How do you match only valid roman numerals with a regular expression?

Question

You can use the following regex for this:

^M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$

Breaking it down, M{0,4} specifies the thousands section and basically restrains it to between 0 and 4000. It's a relatively simple:

   0: <empty>  matched by M{0}
1000: M        matched by M{1}
2000: MM       matched by M{2}
3000: MMM      matched by M{3}
4000: MMMM     matched by M{4}

You could, of course, use something like M* to allow any number (including zero) of thousands, if you want to allow bigger numbers.

Next is (CM|CD|D?C{0,3}), slightly more complex, this is for the hundreds section and covers all the possibilities:

  0: <empty>  matched by D?C{0} (with D not there)
100: C        matched by D?C{1} (with D not there)
200: CC       matched by D?C{2} (with D not there)
300: CCC      matched by D?C{3} (with D not there)
400: CD       matched by CD
500: D        matched by D?C{0} (with D there)
600: DC       matched by D?C{1} (with D there)
700: DCC      matched by D?C{2} (with D there)
800: DCCC     matched by D?C{3} (with D there)
900: CM       matched by CM

Thirdly, (XC|XL|L?X{0,3}) follows the same rules as previous section but for the tens place:

 0: <empty>  matched by L?X{0} (with L not there)
10: X        matched by L?X{1} (with L not there)
20: XX       matched by L?X{2} (with L not there)
30: XXX      matched by L?X{3} (with L not there)
40: XL       matched by XL
50: L        matched by L?X{0} (with L there)
60: LX       matched by L?X{1} (with L there)
70: LXX      matched by L?X{2} (with L there)
80: LXXX     matched by L?X{3} (with L there)
90: XC       matched by XC

And, finally, (IX|IV|V?I{0,3}) is the units section, handling 0 through 9 and also similar to the previous two sections (Roman numerals, despite their seeming weirdness, follow some logical rules once you figure out what they are):

0: <empty>  matched by V?I{0} (with V not there)
1: I        matched by V?I{1} (with V not there)
2: II       matched by V?I{2} (with V not there)
3: III      matched by V?I{3} (with V not there)
4: IV       matched by IV
5: V        matched by V?I{0} (with V there)
6: VI       matched by V?I{1} (with V there)
7: VII      matched by V?I{2} (with V there)
8: VIII     matched by V?I{3} (with V there)
9: IX       matched by IX

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Just keep in mind that that regex will also match an empty string. If you don't want this (and your regex engine is modern enough), you can use positive look-behind and look-ahead:

(?<=^)M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})(?=$)

(the other alternative being to just check that the length is not zero beforehand).

Actually, your premise is flawed. 990 IS "XM", as well as "CMXC".

The Romans were far less concerned about the "rules" than your third grade teacher. As long as it added up, it was OK. Hence "IIII" was just as good as "IV" for 4. And "IIM" was completely cool for 998.

(If you have trouble dealing with that... Remember English spellings were not formalized until the 1700s. Until then, as long as the reader could figure it out, it was good enough).

Just to save it here:

(^(?=[MDCLXVI])M*(C[MD]|D?C{0,3})(X[CL]|L?X{0,3})(I[XV]|V?I{0,3})$)

Matches all the Roman numerals. Doesn't care about empty strings (requires at least one Roman numeral letter). Should work in PCRE, Perl, Python and Ruby.

Online Ruby demo: http://rubular.com/r/KLPR1zq3Hj

Online Conversion: http://www.onlineconversion.com/roman_numerals_advanced.htm

To avoid matching the empty string you'll need to repeat the pattern four times and replace each 0 with a 1 in turn, and account for V, L and D:

(M{1,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})|M{0,4}(CM|C?D|D?C{1,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})|M{0,4}(CM|CD|D?C{0,3})(XC|X?L|L?X{1,3})(IX|IV|V?I{0,3})|M{0,4}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|I?V|V?I{1,3}))

In this case (because this pattern uses ^ and $) you would be better off checking for empty lines first and don't bother matching them. If you are using word boundaries then you don't have a problem because there's no such thing as an empty word. (At least regex doesn't define one; don't start philosophising, I'm being pragmatic here!)

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In my own particular (real world) case I needed match numerals at word endings and I found no other way around it. I needed to scrub off the footnote numbers from my plain text document, where text such as "the Red Sea^cl and the Great Barrier Reef^cli" had been converted to the Red Seacl and the Great Barrier Reefcli. But I still had problems with valid words like Tahiti and fantastic are scrubbed into Tahit and fantasti.

Fortunately, the range of numbers is limited to 1..3999 or thereabouts. Therefore, you can build up the regex piece-meal.

<opt-thousands-part><opt-hundreds-part><opt-tens-part><opt-units-part>

Each of those parts will deal with the vagaries of Roman notation. For example, using Perl notation:

<opt-hundreds-part> = m/(CM|DC{0,3}|CD|C{1,3})?/;

Repeat and assemble.

Added: The <opt-hundreds-part> can be compressed further:

<opt-hundreds-part> = m/(C[MD]|D?C{0,3})/;

Since the 'D?C{0,3}' clause can match nothing, there's no need for the question mark. And, most likely, the parentheses should be the non-capturing type - in Perl:

<opt-hundreds-part> = m/(?:C[MD]|D?C{0,3})/;

Of course, it should all be case-insensitive, too.

You can also extend this to deal with the options mentioned by James Curran (to allow XM or IM for 990 or 999, and CCCC for 400, etc).

<opt-hundreds-part> = m/(?:[IXC][MD]|D?C{0,4})/;

import re
pattern = '^M{0,3}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$'
if re.search(pattern, 'XCCMCI'):
    print 'Valid Roman'
else:
    print 'Not valid Roman'

For people who really want to understand the logic, please take a look at a step by step explanation on 3 pages on diveintopython.

The only difference from original solution (which had M{0,4}) is because I found that 'MMMM' is not a valid Roman numeral (also old Romans most probably have not thought about that huge number and will disagree with me). If you are one of disagreing old Romans, please forgive me and use {0,4} version.

In my case, I was trying to find and replace all occurences of roman numbers by one word inside the text, so I couldn't use the start and end of lines. So the @paxdiablo solution found many zero-length matches. I ended up with the following expression:

(?=\b[MCDXLVI]{1,6}\b)M{0,4}(?:CM|CD|D?C{0,3})(?:XC|XL|L?X{0,3})(?:IX|IV|V?I{0,3})

My final Python code was like this:

import re
text = "RULES OF LIFE: I. STAY CURIOUS; II. NEVER STOP LEARNING"
text = re.sub(r'(?=\b[MCDXLVI]{1,6}\b)M{0,4}(?:CM|CD|D?C{0,3})(?:XC|XL|L?X{0,3})(?:IX|IV|V?I{0,3})', 'ROMAN', text)
print(text)

Output:

RULES OF LIFE: ROMAN. STAY CURIOUS; ROMAN. NEVER STOP LEARNING