How To Know If A List Has An Even Or Odd Number Of Elements

How can I find out if there is even, or odd, number of elements in an arbitrary list. I tried list.index() to get all of the indices... but I still don't know how I can tell the pr

Solution 1:

You can use the built in function len() for this.

Python Doc -- len()

Gets the length (# of elements) of any arbitrary list.

myList = [0,1,2,3,4,5]

if len(myList) % 2 == 0:
    print ("even")
else
    print ("odd")

Define function that returns a bool (true or false).

def is_even(myList):

    if len(myList) % 2 == 0:
        return true
    else:
        return false

main():

    myList = [0,1,2,3]
    theListIsEven = is_even(myList)  # will be true in this example
                                     # because 4 items in myList

    if theListIsEven(myList) == True:
        # do something
    else:
        # do something else

    return 0

The modulus operator % gives the remainder.

EX: 7 % 2 = 1

• Closest number to 7 that 2 will divide evenly is 6
• Which is 1 away from 7.
• Thus, remainder of 1 for 7 % 2.

EX: 4 % 2 = 0

• Any even number n will give 0 as the remainder when n % 2
• Because n has divided evenly by 2

Solution 2:

All you need is

len(listName)

Which will give you the length.

I guess you could also do this then

if len(listName) % 2 == 0:
    return True  # the number is even!
else:
    return False # the number is odd!

Solution 3:

your_list = [1,2,3,(4,5)]

# modulo operation finds the remainder of division of one number by another.
if len(your_list) % 2 == 0:
    print "Even Number"
else:
    print"number is odd"

Solution 4:

if len(mylist)%2==0:
     #even
else:
     #odd

Solution 5:

def has_even_length(some_sequence):
    return not len(some_sequence)%2

def has_odd_length(some_sequence):
    return bool(len(some_sequence)%2)