Replace repeated item with the number of times it is repeated

[1v4n0]

Hello, is there a way to replace a character or a string, repeated continuously N times, with the number N?
Say, replace "AA" with "2", "AAA" with "3" and so on, in just one regular expression?


Thank you.

[The_book]

Originally Posted by 1v4n0

Hello, is there a way to replace a character or a string, repeated continuously N times, with the number N?
Say, replace "AA" with "2", "AAA" with "3" and so on, in just one regular expression?


Thank you.

calibre book-edit regex function replace. using python to do so.

Hope someone develops a similar plug-in, or this feature will be also added to Sigil

[1v4n0]

Thank you. And what would the regex be? (just so I don't have to wade through the whole user manual )

[Doitsu]

Originally Posted by 1v4n0

Thank you. And what would the regex be? (just so I don't have to wade through the whole user manual )

In the Calibre Editor, press CTRL+F, select the regex function mode, click Create/edit, paste the following code into the editor,

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    return str(len(match.group()))

give the function a name, e.g. CharLen and enter (A+) as the search expression.

If your text contains AAA and AAAA, they should be replaced with 3 and 4.