Find the index of the n'th item in a list

Question

I want to find the index of the n\'th occurrence of an item in a list. e.g.,

x=[False,True,True,False,True,False,True,False,False,False,True,False,True]
         


        

Answer 1

A solution that first creates a list object and returns the nth-1 element of this list : function occurence()

And a solution that fulfill functional programmers'dreams too, I think, using generators, because I love them : function occur()

S = 'stackoverflow.com is a fantastic amazing site'
print 'object S is string %r' % S
print "indexes of 'a' in S :",[indx for indx,elem in enumerate(S) if elem=='a']

def occurence(itrbl,x,nth):
    return [indx for indx,elem in enumerate(itrbl)
            if elem==x ][nth-1] if x in itrbl \
           else None

def occur(itrbl,x,nth):
    return (i for pos,i in enumerate(indx for indx,elem in enumerate(itrbl)
                                     if elem==x)
            if pos==nth-1).next() if x in itrbl\
            else   None

print "\noccurence(S,'a',4th) ==",occurence(S,'a',4)
print "\noccur(S,'a',4th) ==",occur(S,'a',4)

result

object S is string 'stackoverflow.com is a fantastic amazing site'
indexes of 'a' in S : [2, 21, 24, 27, 33, 35]

occur(S,'a',4th) == 27

occurence(S,'a',4th) == 27

The second solution seems complex but it isn't really. It doesn't need to run completely through the iterable: the process stops as soon as the wanted occurence is found.