JS/RegEx: Split string into words but using exception regex

Question

I need to split up a string into single words, but there are some cases which should not be splitted.

An example for type I string
An example for degree II string

So every type | degree + I | II | III | IV | V should be kept as a string

The result of the example strings should be

['An', 'example', 'for', 'type I', 'string']
['An', 'example', 'for', 'degree II', 'string']

In my regex I have to search for type or degree, followed by space, followed by a string with characters I or V with maximum length of 3. Those matches should not be splited.

consr regex = '/(type|degree)\s(I{1,3}|V{1})/' // <-- regEx is wrong as it is not working
const result = string.split(' ')

I'm not quite sure how to use the regex in combination with splitting in a way, that all matches are exceptions for splitting by space character.

(I{1,3}|V{1}) means “either (from one to three I) or (exactly one V)” — spectras
– spectras, Commented Sep 20, 2017 at 8:19
You might want to support all Roman numbers - regex demo (I shortened this Roman number regex a bit). — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Sep 20, 2017 at 8:21
@WiktorStribiżew That regex is great. Could you provide an answer how to use this with split()? — user3142695
– user3142695, Commented Sep 20, 2017 at 8:24
Wrap with (...). See jsfiddle.net/fu63Lyz5. I think you will also need to match that as whole words, hence, I added \b in the demo. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Sep 20, 2017 at 8:24
Thanks. But please have a look at the example shown in the post. I need all words splitted - the regex should match to the exceptions, which should not be splitted by itself. Your JsFiddle gives an array with three items... — user3142695
– user3142695, Commented Sep 20, 2017 at 8:28

Wiktor Stribiżew · Accepted Answer · 2017-09-20 08:34:37Z

2

You may match the words type and degree followed with any Roman number or any 1+ non-whitespace chars with

var s = "An example for degree II string";
var rx = /\b(?:type|degree)\s+M{0,4}(?:C[MD]|D?C{0,3})(?:X[CL]|L?X{0,3})(?:I[XV]|V?I{0,3})\b|\S+/g;
console.log(s.match(rx));

I borrowed and shortened the Roman number regex from here. The pattern matches

\b - a word boundary
(?:type|degree) - a non-capturing group matching either type or degree substrings
\s+ - 1 or more whitespaces
M{0,4}(?:C[MD]|D?C{0,3})(?:X[CL]|L?X{0,3})(?:I[XV]|V?I{0,3}) - the Roman number regex
\b - a trailing word boundary (this will make sure at least 1 Roman number is present)
| - or
\S+ - 1 or more non-whitespace chars.

Note that in case any symbol or punctuation char is present in front of the degree or type words, it will be matched with \S+ branch, so you need to handle those cases before applying this regex.

answered Sep 20, 2017 at 8:34

Wiktor Stribiżew

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2 Comments

user3142695 Over a year ago

Only for better understanding, as regexes are still hard for me: If I only want to keep matched strings with minimum length of 3 characters, where do I have to add this?

Wiktor Stribiżew Over a year ago

@user3142695: Perhaps, \S{3,} instead of \S+.

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