reduce expression to test sublist
Terry Reedy
tjreedy at udel.edu
Sat Jan 5 16:55:59 EST 2013
On 1/5/2013 1:25 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)
reduce(lambda x, y: x and (y in lst2), lst1, True)
would work, but as other have said, you want to stops at the first
False, which all() does, and if lst2 is a list of hashable items, x in
set for O(1) rather than O(n) check.
> Moreover, for the lists of strings the following for-loop gives
> correct results when the above reduce expression doesn't.
You should include data for testing.
>
> isSublist = True
> for i in lst1:
> isSublist = isSublist and (i in> lst2)
If isSublist remains True, there is no need to rebind it
> if not isSublist:
> isSublist = False
Setting isSublist False when it is False is redundant.
> break
Taking into account the comments:
def is_sublist(a, b):
b = set(b) # optional, depending on contents
for item in a:
if item not in b:
return False
else: # not needed because return above
return True
The code for all() is similar, except that all takes an iterable, so
that the testing is done as part of the input iterable.
def all(iterable):
it = iter(iterable):
for item in it:
if not it:
return False:
else:
return True
def issublist(a, b):
b = set(b)
return all(item in b for item in a)
--
Terry Jan Reedy
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