(Using Python 3.1)
I know this question has been asked many times for the general question of testing if iterator is empty; obviously, there's no neat solution to that (I guess for a reason - an iterator doesn't really know if it's empty until it's asked to return its next value).
I have a specific example, however, and was hoping I can make clean and Pythonic code out of it:
#lst is an arbitrary iterable
#f must return the smallest non-zero element, or return None if empty
def f(lst):
flt = filter(lambda x : x is not None and x != 0, lst)
if # somehow check that flt is empty
return None
return min(flt)
Is there any better way to do that?
EDIT: sorry for the stupid notation. The parameter to the function is indeed an arbitrary iterable, rather than a list.
5 Answers
def f(lst):
flt = filter(lambda x : x is not None and x != 0, lst)
try:
return min(flt)
except ValueError:
return None
min throws ValueError when the sequence is empty. This follows the common "Easier to Ask for Forgiveness" paradigm.
EDIT: A reduce-based solution without exceptions
from functools import reduce
def f(lst):
flt = filter(lambda x : x is not None and x != 0, lst)
m = next(flt, None)
if m is None:
return None
return reduce(min, flt, m)
t = [1,2,3]
if any(filter(lambda x: x == 10, t)):
print("found 10")
def f(lst):
# if you want the exact same filtering as the original, you could use
# lst = [item for item in lst if (item is not None and item != 0)]
lst = [item for item in lst if item]
if lst: return min(lst)
else: return None
the list comprehension only allows items that don't evaluate to boolean false (which filters out 0 and None)
an empty list i.e. [] will evaluate to False, so "if lst:" will only trigger if the list has items
you can go for reduce expression too return reduce(lambda a,b: a<b and a or b,x) or None
If you just want to check if the return of filter is empty, you might do (Python3)
len(list(filter(lambda e : e == 2, [1,2,3])))
But notice, hence filter is a generator if you this test twice, second time, you will receive a diferent result:
len(list(filter(lambda e : e == 2, [1,2,3])))
len(list(filter(lambda e : e == 2, [1,2,3])))
>>> 1
>>> 1
But:
f = filter(lambda e : e == 2, [1,2,3])
len(list(f))
len(list(f))
>>> 1
>>> 0
