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It is a practice from the site edabit.com and you have to check if the input is valid. That is the original task:

Given an RGB(A) CSS color, determine whether its format is valid or not. Create a function that takes a string (e.g. "rgb(0, 0, 0)") and return True if it's format is correct, otherwise return False.

The rules are shown in the testing code:

# True Tests
Test.assert_equals(valid_color('rgb(0,0,0)'), True, 'rgb lowest valid numbers')
Test.assert_equals(valid_color('rgb(255,255,255)'), True, 'rgb highest valid numbers')
Test.assert_equals(valid_color('rgba(0,0,0,0)'), True, 'rgba lowest valid numbers')
Test.assert_equals(valid_color('rgba(255,255,255,1)'), True, 'rgba highest valid numbers')
Test.assert_equals(valid_color('rgba(0,0,0,0.123456789)'), True, 'alpha can have many decimals')
Test.assert_equals(valid_color('rgba(0,0,0,.8)'), True, 'in alpha the number before the dot is optional')
Test.assert_equals(valid_color('rgba(   0 , 127 , 255 , 0.1 )'), True, 'whitespace is allowed around numbers (even tabs)')
Test.assert_equals(valid_color('rgb(0%,50%,100%)'), True, 'numbers can be percentages')

# False Tests
Test.assert_equals(valid_color('rgb(0,,0)'), False, 'INVALID: missing number')
Test.assert_equals(valid_color('rgb (0,0,0)'), False, 'INVALID: whitespace before parenthesis')
Test.assert_equals(valid_color('rgb(0,0,0,0)'), False, 'INVALID: rgb with 4 numbers')
Test.assert_equals(valid_color('rgba(0,0,0)'), False, 'INVALID: rgba with 3 numbers')
Test.assert_equals(valid_color('rgb(-1,0,0)'), False, 'INVALID: numbers below 0')
Test.assert_equals(valid_color('rgb(255,256,255)'), False, 'INVALID: numbers above 255')
Test.assert_equals(valid_color('rgb(100%,100%,101%)'), False, 'INVALID: numbers above 100%')
Test.assert_equals(valid_color('rgba(0,0,0,-1)'), False, 'INVALID: alpha below 0')
Test.assert_equals(valid_color('rgba(0,0,0,1.1)'), False, 'INVALID: alpha above 1')

I fear that my solution is not very elegant, efficent or short:

def create_rgbA_tuple(color):
    indexBracket1 = color.rfind("(")
    save = color[0] + color[1] + color[2] + color[3]
    if save != 'rgb(' and save != 'rgba':
        return False
    indexBracket2 = color.rfind(")")
    bracketArea = ""
    for i in range(indexBracket1 + 1, indexBracket2):
        bracketArea = bracketArea + color[i]

    bracketArea = bracketArea.split(',')  # bracketArea is now an Array (r,g,b/A)
    if len(bracketArea) == 4 and save != 'rgba' or len(bracketArea) == 3 and save != 'rgb(':
        return False
    try:
        for i in range(len(bracketArea)):
            bracketArea[i] = float(bracketArea[i])
    except:
        try:
            for i in range(len(bracketArea)):
                if bracketArea[i][-1] == "%":
                    bracketArea[i] = int(bracketArea[i].strip('%')) / 100 * 255
                else:
                    return False
        except:
            return False
    return bracketArea


def valid_color(color):
    rgb = create_rgbA_tuple(color)
    if not rgb:
        return False
    for i in range(3):
        if int(rgb[i]) < 0 or int(rgb[i]) > 255:
            return False

    if color[3] == "a":  # if rgbA
        if rgb[3] < 0 or rgb[3] > 1:
            return False
    return True
```
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1 Answer 1

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One thing the tests don't indicate is whether space is allowed before/after the color string. I'm inclined to be generous, so I'd suggest just discarding any space before you start:

color = color.strip()

Next, you are doing a lot of work to replace some python built-in functions. Before you do anything else, you should probably read the String Methods documentation to see what tools are available to you.

Try something like:

is_rgb = color.startswith('rgb(')
is_rgba = color.startswith('rgba(')

if not is_rgb and not is_rgba:
    return False

if not color.endswith(')'):
    return False

# strip off prefix "...(" and suffix ")"   
color = color[color.index('(') + 1:-1]

You can use .endswith to check for percent signs, also.

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  • \$\begingroup\$ Thank you for your advise! You are right these inbuilt functions are far more elegant, I guess I'll start learning some. \$\endgroup\$
    – Jonas
    Commented Aug 7, 2020 at 21:19

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