First uncategorized solution for Roman Numerals by silg2 - python coding challenges

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First solution in Uncategorized category for Roman Numerals by silg2

def checkio(number):
    'return roman numeral using the specified integer value from range 1...3999'
    
    bases=[1000,900,500,400,100,90,50,40,10,9,5,4,1]
    letters=['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I']
    
    string=''
    
    for base, letter in list(zip(bases,letters)):
        root=number//base
        string+=''.join(letter for i in range(0,root))
        number-=root*base
    
    return string
    
if __name__ == '__main__':
    assert checkio(6) == 'VI', 'First'
    assert checkio(76) == 'LXXVI', 'Second'
    assert checkio(499) == 'CDXCIX', 'Third'
    assert checkio(3888) == 'MMMDCCCLXXXVIII', 'Fourth'
    print('All ok')

Nov. 10, 2012

Comments:

calops on July 5, 2013, 1:58 p.m.
<pre class='brush: python'>string+=''.join(letter for i in range(0,root)) </pre> is equivalent to (but less efficient): <pre class='brush: python'>string += letter * root </pre> Otherwise, decent solution.

nobrains on Sept. 14, 2013, 2:12 p.m.
<pre class='brush: python'> assert checkio(6) == 'VI', 'First' list(zip(bases,letters)) </pre> zip returns iterator so you don't need convert it to list

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