Python odd number diamond

I worked up a quick diamond program that takes only odd numbers while returning an error with even numbers or 0.

The code looks rather ugly to me, especially with the recurring int(). I haven't found out how to clean that up with my ENTER to quit.

while True:
    number = raw_input("Enter an odd number or hit ENTER to quit program: ")
    if number == "":
        print "\nQuitting program...\n"
        break
    elif int(number) == 0:
        print "\nCannot print symbols for a zero value, try again.\n"
    elif int(number) % 2 == 0:
        print "\nThat's an even number, try again.\n"
    else:   
        for i in range((int(number))+1):
            if i % 2 == 0:
                None
            else:
                print ("*" * i).center(int(number))

        for i in range((int(number))-1):
            if i % 2 == 0:
                None
            else: 
                print ("*" * ((int(number)) - (i + 1))).center(int(number))

• You can test for emptiness of number by just if not number

• After testing for emptiness, you could go straight to an int by adding number = int(number)

• Inside the loops, when you test for i % 2 == 0:, and then put None as the loop body (that should really be pass), you could just test for if i % 2 != 0:, and replace None with the else body, and remove the else.

• 0 is even, so you can remove that test.

• You can replace int(number) % 2 == 0 with not int(number) % 2.

• Looking at the for loops, the range could be changed to range(1, number-1, 2), removing the need for an if statement.

• You can also invert the last range, so the subtracting from number and 1 is not needed.

• Python automatically adds newlines, so the \ns aren't really needed.

That leaves us with (after removing a few useless brackets):

while True:
    number = raw_input("Enter an odd number or hit ENTER to quit program: ")
    if not number:
        print "Quitting program..."
        break
    number = int(number)
    if not number % 2:
        print "That's an even number, try again."
    else:   
        for i in range(1, number + 1, 2):
            print ("*" * i).center(number)
        for i in range(number - 2, 0, -2):
            print ("*" * i).center(number)

If you want to go functional, you can move your user input checking into a function. In this function you could also check if the user inputs a non-number, such as abc, by catching the exception or using str.isdigit:

def get_input():
    while True:
        number = raw_input("Enter an odd number or hit ENTER to quit program: ")
        if not number:
            return  # return None when the user wants to quit
        if number.isdigit() and int(number) % 2:
            return int(number)  # return the number, as its odd
        print "That's not a valid input (must be an odd integer), try again."

With this function, the rest of your program reduces to:

for number in iter(get_input, None):
    for i in range(1, number + 1, 2):
        print ("*" * i).center(number)
    for i in range(number - 2, 0, -2):
        print ("*" * i).center(number)

print "Quitting program..."

iter calls the function and gives its value until the function returns None (the sentinel value).

On a stylistic note:

• You should always have spaced around operators, so range(int(number)+1) should be range(int(number) + 1) (that's PEP8).

• You may want to add a big fat comment at the start of your program describing what it does.

Also, nice use of str.center, it avoided having too many loops!

StackExchange Code Review Q#68365, answer score: 11