Need help using range for loop

It's important to know if n is greater than m and invert situation if so. Other than that, you need to know if the smallest number is odd or even and begin the two ranges accordingly:

n = int(input("Please enter a number: "))
m = int(input("Please enter another number: "))

# n will always be the smaller one
if n > m:
    n, m = m, n

n_is_odd = n % 2 # Gives 1 if n is odd

n_even = n + n_is_odd # Sum 1 if n is odd
n_odd = n + (not n_is_odd) # Sum 1 if n is even

print("the total sum of even numbers is %d" % sum(range(n_even, m+1, 2)) )
print("the total sum of odd numbers is %d" % sum(range(n_odd, m+1, 2)) )

Input validation is a big part of good coding. A good overview can be found here:

• Asking the user for input until they give a valid response

To make the validation it reusable I put the validation in a function that only accept integers and (if a minval is provided, makes sure that the input is bigger that the minval.

def while_invalid_ask_input_return_integer(text, minval = None):
    """Aks for input until a number is given that is > minval if minval not None
    returns an integer."""
    while True:
        c = input (text)
        try:
            c = int(c)
            if minval is not None and c < minval:
                raise ValueError  # its too small, raise an erros so we jump to except:
            return c
        except ValueError:
            if minval is not None:
                print("must be a number and greater ", minval)
            else:
                print("not a number")

I use it to get the first number, and the second number gets the first one as "constraint" so it will be bigger. For summation I just use the range starting once with n once with n+1 till m and a range step of 2. I check what even/oddness n has and print text accordingly:

n = while_invalid_ask_input_return_integer("please enter a number ")
m = while_invalid_ask_input_return_integer("enter number bigger then {}".format(n),n)

print( "Odd sum:" if n % 2 == 1 else "Even sum:", sum(range(n,m+1,2)) )
print( "Even sum:" if n % 2 == 1 else "Odd sum:", sum(range(n+1,m+1,2)) )

Output:

please enter a number k
not a number
please enter a number 55
enter number bigger then 55 2
must be a number and greater  55
enter number bigger then 55 150
Odd sum: 4896
Even sum: 4944

Doku:

• sum(iterable)
• try: except: error handling
• python ternary operator (thats the thing @ "Odd sum:" if n % 2 == 1 else "Even sum:" in the print statement)

Here's a function I think fits into the description of what you asked above. It returns None if the user doesn't enter the type of query he or she wants.

So query can either be odd or even and depending on this, it calculates the sum that you want. The function makes use of list comprehension which is super cool too.

def calculate_odd_or_even_sum(query):
    start = int(input(" please enter a number"))
    end = int(input(" please enter another number"))
    count = 0

    if query == 'even':
        return sum([x for x in range(start, end) if x % 2 == 0])
    elif query == 'odd':
        return sum([x for x in range(start, end) if x % 2 != 0])
    else:
        return 0