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]

iflen(myList) % 2 == 0:
    print ("even")
elseprint ("odd")

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

def is_even(myList):

    iflen(myList) % 2 == 0:
        returntrueelse:
        returnfalsemain():

    myList = [0,1,2,3]
    theListIsEven = is_even(myList)  # will be true in this example# because 4 items in myListiftheListIsEven(myList) == True:
        # do somethingelse:
        # do something elsereturn0

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

iflen(listName) % 2 == 0:
    returnTrue# the number is even!else:
    returnFalse# 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.iflen(your_list) % 2 == 0:
    print"Even Number"else:
    print"number is odd"

Solution 4:

iflen(mylist)%2==0:
     #evenelse:
     #odd

Solution 5:

defhas_even_length(some_sequence):
    returnnotlen(some_sequence)%2defhas_odd_length(some_sequence):
    returnbool(len(some_sequence)%2)