Module `stikpetP.effect_sizes.eff_size_jbm_r`

Expand source code

import pandas as pd
from scipy.special import comb

def es_jbm_r(data, success=None):
    '''
    Berry-Johnston-Mielke R
    -----------------------
    A chance-corrected version of eta-squared, as an effect size measure for a Cochran Q test.
    
    Parameters
    ----------
    data : dataframe
        dataframe with a column for each category
    success : object, optional
        the value that represents a 'success'. If None (default) the first value found will be used as success.
        
    Returns
    -------
    r : float
        the effect size measure value
        
    Notes
    -----
    The formula used (Berry et al., 2007, pp. 1237,1239; Alexis, 2014):
    $$\\Re = 1 - \\frac{\\delta}{\\mu_{\\delta}}$$
    
    With:
    $$\\delta = \\left[k\\times\\binom{n}{2}\\right]^{-1}\\times\\sum_{c=1}^k\\sum_{i=1}^{n-1}\\sum_{j=i+1}^n\\left|x_{i,c}-x_{j,c}\\right|$$
    $$\\mu_{\\delta}=\\frac{2}{n\\times\\left(n-1\\right)}\\times\\left(\\left(\\sum_{i=1}^n p_i\\right)\\times\\left(n-\\sum_{i=1}^n p_i\\right)-\\sum_{i=1}^n p_i\\times\\left(1-p_i\\right)\\right)$$
    $$\\sum_{i=1}^n p_i = \\frac{\\sum_{j=1}^k x_{i,j}}{k}$$
    
    Note that the function uses an equivelant for the calculation of \\(\\delta\\), using:
    $$\\sum_{c=1}^k\\sum_{i=1}^{n-1}\\sum_{j=i+1}^n\\left|x_{i,c}-x_{j,c}\\right|=\\sum_{j=1}^k C_j\\times\\left(n-C_j\\right)$$
    
    *Symbols used*
    
    * \\(k\\), the number of categories (factors)
    * \\(x_{i,j}\\), the i-th score in category j, 1 = success, 0 = no success.
    * \\(n\\), the number of cases
    * \\(C_j\\), the number of successes in category j
    
    The original article has in the equation for \\(\\mu_{\\delta}\\) the first factor written as \\(\\frac{2}{k\\times\\left(k - 1\\right)}\\). In personal communication with one of the authors Alexis (2014) indicated this was wrong and \\(n\\) should be used.
    
    References
    ----------
    Alexis. (2014, September 7). Answer to “Effect size of Cochran’s Q.” Cross Validated. https://stats.stackexchange.com/a/114649
    
    Berry, K. J., Johnston, J. E., & Mielke, P. W. (2007). An alternative measure of effect size for Cochran’s Q test for related proportions. *Perceptual and Motor Skills, 104*(3_suppl), 1236–1242. doi:10.2466/pms.104.4.1236-1242
    
    Author
    ------
    Made by P. Stikker
    
    Companion website: https://PeterStatistics.com  
    YouTube channel: https://www.youtube.com/stikpet  
    Donations: https://www.patreon.com/bePatron?u=19398076
    
    '''
    #Remove rows with missing values and reset index
    data = data.dropna()    
    data = data.reset_index(drop=True)
    
    n = len(data)
    k = len(data.columns)
    
    if success is None:
        suc = data.iloc[0,0]
    else:
        suc = success
    
    isSuc = data == suc
    cj = isSuc.sum()
    ri = isSuc.sum(axis=1)
    ns = cj.sum()
    
    pi = ri/k
    
    delta = 1/(k*comb(n, 2)) * (cj*(n - cj)).sum()
    mu = 2/(n*(n - 1)) * (pi.sum()*(n - pi.sum()) - (pi*(1 - pi)).sum())
    r = 1 - delta/mu
    
    return r

Functions

def es_jbm_r(data, success=None)

Berry-Johnston-Mielke R

A chance-corrected version of eta-squared, as an effect size measure for a Cochran Q test.

Parameters

data : dataframe: dataframe with a column for each category
success : object, optional: the value that represents a 'success'. If None (default) the first value found will be used as success.

Returns

r : float: the effect size measure value

Notes

The formula used (Berry et al., 2007, pp. 1237,1239; Alexis, 2014): $\Re = 1 - \frac{\delta}{\mu_{\delta}}$

With: $\delta = \left[k\times\binom{n}{2}\right]^{-1}\times\sum_{c=1}^k\sum_{i=1}^{n-1}\sum_{j=i+1}^n\left|x_{i,c}-x_{j,c}\right|$ $\mu_{\delta}=\frac{2}{n\times\left(n-1\right)}\times\left(\left(\sum_{i=1}^n p_i\right)\times\left(n-\sum_{i=1}^n p_i\right)-\sum_{i=1}^n p_i\times\left(1-p_i\right)\right)$ $\sum_{i=1}^n p_i = \frac{\sum_{j=1}^k x_{i,j}}{k}$

Note that the function uses an equivelant for the calculation of $\delta$ , using: $\sum_{c=1}^k\sum_{i=1}^{n-1}\sum_{j=i+1}^n\left|x_{i,c}-x_{j,c}\right|=\sum_{j=1}^k C_j\times\left(n-C_j\right)$

Symbols used

$k$ , the number of categories (factors)
$x_{i,j}$ , the i-th score in category j, 1 = success, 0 = no success.
$n$ , the number of cases
$C_j$ , the number of successes in category j

The original article has in the equation for $\mu_{\delta}$ the first factor written as $\frac{2}{k\times\left(k - 1\right)}$ . In personal communication with one of the authors Alexis (2014) indicated this was wrong and $n$ should be used.

References

Alexis. (2014, September 7). Answer to “Effect size of Cochran’s Q.” Cross Validated. https://stats.stackexchange.com/a/114649

Berry, K. J., Johnston, J. E., & Mielke, P. W. (2007). An alternative measure of effect size for Cochran’s Q test for related proportions. Perceptual and Motor Skills, 104(3_suppl), 1236–1242. doi:10.2466/pms.104.4.1236-1242

Author

Made by P. Stikker

Companion website: https://PeterStatistics.com
YouTube channel: https://www.youtube.com/stikpet
Donations: https://www.patreon.com/bePatron?u=19398076

Expand source code

def es_jbm_r(data, success=None):
    '''
    Berry-Johnston-Mielke R
    -----------------------
    A chance-corrected version of eta-squared, as an effect size measure for a Cochran Q test.
    
    Parameters
    ----------
    data : dataframe
        dataframe with a column for each category
    success : object, optional
        the value that represents a 'success'. If None (default) the first value found will be used as success.
        
    Returns
    -------
    r : float
        the effect size measure value
        
    Notes
    -----
    The formula used (Berry et al., 2007, pp. 1237,1239; Alexis, 2014):
    $$\\Re = 1 - \\frac{\\delta}{\\mu_{\\delta}}$$
    
    With:
    $$\\delta = \\left[k\\times\\binom{n}{2}\\right]^{-1}\\times\\sum_{c=1}^k\\sum_{i=1}^{n-1}\\sum_{j=i+1}^n\\left|x_{i,c}-x_{j,c}\\right|$$
    $$\\mu_{\\delta}=\\frac{2}{n\\times\\left(n-1\\right)}\\times\\left(\\left(\\sum_{i=1}^n p_i\\right)\\times\\left(n-\\sum_{i=1}^n p_i\\right)-\\sum_{i=1}^n p_i\\times\\left(1-p_i\\right)\\right)$$
    $$\\sum_{i=1}^n p_i = \\frac{\\sum_{j=1}^k x_{i,j}}{k}$$
    
    Note that the function uses an equivelant for the calculation of \\(\\delta\\), using:
    $$\\sum_{c=1}^k\\sum_{i=1}^{n-1}\\sum_{j=i+1}^n\\left|x_{i,c}-x_{j,c}\\right|=\\sum_{j=1}^k C_j\\times\\left(n-C_j\\right)$$
    
    *Symbols used*
    
    * \\(k\\), the number of categories (factors)
    * \\(x_{i,j}\\), the i-th score in category j, 1 = success, 0 = no success.
    * \\(n\\), the number of cases
    * \\(C_j\\), the number of successes in category j
    
    The original article has in the equation for \\(\\mu_{\\delta}\\) the first factor written as \\(\\frac{2}{k\\times\\left(k - 1\\right)}\\). In personal communication with one of the authors Alexis (2014) indicated this was wrong and \\(n\\) should be used.
    
    References
    ----------
    Alexis. (2014, September 7). Answer to “Effect size of Cochran’s Q.” Cross Validated. https://stats.stackexchange.com/a/114649
    
    Berry, K. J., Johnston, J. E., & Mielke, P. W. (2007). An alternative measure of effect size for Cochran’s Q test for related proportions. *Perceptual and Motor Skills, 104*(3_suppl), 1236–1242. doi:10.2466/pms.104.4.1236-1242
    
    Author
    ------
    Made by P. Stikker
    
    Companion website: https://PeterStatistics.com  
    YouTube channel: https://www.youtube.com/stikpet  
    Donations: https://www.patreon.com/bePatron?u=19398076
    
    '''
    #Remove rows with missing values and reset index
    data = data.dropna()    
    data = data.reset_index(drop=True)
    
    n = len(data)
    k = len(data.columns)
    
    if success is None:
        suc = data.iloc[0,0]
    else:
        suc = success
    
    isSuc = data == suc
    cj = isSuc.sum()
    ri = isSuc.sum(axis=1)
    ns = cj.sum()
    
    pi = ri/k
    
    delta = 1/(k*comb(n, 2)) * (cj*(n - cj)).sum()
    mu = 2/(n*(n - 1)) * (pi.sum()*(n - pi.sum()) - (pi*(1 - pi)).sum())
    r = 1 - delta/mu
    
    return r