如何检查Iterable中是否有任何奇数/偶数（例如列表/元组）？

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

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