22nd July To 28th July 2019

[Yoriz]

The purpose of this thread is to compile a list of posts made between Mon 22nd July To Sun 28th July that have well established posts to be nominated top pick of the week.
Well established post meaning it is detailed, instructional, above and beyond the normal answer, etc.

Please make 1 link in the post (by copying the post ID # in the top right area of the post). If you have more than post to nominate, make a separate post for each. You may make an additional comment regarding the post you are nominating.

Use the like button to cast your votes. you may choose as many as you want. The winner will be the one with the most liked at the end of the week.

[Yoriz]

https://python-forum.io/Thread-Help-deci...2#pid86992

[Yoriz]

Congratulations scidam your post is this weeks winner.

I don't understand this code too, but I made some comments, hope that helps:

def calcPrior(n, N):
    # this line produces a list [N/1, N/2, ... , 1], i.e. a part of harmonic sequence.
    # in Python3 would consist of floats, in Python2 it would consist of integers
    l = [N/i for i in range(1, N+1)]
    
    # this is sum of the harmonic sequence 
    tot = sum(l) # tot is a number
    
    #  each value in harmonic series was divided onto
    # the sum of all values; l variable is overridden
    l = [i/tot for i in l]
    
    # this line assumes that 0<=n<=1; e.g. if N = 10, n=0.5, then
    # we return l[int(10*0.5)] = l[int(5.0)] = l[5], i.e. fifth
    # element of the array l. int(....) just drops mantissa of a float number, e.g.
    # int(1.919) = 1;
    return l[int(n*N)]
 
def calcL(s, initialF, finalF):
    N = 180  
    eps = 0.0005  # some magic numbers here
    mu = initialF  # this string wouldn't be needed, if finalF was declared as mu. (this is for convenience, mu is shorter than finalF)
    sd = (mu*(1-mu))**0.5#/N
    
    # This is likely iterative method of finding a solution of some equation (system of two equations)
    for i in range(50):
        old_mu = mu  
        mu = (1+s)*mu/(1+s*mu)
        sd = (1.0/N)*(mu*(1-mu)) + (((1+s)/(1+s*old_mu))**2)*sd
    # these lines could be rewritten, e.g. as
    # for i in range(50): # 50 is another magic number
    #    mu, sd =(1+s)*mu/(1+s*mu), (1.0/N)*(mu*(1-mu)) + (((1+s)/(1+s*old_mu))**2)*sd
    # e.g. without introducing old_mu
        
    
    # As we found a solution of equations, we integrate probability density function of normal
    # distribution over interval (finalF-eps, finalF+eps)
    #I completely don't understand why we do so. integral of pdf is cdf, we could just use
    # cdf instead. eps seems to be small, so this integral is almost equal to 
    #something like `norm(loc=mu, scale=sd).cdf(finalF)*2*eps`
    prob = quad(norm(loc=mu, scale=sd).pdf, finalF-eps, finalF+eps)
    # ok, integration is done, as a result we got some probability
    
    # Finally, we returns that probability being multiplied on [initialF*N] - index of normalized harmonic sequence
    return prob[0]*calcPrior(initialF, N)

# to get this code worked some import should be included at the beginning of the script,
# import numpy as np
# from scipy.stats import norm, chi2
# from scipy.integrate import quad
# function findS is not defined in this code., start variable is not defined too.
p value of s=0.5
p = 1-chi2.cdf(2*(np.log10(calcL(0.5, start, 1))-np.log10(findS(0, start, 1))), 1)