Delete the entire content of all files with less than 100 words

Paul Wormer

rodica F

@guy038 said in Delete the entire content of all files with less than 100 words:
(?s)\A[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,98}[^[:space:]]+[[:space:]]*\z|\A[[:space:]]+\z

One more question I have for @guy038 I want to use one of your GENERIC S/R for this case. SO I need to delete the content of a file that have less then 10 words between section <START> and <FINAL>

<START>

The first, thing to note when

<FINAL>

So, I test with all your GENERIC regex formulas you done a long time ago.

BSR = <START>
ESR = <FINAL>
FR = (?s)\A[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,10}[^[:space:]]+[[:space:]]*\z|\A[[:space:]]+\z

REGEX:

(?-si:BSR|(?!\A)\G)(?s-i:(?!ESR).)*?\x20\K(FR)

(?-si:BSR|(?!\A)\G)(?s-i:(?!ESR).)*?\x20\KFR(?=\x20)

(?-si:BSR|(?!\A)\G)(?s-i:(?!ESR).)*?\x20\KFR

(?-si:BSR|(?!\A)\G)(?s-i:(?!ESR).)*?\x20\KFR(?=\x20)

(?-i:BSR|\G(?!^))(?s:(?!ESR).)*?\K(?-i:FR)

(?-i:BSR|(?!\A)\G)(?s:(?!ESR).)*?\K(?-i:FR)

(?-i:BSR|(?!^)\G)(?s:(?!ESR).)*?\K(?-i:FR)

(?-i:BSR|(?!\A)\G)(?s:(?!ESR).)*?\K(?-i:FR)

It is not working, in any of the cases. I get the same message on F/R: “Cannot find the text…”

guy038

Hi, @rodica-f and All,

EDIT : The regexes, below, are incomplete. See the correct solution in my next post

You do not need to use these generic regexes at all !

Simply, replace \A by <START> and \z by <FINAL> and, of course, change the value of the quantifier of the non-capturing group from 98 to 8, giving the functional regex S/R below :

SEARCH (?s)<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,8}[^[:space:]]+[[:space:]]*<FINAL>|<START>[[:space:]]+<FINAL>

REPLACE Leave EMPTY

So, the general formula for deleting all file contents, if there are less than N words between the two boundaries <START> and <FINAL>, is :

SEARCH (?s)<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,N-2}[^[:space:]]+[[:space:]]*<FINAL>|<START>[[:space:]]+<FINAL>

REPLACE Leave EMPTY

BR

guy038

rodica F

@guy038 correct me if I’m wrong. The GENERIC formula in this case will be:

(?s)BSR(FR)*ESR|BSR+ESR

I think I’m wrong somewhere.

rodica F

@guy038 by the way I test your generic formula you done for me.

(?s)<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,8}[^[:space:]]+[[:space:]]*<FINAL>|<START>[[:space:]]+<FINAL>

In the context below, delete only everything that is framed in <START> and <FINAL>

But does not delete the entire file, I mean the other words around it.

blah blah     blah


<START>

The first, thing to note when

<FINAL>

   blah blah

guy038

Hello, @rodica-f and All,

Oh… Yes ! I was wrong about it ! The correct regex S/R is, of course :

SEARCH (?s)\A.*<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,8}[^[:space:]]+[[:space:]]*<FINAL>.*\z|\A.*<START>[[:space:]]+<FINAL>.*\z

REPLACE Leave EMPTY

And the general formula for deleting all file contents, if there are less than N words between the two boundaries <START> and <FINAL>, becomes :

SEARCH (?s)\A.*<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,N-2}[^[:space:]]+[[:space:]]*<FINAL>.*\z|\A.*<START>[[:space:]]+<FINAL>.*\z

REPLACE Leave EMPTY

This regex will delete all file contents in all these cases :

If there no non-space char ( 0 word ), and only some space chars => the regex is \A.*<START>[[:space:]]+<FINAL>.*\z ( the part after the | symbol )
If there are several non-space chars ( one word ), possibly surrounded with space chars => quantifier = 0 and the regex becomes (?s)\A.*<START>[[:space:]]*[^[:space:]]+[[:space:]]*<FINAL>.*\z
If there are several non-space chars followed with space chars, twice ( so two words) => quantifier = 1 and the regex becomes (?s)\A.*<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+)[^[:space:]]+[[:space:]]*<FINAL>.*\z
If there are several non-space chars followed with space chars, third times ( so three words) => quantifier = 2 and the regex becomes (?s)\A.*<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){2}[^[:space:]]+[[:space:]]*<FINAL>.*\z

and so on… till :

If there are several non-space chars followed with space chars, ninth times ( so nine words) => quantifier = 8 and the regex becomes (?s)\A.*<START>[[:space:]]*(?:[^[:space:]]+[[:space:]]+){8}[^[:space:]]+[[:space:]]*<FINAL>.*\z

Now, to answer your question, I would say :

SEARCH (?s)\A.*BSR(FR)ESR.*\z

where FR = [[:space:]]*(?:[^[:space:]]+[[:space:]]+){0,N-2}[^[:space:]]+[[:space:]]* OR FR = [[:space:]]+ ( case no word )

Best Regards,

guy038

rodica F

@guy038 thank you very much !

rodica F

@rodica-f

Delete the entire content of all files with less than 6 words

FIND:
\A(?i)[^\w+]*(?:[\w*]+[^\w*]+){0,5}(?:[\w*]+[^\w+]*)?\z

REPLACE: (LEAVE EMPTY)

guy038

Hi, @rodica-f and All,

I sorry to tell you that your last regex does not meet exactly the previous rules and is rather erroneous !

First, and just anecdotal, the (?i) modifier is useless as no range of letters occurs in your regex

Secondly, this regex will delete all file contents if more than 0 word char and less than 7 word chars

Thirdly, let’s consider this somple phrase :

let abc - xyz

It contains 4 non-space expressions ( let, abc, - and xyz )

Your regex seems OK as it correctly select all text which contains less than 7 words

Now, change the - sign by a + sign :

let abc + xyz

This time, your regex does not match anything although there are, still, 4 non-space expressions :((

Why this behaviour occurs ? Well, the different sub-expressions, that you used in your regex, are erroneous !

[^\w+]* means “find a a char different from a word char and different from the + sign”, repeated from 0 to any

[\w*]+ means “find a word char or a * symbol”, repeated from 1 to any

[^\w*]+ means “find a char different from a word char and different from the * symbol”, repeated from 1 to any

So, an almost-correct solution would be \A[^\w]*(?:\w+[^\w]+){0,4}(?:\w+[^\w]*)?\z. However, note that it also matches a true empty file which does not need any replacement as already empty !!

Now, the important drawback of using word chars \w and non-word chars [^\w], is that any symbol, met in text, will increase the number of words !. For instance, see the difference betwen :

This is a simple example

and :

This is a sim-ple example

If I use my last “word” version \A[^\w]*(?:\w+[^\w]+){0,4}(?:\w+[^\w]*)?\z, it matches the text This is a simple example and not the text This is a sim-ple example ! Because, in the former case, it counts 5 words and, in the later case, it counts 6 words

That’s why my previous and @terry-r’s version, using non-space characters [[:^space:]] and space chars [[:space:]], seems more rigorous and practical ;-))

Best Regards

guy038

rodica F

@guy038 said in Delete the entire content of all files with less than 100 words:

\A[^\w]*(?:\w+[^\w]+){0,4}(?:\w+[^\w]*)?\z

My joy is that, thanks to my regex, an alternative method has been discovered, quite good.

thank you @guy038