How to check if there are even / even numbers in Iterable (e.g. list / tuple)?

Question

Suppose we check for any odd numbers in list . The most direct way:

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

The has_odd function checks for any odd numbers in list , once an odd number is found, it returns True . But that seems a bit verbose. A more concise way to use reduce is as follows:

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

But it will be an iteration over all elements and is not needed, because as soon as an odd number is found, the result is certainly True .

So, are there other ways to do this?

Answer 1

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

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

Note, however, any() almost the same as you originally just generalized , so don't expect speed improvements:

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

Answer 2

first of all write a small indicator function for "oddness", for example,

 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 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 

Answer 3

Another way. you can use not all()

 >>> l = [2, 4, 5, 6, 7, 8, 9, 10] >>> not all(n%2==1 for n in l) True