How to find, and standardize the spelling of chapter and page references (Adobe InDesign)?

Question

I want to find digits followed by "f", "ff", "f." or "ff." to standardize the spelling following given conventions/rules.

I already tried some regular expressions, but unfortunately I did not find an universal expression grabbing all of the cases above (f, ff, f., ff.).

In spoken words it seems easy:

• find digits
• followed by an optional whitespace
• then followed by f, ff, f. or ff.
• only whitespaces or NOT word boundaries are allowed before and after the expression

The beginning of the regex is quite easy, but I can’t figure out how to handle the different "f"-cases and the NOT boundaries following.

---

My best guess yet is:

(?<=\b)(\d+(\h|\b)?f{1,2})\.?

but then still the stings followed by a word character are found.

---

When I extend the regex to:

(?<=\b)(\d+(\h|\b)?f{1,2})\.?(\W)

the numbered of "false funds" are decreasing, but still it is not the solution

---

I prepared lines for testing. The lines containing a plus "+" should be found, at the same time the ones with a minus "-" should not be found.

00f aaa +
00f. aaa +
00ff aaa +
00ff. aaa +
00 f aaa + 
00 f. aaa +
00 ff aaa +
00 ff. aaa +
+ aaa 00f aaa +
+ aaa 00f. aaa +
+ aaa 00ff aaa +
+ aaa 00ff. aaa +
+ aaa 00 f aaa + 
+ aaa 00 f. aaa +
+ aaa 00 ff aaa +
+ aaa 00 ff. aaa +
+ aaa 00f
+ aaa 00f.
+ aaa 00ff
+ aaa 00ff.
+ aaa 00 f 
+ aaa 00 f.
+ aaa 00 ff
+ aaa 00 ff.

00 faaa -
00 f.aaa -
00 ffaaa -
00 ff.aaa -
00af aaa - 
00af. aaa -
00aff aaa -
00aff. aaa -
- aaa 00 faaa -
- aaa 00 f.aaa -
- aaa 00 ffaaa -
- aaa 00 ff.aaa -
- aaa 00af aaa - 
- aaa 00af. aaa -
- aaa 00aff aaa -
- aaa 00aff. aaa -
- aaa00f
- aaa00f.
- aaa00ff
- aaa00ff.
- aaa 00af 
- aaa 00af.
- aaa 00aff
- aaa 00aff.

00faaa -
00f.aaa -
00ffaaa -
00ff.aaa -
00af aaa - 
00af. aaa -
00aff aaa -
00aff. aaa -
- aaa00 faaa -
- aaa00 f.aaa -
- aaa00 ffaaa -
- aaa00 ff.aaa -
- aaa00af aaa - 
- aaa00af. aaa -
- aaa00aff aaa -
- aaa00aff. aaa -
- aaa00af 
- aaa00af.
- aaa00aff
- aaa00aff.

Further, the aim is to group the digits anf "f"-cases in a manner, so that they can be uses in a replacement-expression to standardize the spelling to one of those cases:

• 123 ff. (with whitespace, with dot)
• 123 ff (with whitespace, without dot)
• 123ff. (without whitespace, with dot)
• 123ff (without whitespace, without dot)

Crazy

would be helpful if you add language you are working with to your tags – Knavish

Answer 1

I suggest

\b(\d+)(\s?)(f{1,2})(?:(\.)\B|\b(?!\.))

See the regex demo

Details

• \b - word boundary
• (\d+) - Group 1: 1+ digits
• (\s?) - Group 2: an optional whitespace
• (f{1,2}) - Group 3: 1 or 2 fs
• (?:(\.)\B|\b(?!\.)) - either of the two:
  • (\.)\B - a . captured in Group 4 if not followed with a word char
  • | - or
  • \b(?!\.) - a word boundary not followed with a dot.

Then, replacing is easy with:

• 123 ff.: $1 $3.
• 123 ff : $1 $3
• 123ff. : $1$3.
• 123ff : $1$3

If the whitespace and dot are not necessary in replacement patterns, remove the groupings and adjust the IDs in the replacement backreferences.

Her

Hi, whilst looking at my own solution and checking yours, e.g. 00 f.!aaa aaa + would still match, but this doesn't adhere to requirements, since only whitespace/boundry at the end is allowed...? – Knavish

@Knavish No idea why you think so. ff. is followed with a non-word boundary and it does meet the requirements. – Sybille

@WiktorStribiżew - Interestingly, running your regexp via Adobe InDesign's Find/Change feature yields the exact same matches as your regexp with the flavor set to ECMAScript, i.e. several wanted matches are not found. – Alejandroalejo

@Alejandroalejo I wrote: *(\h?) - Group 2: an optional horizontal whitespace (use \s if any whitespace is allowed here). My \b(\d+)(\s?)(f{1,2})(?:(\.)\B|\b(?!\.)) regex works. Edited to get rid of \h. – Sybille

@WiktorStribiżew - Yes using \s instead does rectify the issue. – Alejandroalejo

Answer 2

What about something like this?

\b\d+\s?(?:ff|f)+\.?(?=\s)

Regex Demo

\b          start with word boundary
\d+         match all digits
\s?         match optional whitespace
(?:ff|f)+   non-capturing group, match either ff or f
\.?         match optional dot (basically checking for ff. or ff or f. or f)
(?=\s)      match if followed by whitespace, without making the whitespace part of the match

With groups, same expression looks like:

\b(\d+)\s?((?:ff|f)+\.?)(?=\s)

Regex Demo

Replacement can be achieved via different combinations of the $1 and $2 groups.

Knavish