正则表达式删除少于3个字符的“字符组”

Question

I am trying to remove any 'groups of characters' with less than 3 characters.

This is the source:

1.29 Cancels part plan C/5879 2030. in i i.r e9g6Pop Iatian Area ProcH 22.4.93 Suburban Lands n f 53dv 3 N014 3.5.98. PLAN or any from 01 53 under M R.5I B.L.1laY98 E35. P0 RT I 0 N S At Maroubrajuncti p /I .z. .0 / .L .I. .I

Settings bounds for word characters with repetition between 1 and 3 e.g. /b\w{1,3}\b/ does not work as "C/5879" would become "5879".

The desired output would be as follows:

1.29 Cancels part plan C/5879 2030. e9g6Pop Iatian Area ProcH 22.4.93 Suburban Lands 53dv N014 3.5.98. PLAN from under R.5I B.L.1laY98 E35. Maroubrajuncti

An alternative which could also work would be to create larger 'groups of characters' by joining 'groups of characters' with 2 or less characters delimited by a whitespace.

For example:

1.29 Cancels part plan C/5879 2030. inii.r e9g6Pop Iatian Area ProcH 22.4.93 Suburban Lands nf 53dv 3N014 3.5.98. PLAN orany from 0153 under MR.5I B.L.1laY98 E35. P0RTI0NS AtMaroubrajuncti p/I.z. .0/.L.I..I

I would be open to either solution to rescue me from Regex Hell.

Answer 1

Your definition of "words" is "whitespace delimited", which differ from regex's defitionition of "word to non-word", so use look arounds:

\s+\S{1,3}(?=\s)

Note that the expression includes (captures) leading spaces, so removing matches will not leave double spaces in the result.

When tested on regextester result is:

1.29 Cancels part plan C/5879 2030. e9g6Pop Iatian Area ProcH 22.4.93 Suburban Lands 53dv N014 3.5.98. PLAN from under R.5I B.L.1laY98 E35. Maroubrajuncti .I