How do I quantify the uniformity of sampling time?

https://stats.stackexchange.com/questions/how-do-i-quantify-the-uniformity-of-sampling-time



In [1]:

    
%pylab inline
from __future__ import division
import scipy as sp
import seaborn as sns
sns.set_style('whitegrid')
sns.set_context('paper', font_scale=2)
np.random.seed(42)









    



Populating the interactive namespace from numpy and matplotlib



In [2]:

    
def calculate_shannon_entropy(p):
    ''' 
    Parameters
    ----------
    p: list
        list of probability values such that sum(p) = 1
    Returns
    -------
    entropy: float
        Shannon Entropy        
    '''
    assert np.allclose(sum(p), 1)
    entropy = -np.nansum(np.array(p) * np.log2(np.array(p)).T)
    return entropy

def calculate_uniformity_index(x, a, b):
    '''
    Parameters
    ----------
    '''

Entropy is maximal for a uniform distribution



In [3]:

    
plt.plot(np.arange(0, 1.1, 0.1), [calculate_shannon_entropy([x, 1-x]) for x in np.arange(0, 1.1, 0.1)])









    



/home/saket/anaconda2/lib/python2.7/site-packages/ipykernel/__main__.py:13: RuntimeWarning: divide by zero encountered in log2
/home/saket/anaconda2/lib/python2.7/site-packages/ipykernel/__main__.py:13: RuntimeWarning: invalid value encountered in multiply






    Out[3]:





[<matplotlib.lines.Line2D at 0x7efd7ec563d0>]

Entropy increase monotonically with sample size for a uniform distribution



In [4]:

    
entropies = []

for n in range(1, 100):
    p_vec = [1/n] * n 
    entropies.append(calculate_shannon_entropy(p_vec))

plt.plot(entropies)









    Out[4]:





[<matplotlib.lines.Line2D at 0x7efd7eb52250>]



In [31]:

    
## Uniform index
hist_bins = 20

s = np.random.uniform(1,10,1000)
count, bins, ignored = plt.hist(s, 10, normed=True)

count, boundaries = np.histogram(s, bins=1000)
normalized_count = count/np.sum(count)

U = np.sqrt(1+(10-1)**2)/np.sum(np.sqrt(1+normalized_count**2))
(U-0.707)/(1-0.707)









    Out[31]:





-2.3820635625174496

Uniformity index

https://arxiv.org/pdf/1508.01146.pdf defines uniformity index as : $$ \mathcal{U}(f_X) = \frac{\sqrt{1+(b-a)^2}}{\sqrt{1+[f_X(t)]^2}dt} $$



In [43]:

    
count, boundaries = np.histogram(s, bins=1000, density=True)



In [48]:

    
sum(np.diff(boundaries)*count)









    Out[48]:





1.0



In [50]:

    
#denom = np.trapz(np.sqrt(1+count**2), np.diff(boundaries))
denom = sum(np.diff(boundaries)*np.sqrt(1+count**2))
denom









    Out[50]:





9.0467435677217765



In [51]:

    
U = np.sqrt(1+(10-1)**2)/denom
(U-0.707)/(1-0.707)









    Out[51]:





1.003260113892593



In [52]:

    
s[np.where( (s>7) & (s<9) )] = 0



In [54]:

    
#count = np.histogram(s)
#normalized_count = count[0]/np.sum(count[0])
count, boundaries = np.histogram(s, bins=1000, density=True)

U = np.sqrt(1+(10-1)**2)/np.sum(np.sqrt(1+normalized_count**2))



In [56]:

    
denom = sum(np.diff(boundaries)*np.sqrt(1+count**2))

U = np.sqrt(1+(10-1)**2)/denom
(U-0.707)/(1-0.707)









    Out[56]:





0.59534182460272966



In [ ]:

    
len(count)



In [58]:

    
s[np.where( (s>1) )] = 0
count, bins, ignored = plt.hist(s, 10, normed=True)



In [64]:

    
count, boundaries = np.histogram(s, bins=10, density=True)
denom = sum(np.diff(boundaries)*np.sqrt(1+count**2))

U = np.sqrt(1+(2-1)**2)/denom
(U-0.707)/(1-0.707)









    Out[64]:





0.12073095937549529



In [60]:

    
denom









    Out[60]:





1.9990004999998734



In [62]:

    
count, bins, ignored = plt.hist(s, 10, normed=True)



In [ ]: