the one that gives you the highest log likelihood.

2, the one that gives you the smallest AIC, BIC or BICc values (see wiki: http://en.wikipedia.org/wiki/Akaike_information_criterion, basically can be viewed as log likelihood adjusted for number of parameters, as distribution with more parameters are expected to fit better)

3, the one that maximize the Bayesian posterior probability. (see wiki: http://en.wikipedia.org/wiki/Posterior_probability)

Of course, if you already have a distribution that should describe you data (based on the theories in your particular field) and want to stick to that, you will skip the step of identifying the best fit distribution.

scipy does not come with a function to calculate log likelihood (although MLE method is provided), but hard code one is easy: see Is the build-in probability density functions of `scipy.stat.distributions` slower than a user provided one?

Answered By: CT Zhu

Answer #4:

Try the distfit library.

pip install distfit

# Create 1000 random integers, value between [0-50]
X = np.random.randint(0, 50,1000)
# Retrieve P-value for y
y = [0,10,45,55,100]
# From the distfit library import the class distfit
from distfit import distfit
# Initialize.
# Set any properties here, such as alpha.
# The smoothing can be of use when working with integers. Otherwise your histogram
# may be jumping up-and-down, and getting the correct fit may be harder.
dist = distfit(alpha=0.05, smooth=10)
# Search for best theoretical fit on your empirical data
dist.fit_transform(X)
> [distfit] >fit..
> [distfit] >transform..
> [distfit] >[norm      ] [RSS: 0.0037894] [loc=23.535 scale=14.450]
> [distfit] >[expon     ] [RSS: 0.0055534] [loc=0.000 scale=23.535]
> [distfit] >[pareto    ] [RSS: 0.0056828] [loc=-384473077.778 scale=384473077.778]
> [distfit] >[dweibull  ] [RSS: 0.0038202] [loc=24.535 scale=13.936]
> [distfit] >[t         ] [RSS: 0.0037896] [loc=23.535 scale=14.450]
> [distfit] >[genextreme] [RSS: 0.0036185] [loc=18.890 scale=14.506]
> [distfit] >[gamma     ] [RSS: 0.0037600] [loc=-175.505 scale=1.044]
> [distfit] >[lognorm   ] [RSS: 0.0642364] [loc=-0.000 scale=1.802]
> [distfit] >[beta      ] [RSS: 0.0021885] [loc=-3.981 scale=52.981]
> [distfit] >[uniform   ] [RSS: 0.0012349] [loc=0.000 scale=49.000]
# Best fitted model
best_distr = dist.model
print(best_distr)
# Uniform shows best fit, with 95% CII (confidence intervals), and all other parameters
> {'distr': <scipy.stats._continuous_distns.uniform_gen at 0x16de3a53160>,
>  'params': (0.0, 49.0),
>  'name': 'uniform',
>  'RSS': 0.0012349021241149533,
>  'loc': 0.0,
>  'scale': 49.0,
>  'arg': (),
>  'CII_min_alpha': 2.45,
>  'CII_max_alpha': 46.55}
# Ranking distributions
dist.summary
# Plot the summary of fitted distributions
dist.plot_summary()

enter image description here

# Make prediction on new datapoints based on the fit
dist.predict(y)
# Retrieve your pvalues with 
dist.y_pred
# array(['down', 'none', 'none', 'up', 'up'], dtype='<U4')
dist.y_proba
array([0.02040816, 0.02040816, 0.02040816, 0.        , 0.        ])
# Or in one dataframe
dist.df
# The plot function will now also include the predictions of y
dist.plot()

Best fit

Note that in this case, all points will be significant because of the uniform distribution. You can filter with the dist.y_pred if required.

Answered By: erdogant

Answer #5:

AFAICU, your distribution is discrete (and nothing but discrete). Therefore just counting the frequencies of different values and normalizing them should be enough for your purposes. So, an example to demonstrate this:

In []: values= [0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4]
In []: counts= asarray(bincount(values), dtype= float)
In []: cdf= counts.cumsum()/ counts.sum()

Thus, probability of seeing values higher than 1 is simply (according to the complementary cumulative distribution function (ccdf):

In []: 1- cdf[1]
Out[]: 0.40000000000000002

Please note that ccdf is closely related to survival function (sf), but it’s also defined with discrete distributions, whereas sf is defined only for contiguous distributions.

Answered By: eat

Answer #6:

It sounds like probability density estimation problem to me.

from scipy.stats import gaussian_kde
occurences = [0,0,0,0,..,1,1,1,1,...,2,2,2,2,...,47]
values = range(0,48)
kde = gaussian_kde(map(float, occurences))
p = kde(values)
p = p/sum(p)
print "P(x>=1) = %f" % sum(p[1:])

Also see http://jpktd.blogspot.com/2009/03/using-gaussian-kernel-density.html.

Answered By: emre

Answer #7:

While many of the above answers are completely valid, no one seems to answer your question completely, specifically the part:

I don’t know if I am right, but to determine probabilities I think I need to fit my data to a theoretical distribution that is the most suitable to describe my data. I assume that some kind of goodness of fit test is needed to determine the best model.

The parametric approach

This is the process you’re describing of using some theoretical distribution and fitting the parameters to your data and there’s some excellent answers how to do this.

The non-parametric approach

However, it’s also possible to use a non-parametric approach to your problem, which means you do not assume any underlying distribution at all.

By using the so-called Empirical distribution function which equals:
Fn(x)= SUM( I[X<=x] ) / n. So the proportion of values below x.

As was pointed out in one of the above answers is that what you’re interested in is the inverse CDF (cumulative distribution function), which is equal to 1-F(x)

It can be shown that the empirical distribution function will converge to whatever ‘true’ CDF that generated your data.

Furthermore, it is straightforward to construct a 1-alpha confidence interval by:

L(X) = max{Fn(x)-en, 0}
U(X) = min{Fn(x)+en, 0}
en = sqrt( (1/2n)*log(2/alpha)

Then P( L(X) <= F(X) <= U(X) ) >= 1-alpha for all x.

I’m quite surprised that PierrOz answer has 0 votes, while it’s a completely valid answer to the question using a non-parametric approach to estimating F(x).

Note that the issue you mention of P(X>=x)=0 for any x>47 is simply a personal preference that might lead you to chose the parametric approach above the non-parametric approach. Both approaches however are completely valid solutions to your problem.

For more details and proofs of the above statements I would recommend having a look at
‘All of Statistics: A Concise Course in Statistical Inference by Larry A. Wasserman’. An excellent book on both parametric and non-parametric inference.

EDIT:
Since you specifically asked for some python examples it can be done using numpy:

import numpy as np
def empirical_cdf(data, x):
    return np.sum(x<=data)/len(data)
def p_value(data, x):
    return 1-empirical_cdf(data, x)
# Generate some data for demonstration purposes
data = np.floor(np.random.uniform(low=0, high=48, size=30000))
print(empirical_cdf(data, 20))
print(p_value(data, 20)) # This is the value you're interested in
Answered By: Martin Skogholt

Answer #8:

Forgive me if I don’t understand your need but what about storing your data in a dictionary where keys would be the numbers between 0 and 47 and values the number of occurrences of their related keys in your original list?
Thus your likelihood p(x) will be the sum of all the values for keys greater than x divided by 30000.

Answered By: PierrOz

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