To make a straightforward implementation for it in Python we first do a bit of math. For a 2x2 contingency table:
$$$ G = 2 \Bigg({O_{1,1} \ln(\frac{O_{1,1}}{E_{1,1}})} + {O_{1,2} \ln(\frac{O_{1,2}}{E_{1,2}})} + {O_{2,1} \ln(\frac{O_{2,1}}{E_{2,1}})} + {O_{2,2} \ln(\frac{O_{2,2}}{E_{2,2}})} \Bigg) $$$
Expand via log equivalences:
| $$$ G = 2 \Bigg( $$$ | $$$ \bigg( {O_{1,1} \ln(O_{1,1}) - O_{1,1} \ln(E_{1,1})} \bigg) + \bigg( {O_{1,2} \ln(O_{1,2}) - O_{1,2} \ln(E_{1,2})} \bigg) $$$ |
| $$$ + \bigg( {O_{2,1} \ln(O_{2,1}) - O_{2,1} \ln(E_{2,1})} \bigg) + \bigg( {O_{2,2} \ln(O_{2,2}) - O_{2,2} \ln(E_{2,2})} \bigg) \Bigg) $$$ |
Group Observed vs Expected together:
| $$$ G = 2 \Bigg( $$$ | $$$ \bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg)$$$ |
| $$$ - \bigg( O_{1,1} \ln(E_{1,1}) + O_{1,2} \ln(E_{1,2}) + O_{2,1} \ln(E_{2,1}) + O_{2,2} \ln(E_{2,2}) \Bigg) $$$ |
For a 2x2 contingency table; $$$E_{i,j} = \frac{T_{i} \cdot T_{j}}{T} $$$, where $$$T_{i}$$$ and $$$T_{j}$$$ are the totals for the row and column of cell $$$E_{i,j}$$$ and $$$T$$$ is the total for all cells in the table. Specifically, $$$$ \ln(E_{1,1}) = \ln \Bigg( \frac{ (O_{1,1}+O_{1,2}) \cdot (O_{1,1}+O_{2,1}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Bigg) $$$$ $$$$ \ln(E_{1,2}) = \ln \Bigg( \frac{ (O_{1,1}+O_{1,2}) \cdot (O_{1,2}+O_{2,2}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Bigg) $$$$ $$$$ \ln(E_{2,1}) = \ln \Bigg( \frac{ (O_{2,1}+O_{2,2}) \cdot (O_{1,1}+O_{2,1}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Bigg) $$$$ $$$$ \ln(E_{2,2}) = \ln \Bigg( \frac{ (O_{2,1}+O_{2,2}) \cdot (O_{1,2}+O_{2,2}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Bigg) $$$$
We then substitute these into our equation for G.
| $$$ G = 2 \Bigg( $$$ | $$$\bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg) - $$$ |
| $$$ \bigg( O_{1,1} \ln \Big( \frac{ (O_{1,1}+O_{1,2}) \cdot (O_{1,1}+O_{2,1}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Big) + O_{1,2} \ln \Big( \frac{ (O_{1,1}+O_{1,2}) \cdot (O_{1,2}+O_{2,2}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Big) + $$$ | |
| $$$ O_{2,1} \ln \Big( \frac{ (O_{2,1}+O_{2,2}) \cdot (O_{1,1}+O_{2,1}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Big) + O_{2,2} \ln \Big( \frac{ (O_{2,1}+O_{2,2}) \cdot (O_{1,2}+O_{2,2}) }{ O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} } \Big) \bigg) \Bigg) $$$ |
Another log equivalence gives us this:
| $$$ G = 2 \Bigg( $$$ | $$$\bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg) - $$$ |
| $$$ \bigg( O_{1,1} \ln ( (O_{1,1}+O_{1,2}) \cdot (O_{1,1}+O_{2,1}) ) - O_{1,1} \ln ( O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} ) + $$$ | |
| $$$ O_{1,2} \ln ( (O_{1,1}+O_{1,2}) \cdot (O_{1,2}+O_{2,2}) ) - O_{1,2} \ln ( O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} ) + $$$ | |
| $$$ O_{2,1} \ln ( (O_{2,1}+O_{2,2}) \cdot (O_{1,1}+O_{2,1}) ) - O_{2,1} \ln ( O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} ) + $$$ | |
| $$$ O_{2,2} \ln ( (O_{2,1}+O_{2,2}) \cdot (O_{1,2}+O_{2,2}) ) - O_{2,2} \ln ( O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} ) \bigg) \Bigg) $$$ |
Group common factor of $$$\ln ( O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2} )$$$:
| $$$ G = 2 \Bigg( $$$ | $$$ \bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg) +$$$ |
| $$$ \bigg( (O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \cdot \ln(O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \bigg) - $$$ | |
| $$$ \bigg( O_{1,1} \ln \big( (O_{1,1}+O_{1,2}) \cdot (O_{1,1}+O_{2,1}) \big) + O_{1,2} \ln \big( (O_{1,1}+O_{1,2}) \cdot (O_{1,2}+O_{2,2}) \big) + $$$ | |
| $$$ O_{2,1} \ln \big( (O_{2,1}+O_{2,2}) \cdot (O_{1,1}+O_{2,1}) \big) + O_{2,2} \ln \big( (O_{2,1}+O_{2,2}) \cdot (O_{1,2}+O_{2,2}) \big) \bigg) \Bigg) $$$ |
More log equivalences:
| $$$ G = 2 \Bigg( $$$ | $$$ \bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg) + $$$ |
| $$$ \bigg( (O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \cdot \ln(O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \bigg) -$$$ | |
| $$$ \bigg( O_{1,1} \ln (O_{1,1}+O_{1,2}) + O_{1,1} \ln (O_{1,1}+O_{2,1}) + O_{1,2} \ln (O_{1,1}+O_{1,2}) + O_{1,2} \ln (O_{1,2}+O_{2,2}) + $$$ | |
| $$$ O_{2,1} \ln (O_{2,1}+O_{2,2}) + O_{2,1} \ln (O_{1,1}+O_{2,1}) + O_{2,2} \ln (O_{2,1}+O_{2,2}) + O_{2,2} \ln (O_{1,2}+O_{2,2}) \bigg) \Bigg) $$$ |
Group common factors of $$$\ln (O_{1,1}+O_{1,2})$$$, $$$\ln (O_{1,1}+O_{2,1})$$$ etc. Yielding:
| $$$ G = 2 \Bigg( $$$ | $$$ \bigg( O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) \bigg) $$$ |
| $$$ + \bigg( (O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \cdot \ln(O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \bigg) $$$ | |
| $$$ - \bigg( (O_{1,1} + O_{1,2}) \ln (O_{1,1}+O_{1,2}) + (O_{2,1} + O_{2,2}) \ln (O_{2,1}+O_{2,2}) \bigg) $$$ | |
| $$$ - \bigg( (O_{1,1} + O_{2,1}) \ln (O_{1,1}+O_{2,1}) + (O_{1,2} + O_{2,2}) \ln (O_{1,2}+O_{2,2}) \bigg) \Bigg) $$$ |
We can then summarise the four parts of the above equation as follows:
$$$$ celltotals = O_{1,1} \ln(O_{1,1}) + O_{1,2} \ln(O_{1,2}) + O_{2,1} \ln(O_{2,1}) + O_{2,2} \ln(O_{1,2}) $$$$ $$$$ total = (O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) \ln(O_{1,1} + O_{1,2} + O_{2,1} + O_{2,2}) $$$$ $$$$ rowtotals = (O_{1,1} + O_{1,2}) \ln (O_{1,1}+O_{1,2}) + (O_{2,1} + O_{2,2}) \ln (O_{2,1}+O_{2,2}) $$$$ $$$$ columntotals = (O_{1,1} + O_{2,1}) \ln (O_{1,1}+O_{2,1}) + (O_{1,2} + O_{2,2}) \ln (O_{1,2}+O_{2,2}) $$$$ where: $$$$ G = 2 \big(celltotals + total - (rowtotals + columntotals) \big) $$$$ If we assign $$$a = O_{1,1}, b = O_{1,2}, c = O_{2,1}, d = O_{2,2}$$$ then we can then implement as follows in Python:
import math
def flnf(f):
return f*math.log(f) if f>0.5 else 0
def gtest(a,b,c,d):
row1 = a+b
row2 = c+d
col1 = a+c
col2 = b+d
total = flnf(a+b+c+d)
celltotals = flnf(a)+flnf(b)+flnf(c)+flnf(d)
rowtotals = flnf(row1)+flnf(row2)
coltotals = flnf(col1)+flnf(col2)
return 2*(celltotals + total-(rowtotals+coltotals))
We then probably want to convert our G-value into a P-value. The G-test approximates a chi squared distribution and has the same number of degrees of freedom.
If the calculated chi-squared distribution cumulative distribution function is equal to exactly 1 (i.e. a p-value of exactly 0) then we increase the precision and try again. This means that we can find the correct order of magnitude for our p-value (which is good for when we FDR correct these p-values later on) while minimising the number of calculations we need to do (the majority of conversions don't need 4000 decimal places computed). ($$$g$$$ is the G-value and $$$df$$$ is the degrees of freedom - 1 for a 2x2 table).
import mpmath
def mpmathcdf(g,df,dps=10):
mpmath.mp.dps = dps
x,k = mpmath.mpf(g), mpmath.mpf(df)
cdf = mpmath.gammainc(k/2, 0, x/2, regularized=True)
# floating point precision insufficient, use more precision
if cdf == 1.0:
if dps > 4000:
return cdf # give up after a certain point
else:
cdf = mpmathcdf(g,df,dps*2)
return cdf
def GtoP(g,df):
assert g >= 0, g
return float(1-mpmathcdf(g,df))
Update:
Following on from the comments below I've written up a generalised version of the G-test for any sized contingency table which uses the trick from above. I've not proven explicitly that this is necessarily generalisable for any sized table but I'm pretty sure it is.
def gtest(a):
rowtotals = []
coltotals = [0]*len(a[0])
cells = []
for row in a:
rowtotals.append(sum(row))
for index,col in enumerate(row):
coltotals[index] += col
cells.append(col)
return 2*(math.fsum([flnf(x) for x in cells]) + flnf(sum(cells)) - ( math.fsum([flnf(x) for x in rowtotals]) + math.fsum([flnf(x) for x in coltotals]) ))
For a table with 2 columns and 3 rows this can then be called as follows:
gtest([[268,807],
[199,759],
[42,184]])
> 7.300817075862142
Again, because I've not proven this will always work it should, of course, be used at your own risk. That said, the same technique is used for up to 10 columns and 50 rows in the G-test spreadsheet at the Handbook for Biological Statistics so it's probably safe at least that far.
