reduce expression to test sublist

[Dave Angel]

On 01/05/2013 01:25 PM, Asim wrote:
> Hi All
>
> The following reduce expression checks if every element of list lst1 is present in list lst2.  It works as expected for integer lists but for lists of strings, it always returns False.
>
>    reduce( lambda x,y: (x in lst2) and (y in lst2), lst1)
>
> Moreover, for the lists of strings the following for-loop gives correct results when the above reduce expression doesn't.
>
>    isSublist = True
>    for i in lst1:
>       isSublist = isSublist and (i in lst2)
>       if not isSublist:
>          isSublist = False
>          break
>
>
> Can someone help me understand why?
>
> Asim
reduce only makes sense if the value you're reducing to is of the same
type as the elements of the list you're iterating over.  Since your
lambda expression returns a boolean (True or False), it'll seem to work
on some lists of ints.  That's just a coincidence since the bools are
compatible with ints, and maybe you've got a 0 or 1 in the list.  But if
your list has only strings, then True will never be that list.

If you're trying to make a faster loop, then I suggest you look into set
differences.  Turn both lists into sets, and subtract them.   Something
like (untested):

   result =  not bool( set(lst1) - set(lst2) )



-- 

DaveA