shawn bright wrote:
> Thanks Kent,
>i am going with option A, the helper set, because i also need to
> count the occurances and this seems to be the easiest solution.
If you need the number of unique items that is just the length of the
final list. If you need the number of occurrences of ea
Thanks Kent,
i am going with option A, the helper set, because i also need to count
the occurances and this seems to be the easiest solution.
thanks for your help.
shawn
On 1/28/07, Kent Johnson <[EMAIL PROTECTED]> wrote:
shawn bright wrote:
> lo there all.
>
> i have a list of lists that i
shawn bright wrote:
> lo there all.
>
> i have a list of lists that i want to build, only if an item is not in
> the list already.
>
> kinda like this
> new_list = []
> for item in lists: # item will look something like [var1, var2, var3]
> if item[0] in new_list ( only the first element of
"shawn bright" <[EMAIL PROTECTED]> wrote
> new_list = []
> for item in lists: # item will look something like [var1, var2,
> var3]
>if item[0] in new_list
> basicly, i want to know if item[0] is one of the items[0] in my
> new_list
Your pseudo code is pretty much exactly right.
What mor
