RegEx questions for Acronyms

[phossler]

Current book I'm cleaning up has 100s of instances of acronyms, 2 - 6 letters, with periods between them and I'd like to just have the letters without the periods.

I made 5 pairs of saved searches, one version with a trailing space and one without since that seems to be a common construction for 2, 3, 4, 5, and 6 char acronyms

e.g. A.B.C.D.E.F.space and just A.B.C.D.E.F.

Case 1 - If there's a space, then I figure it's inside a paragraph and I just want ABCDEF blah blah

Case 2 - If there's NOT a space, then I figure it's at the end of a paragraph and I want ABCDEF. Blah blah

So for Case 1

Find: (has trailing space)

([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.

Replace: (also with trailing space

\1\2\3\4\5\6

and for Case 2

Find: (NO trailing space)

([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.([A-Z])\.

Replace: (NO trailing space, but period instead

\1\2\3\4\5\6.

I have to be careful to run them in order:6+space, 6-no space, 5+space, 5-no Space, .... otherwise substrings would get replaced

I figure that there HAS to be a more intelligent way to set up these searches, and hopefully someone can suggest an idea

[hobnail]

Regex has a count/number thing where you can specify how many times it matches. I've never used it and can't remember how it works off the top of my head but I think it might be something like the following for the letters and periods for 1 to 6 repititions:

([A-Z]\.){1,6}

But check the usual helpful regex web sites for how to do it.

[theducks]

Originally Posted by hobnail

([A-Z]\.)\s*{1,6}

0 or 1 space but NOT in the capture

[lomkiri]

You want to see if your acro is at the end of a paragraph (A.B.</p>) or of a sentence (A.B. Then something) or if the is a word in lowercase

I propose you a regex function, working with this regex :

((?:\p{Lu}\.){2,})(?:(<(?:/p|/div|br/|/blockquote)>)|( \p{Lu})|( \p{Ll}))

\p{Lu} is an uppercase letter, \p{Ll} lowercase, (?: is a non-capturing group.
(?:\p{Lu}\.){2,})) will capture in match.group(1) all acro. with at least 2 letters (put 3 if you want to start with acro of 3 letters).

The function will put a period or not,, depending of what is after your acronym. It is possible that it doesn't cover all cases, it's to you to check. It would be wise to polish the book first, to avoid unexpected end of line, or space before </p>, etc.

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    acro = match.group(1).replace('.', '')
    if  end := match.group(2):	# </p> or <br/> etc.
        period = '.'
    elif  end := match.group(3):	# <space>[A-Z]
        period = '.'
    elif  end := match.group(4):	# <space>[a-z]
        period = ''
    else:
        end = ''
        period = ''
  
    return acro + period + end

See if you want to consider other cases. Notice that, e.g., A.B. Lda will give AB. Lda, because of the capital letter of the next word

[hobnail]

Originally Posted by theducks

0 or 1 space but NOT in the capture

Doesn't the asterisk mean 0 or more? I was thinking about the space after I replied and was thinking he could use something like \s{0,1}. But I guess it doesn't matter if there is more than 1 space.

[theducks]

Originally Posted by hobnail

Doesn't the asterisk mean 0 or more? I was thinking about the space after I replied and was thinking he could use something like \s{0,1}. But I guess it doesn't matter if there is more than 1 space.

You are correct (both )

[phossler]

Originally Posted by hobnail

Regex has a count/number thing where you can specify how many times it matches. I've never used it and can't remember how it works off the top of my head but I think it might be something like the following for the letters and periods for 1 to 6 repititions:

([A-Z]\.){1,6}

But check the usual helpful regex web sites for how to do it.

Actually I had thought for the Find

([A-Z]\.){2,6} 

and 

([A-Z]\.){2,6}

where there is a trailing space after the first case

That would Find A.B.C.<space> but I couldn't figure out how to do the Replace (trailing space) since it depends on the number of letters in the acronym, 3 in this case

\1\2\3

[phossler]

Originally Posted by lomkiri

You want to see if your acro is at the end of a paragraph (A.B.</p>) or of a sentence (A.B. Then something) or if the is a word in lowercase

I propose you a regex function, working with this regex :

See if you want to consider other cases. Notice that, e.g., A.B. Lda will give AB. Lda, because of the capital letter of the next word

Thanks

This looks a LOT easier than the brute force way I tried

I can single step through and address the A.B. Lda case

 <!-- CASE 1
      TRAILING SPACE FOLLOWED BY CAPITAL LETTER SO ASSUME END OF SENTENCE
      AND LEAVE LAST PERIOD -->
  <!-- CASE 2
        NO TRAILING SPACE SO ASSUME END OF PARAGRAPH 
        AND LEAVE LAST PERIOD -->
  <!-- CASE 3
        HAS TRAILING SPACE SO ASSUME IN MIDDLE OF SENTENCE WITH NO PERIODS -->

I have 3 cases and your function works great on case 1 and case 2. Doesn't seem to do anything for case 3. Can you tweak it a little for me please?

[lomkiri]

Originally Posted by phossler

I have 3 cases and your function works great on case 1 and case 2. Doesn't seem to do anything for case 3. Can you tweak it a little for me please?

OK, let's modfify the 3rd case from <space>[a-z] to <space>, and add a collector for all anything else (comma, for example, or semicolon). In both case, it removes the period

((?:\p{Lu}\.)+)(?:(</(?:p|div|b/|blockquote)>)|( \p{Lu})|(' ')|(.))
(note : the regex above is wrong and was corrected in msg #13 toward the one below:)
((?:\p{Lu}\.){2,})(?:\s*(<(?:/p|/div|br/|/blockquote)>)|( \p{Lu})|(' ')|(.))

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    acro = match.group(1).replace('.', '')
    if  end := match.group(2):	# </p> or <br/> etc.
        period = '.'
    elif end := match.group(3):	# <space>[A-Z]
        period = '.'
    elif end := match.group(4):	# <space>
        period = ''
    elif end := match.group(5):	# anything else
        period = ''
  
    return acro + period + end

I've made it to be easily red and modified, but it may be shortened this way, it does exactly the same thing :

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    acro = match.group(1).replace('.', '')
    if  end := (match.group(2) or match.group(3)):
        period = '.'
    elif end := (match.group(4) or match.group(5)):
        period = ''
    return acro + period + end

[phossler]

That is really, super-duper, extra special terrific

Works great and SO much cleaner than the way I was trying

Thanks again

PS - I'm going with the wordy version - I like wordy

[hobnail]

Originally Posted by phossler

Actually I had thought for the Find

([A-Z]\.){2,6} 

and 

([A-Z]\.){2,6}

where there is a trailing space after the first case

That would Find A.B.C.<space> but I couldn't figure out how to do the Replace (trailing space) since it depends on the number of letters in the acronym, 3 in this case

\1\2\3

I think it doesn't matter if some of them aren't "filled"? E.g., for a simple example if you were using ([a-z]){1,3} for the search and x\3x for the replacement I'm guessing that you'd get xx when it matched aa. The \3 would be empty.

[phossler]

@lomkiri --

Me again - small problem came up on my first real run

All of a sudden I got lots of errors and tracked it down to the before / after below


"www.w3" and "www.idpf.org" lost their periods

b: <html xmlns="http://www.w3.org/1999/xhtml" xmlns:epub="http://www.idpf.org/2007/ops" xml:lang="en">

a: <html xmlns="http://wwww3.org/1999/xhtml" xmlns:epub="http://wwwidpforg/2007/ops" xml:lang="en">

Not sure why since "www" is lower case and should not be considered. This is the case that threw the errors, but I'm assuming that the situation could be anywhere.

[lomkiri]

You didn't check "Case sensitive". It is mandatory, since you're looking for uppercase letters. Without checking this, then \p{Lu} (or [A-Z]) select any letter, not only uppercase.

Another mistake (mine, this time) : there was a mismatch in the history-box of the searches, and I put an old version of the string, it is not

((?:\p{Lu}\.)+)(?:(</(?:p|div|b/|blockquote)>)|( \p{Lu})|(' ')|(.))

but (as I put in my first msg):

((?:\p{Lu}\.){2,})(?:\s*(<(?:/p|/div|br/|/blockquote)>)|( \p{Lu})|(' ')|(.))

Both conditions gave this result. Terribly sorry for this !

[phossler]

Never mind -- I had the wrong Case Sensitive box checked

Everything works very well

Thanks again