On Thu, Sep 9, 2010 at 3:11 PM, Roelof Wobben <rwob...@hotmail.com> wrote:
> > > ------------------------------ > From: anand.shash...@gmail.com > Date: Thu, 9 Sep 2010 15:08:10 +0530 > Subject: Re: [Tutor] recursive problem > To: rwob...@hotmail.com > CC: tutor@python.org > > > > On Thu, Sep 9, 2010 at 2:21 PM, Roelof Wobben <rwob...@hotmail.com> wrote: > > Hello, > > I have this : > > def recursive_count(target, nested_num_list): > """ > >>> recursive_count(2, [2, 9, [2, 1, 13, 2], 8, [2, 6]]) > 4 > >>> recursive_count(7, [[9, [7, 1, 13, 2], 8], [7, 6]]) > 2 > >>> recursive_count(15, [[9, [7, 1, 13, 2], 8], [2, 6]]) > 0 > >>> recursive_count(5, [[5, [5, [1, 5], 5], 5], [5, 6]]) > 6 > """ > for element in nested_num_list: > if type(element) == type([]): > test = recursive_count(target, element) > print element, target > if element == target : > count = count + 1 > return count > > if __name__ == "__main__": > import doctest > doctest.testmod() > > > Now I get this message : > > UnboundLocalError: local variable 'count' referenced before assignment > > > It is because you are doing count = count + 1 > But where is count defined. > > > > But if I do this : > > def recursive_count(target, nested_num_list): > """ > >>> recursive_count(2, [2, 9, [2, 1, 13, 2], 8, [2, 6]]) > 4 > >>> recursive_count(7, [[9, [7, 1, 13, 2], 8], [7, 6]]) > 2 > >>> recursive_count(15, [[9, [7, 1, 13, 2], 8], [2, 6]]) > 0 > >>> recursive_count(5, [[5, [5, [1, 5], 5], 5], [5, 6]]) > 6 > """ > count = 0 > for element in nested_num_list: > if type(element) == type([]): > test = recursive_count(target, element) > print element, target > if element == target : > count = count + 1 > return count > > if __name__ == "__main__": > import doctest > doctest.testmod() > > The count will always be 0 if a nested list is being found. > > What's the python way to solve this > > > I am not sure what do you want to achieve by this ? > What is the problem statement ? > > The problem statement is that I must count how many times the target is in > the nested_list. > So I thougt that every nested list the function is called again so this > list is also iterated. > Let's say n, l = 2, [2, 9, [2, 1, 13, 2], 8, [2, 6]] Simply Flatten the list l, which will be then be; flatten(l) = [ 2, [2, 9, 2, 1,13, 2, 8, 2, 6 ] Now count for n; flatten.count(n) -- ~l0nwlf
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