Log base 2 calculation in python

Question

I am trying to calculate the average disorder in ID trees. My code is below:

Republican_yes = yes.count('Republican')
Democrat_yes = yes.count('Democrat')
Republican_no = no.count('Republican')
Democrat_no = no.count('Democrat')
Indep_yes = yes.count('Independent')
Indep_no = no.count('Independent')

disorder_yes= Republican_yes/len(yes)*(math.log(float(Republican_yes)/len(yes),2))+   Democrat_yes/len(yes)*(math.log(float(Democrat_yes)/len(yes),2))+Indep_yes/len(yes)*(math.log(float(Indep_yes)/len(yes),2))

disorder_no= Republican_no/len(no)*(math.log(float(Republican_no)/len(no),2))+Democrat_no/len(no)*(math.log(float(Democrat_no)/len(no),2))+Indep_no/len(no)*(math.log(float(Indep_no)/len(no),2))

avgdisorder = -len(yes)/(len(yes)+len(no))*disorder_yes - len(no)/(len(yes)+len(no))*disorder_no

return avgdisorder

why do I keep getting math domain error?

Have you tried checking to make sure none of the variable values are zero? That is the most likely problem. — user1613254
– user1613254, Commented Nov 18, 2013 at 4:46
Thanks! How do I only choose non-zero values for the formula? — user3003329
– user3003329, Commented Nov 18, 2013 at 5:19

shad0w_wa1k3r · Accepted Answer · 2013-11-18 05:57:29Z

1

Check if the lengths are 0 or not, else you will get MathError.

if len(yes):
    disorder_yes= Republican_yes/len(yes)*(math.log(float(Republican_yes)/len(yes),2))+   Democrat_yes/len(yes)*(math.log(float(Democrat_yes)/len(yes),2))+Indep_yes/len(yes)*(math.log(float(Indep_yes)/len(yes),2))

if len(no):
    disorder_no= Republican_no/len(no)*(math.log(float(Republican_no)/len(no),2))+Democrat_no/len(no)*(math.log(float(Democrat_no)/len(no),2))+Indep_no/len(no)*(math.log(float(Indep_no)/len(no),2))

if len(yes) or len(no):
    avgdisorder = -len(yes)/(len(yes)+len(no))*disorder_yes - len(no)/(len(yes)+len(no))*disorder_no

If you want, you can always add the else clause for all 3 if statements as per your requirement.

answered Nov 18, 2013 at 5:57

shad0w_wa1k3r

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