Python filter / max combo - checking for empty iterator

Question

(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 t

Answer 1

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

Answer 2

you can go for reduce expression too return reduce(lambda a,b: a<b and a or b,x) or None

Answer 3

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)

Answer 4

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