How to check if there's any odd/even numbers in an Iterable (e.g. list/tuple)?

Question

Suppose we're checking if there's any odd numbers in a list. The most direct way is:

def has_odd(L):
    for v in L:
        if v % 2 == 1:
            return True
    return False

The has_odd function checks if there's any odd numbers in a list, once an odd number is found, it returns True. But this seems a bit verbose. A more concise way using reduce is as follow:

reduce(lambda res, v: res or bool(v), L, False)

But this will iterate through all elements and is unnecessary, because once an odd number is found the result is surely True.

So, are there any other ways to do this?

Answer 1

You can use the any() function to reduce the verbosity:

>>> l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> any(n % 2 == 1 for n in l)
True
>>>

Note however, any() is pretty much the same as you had originally just generalized, so don't expect a speed improvement:

def any(iterable):
    for element in iterable:
        if element:
            return True
    return False

Answer 2

first of all let's write small indicator function for "oddness" like

def is_odd(number):
    return number % 2

then we can write our indicator for "has at least one odd number" using any with imap/map

• Python 2.*

  from itertools import imap
  
  
  def has_odd_number(iterable):
      return any(imap(is_odd, iterable))
  

• Python 3.*

  def has_odd_number(iterable):
      return any(map(is_odd, iterable))
  

or with generator expression

def has_odd_number(iterable):
    return any(is_odd(number) for number in iterable)

Examples:

>>> has_odd_number([0])
False
>>> has_odd_number([1])
True