How can I sort elements of array in function?

July 5, 2023

Why I got: None
None

def odd_even(*number):
    even = []
    odd = []
    for i in number:
        if i % 2 == 0:
            even.append(i)
        else:
            odd.append(i)
    print(even.sort())
    print(odd.sort())

odd_even(1,5,7,8,4,3,26,59,48,7,5,45)

I just tried sorting elements of array
But result is

None
None

>Solution :

you can use the sorted() function which returns a new sorted list and leaves the original list unaffected:

def odd_even(*number):
    even = []
    odd = []
    for i in number:
        if i % 2 == 0:
            even.append(i)
        else:
            odd.append(i)
    print(sorted(even))
    print(sorted(odd))

odd_even(1,5,7,8,4,3,26,59,48,7,5,45)
#output
#[4, 8, 26, 48]
#[1, 3, 5, 5, 7, 7, 45, 59]

Other Method (Fix your solution):

If you want to print the sorted lists, you should sort the lists first and then print them:

def odd_even(*number):
    even = []
    odd = []
    for i in number:
        if i % 2 == 0:
            even.append(i)
        else:
            odd.append(i)
    even.sort()
    odd.sort()
    print(even)
    print(odd)

odd_even(1,5,7,8,4,3,26,59,48,7,5,45)

Other Methode:

In this code, we use a list comprehension to filter out even and odd numbers from number and sort them directly as they are created. Then we print the even and odd lists which are already sorted.

def odd_even(*number):
    even = sorted([i for i in number if i % 2 == 0])
    odd = sorted([i for i in number if i % 2 != 0])
    print(even)
    print(odd)
odd_even(1,5,7,8,4,3,26,59,48,7,5,45)