How to sum even and odd values with one for-loop and no if-condition?

Purely for educational purposes (and a bit of fun), here is a solution that does not use any for loops at all. (Granted, in the underlying logic of the functions, there are at least five loops.)

num = list(range(int(input('Enter number: '))))

even = num[::2]
odd = num[1::2]

print('Even list:', even)
print('Odd list:', odd)

print('Even:', sum(even))
print('Odd:', sum(odd))

Output:

Enter number: 10
Even list: [0, 2, 4, 6, 8]
Odd list: [1, 3, 5, 7, 9]
Even: 20
Odd: 25

How does it work?

• The input() function returns a str object, which is converted into an integer using the int() function.
• The integer is wrapped in the range() and list() functions to convert the given number into a list of values within that range.
  • This is a convention you will use/see a lot through your Python career.
• List slicing is used to get every second element in the list. Given the list is based at zero, these will be even numbers.
• Slice the same list again, starting with the second element, and get every second element ... odd numbers.
  • Link to a nice SO answer regarding slicing in Python.
• The simply use the sum() function to get the sums.

for n in range(number):
    x += (1 - n % 2) * n
    y += (n % 2) * n

You asked for a solution with one loop, but how about a solution with no loop?

It is well known that the sum of the numbers from 1 to n is (n+1)*n/2. Thus, the sum of even numbers is 2 * (m+1)*m/2 with m = n//2 (i.e. floor(n/2)). The sum of odd can then be calculated by the sum of all numbers minus the sum of even numbers.

n = 12345
m = n // 2
e = (m+1)*m
o = (n+1)*n//2 - e

Verification:

>>> e, e==sum(i for i in range(n+1) if i % 2 == 0)
38112102 True
>>> o, o==sum(i for i in range(n+1) if i % 2 == 1)
38105929 True

Note: This calculates the sums for number up to and including n.

for n in range(1,number,2):
  x += n
  y += n-1
print(y)
print(x)

This code has the same output with the example.

Ternary operator:

for n in range(number):
  x += (n,0)[n%2]
  y += (0,n)[n%2]

I think you are a beginner. I wouldn't like to confuse you with slicing operators complex implementation.

As you mentioned

The purpose of the code is to find the sum of the even and the sum of the odd numbers from zero to the number inputted and print it to the screen.

There is no need to find the initial number is odd/even And your program is wrong if you want to include the input number in calculating the even/odd sum.

Example

Input

5

Expected Output

6 9

Explanation

Even Sum : 2+4 = 6

Odd Sum : 1+3+5 = 9

Your Output

6 4 (wrong output)

The range() function will exclude the number. It will only iterate from 0 to 4 while the input is 5. so if you want to include 5, you should add 1 to the number while passing it in the range() function.

number = int(input())
x = 0
y = 0
for n in range(number+1):
    x += (1 - n % 2) * n  #this will add 0 if not even
    y += (n % 2) * n      #this will add 0 if not odd
print(x)
print(y)

There is also mathematical way:

num = int(input("Enter number:"))
odd = ((num+1)/2)**2
even = num*(num+1)/2 - odd

The sum of the first n odd numbers is n^2. To get count of odd numbers we use (num+1)/2. To get sum of even numbers, we could use similar approach, but I preferred, subtracting odd from the sum of the first n numbers, which is n*(n+1)/2.

Here is my 2cents if we are allowed to use numpy.

import numpy as np
number = int(input())

l = np.array(range(number)) 
print('odd:',sum(l % 2 * l))
print('even:', sum((1- l % 2) * l))

If you're allowed to use a list

number = int( input() )
counts = [ 0, 0 ]
for n in range( number ):
    counts[ n % 2 ] += n
print( counts[ 0 ] )
print( counts[ 1 ] )

There's another way which sums the odd and even indices together in the for loop based on the remainder modulo 2:

number = int(input())
odd = 0
even = 0
for i in range(len(number)):
    odd += i * (i % 2)
    even += i * ((i + 1) % 2)

print (odd, even)

number = 1000000
x = 0
y = 0
[(x:=x+n, y:=y+(n+1)) for n in range(0,number,2)]
print(f'{x}, {y}')

This uses a list comprehension and the new Python assignment operator.

Sum of first n numbers is n(n+1)/2 (Mathematically derived). So if we know the value of n then we can find the sum of all numbers from 1 to n.

If we find the sum of all odd numbers before n and subratract it from the sum of first n we get he sum of all even numbers before n.

Here's the code:

n = int(input("Enter a number: "))
odd = 0
for i in range(1,n+1,2):
    odd += i
even = int(n*(n+1)/2) - odd
print("even:",even,"odd:",odd)