Python regular expression replace?

Use a lookahead assertion p = re. Compile('\r\n(?=a-z)') res = p. Sub(' ', data).

Use a lookahead assertion. P = re. Compile('\r\n(?=a-z)') res = p.

Sub(' ', data).

That does not seem to find the substring. P. Match(data) returns nothing – foosion Oct 16 at 0:17 @foosion: match searches from the start of the string.

Try p. Search(data). – Peter Graham Oct 16 at 0:27.

I think you should use the flag re. DOTALL, which means it will "see" nonprintable characters, such as linebreaks, as if they were regular characters. So, first line of your code would become : p = re.

Compile('..(a-z)', re. DOTALL) (not the two unescaped dots instead of the linebreak). Actually, there is also re.

MULTILINE, everytime I have a problem like this one of those end up solving the problem. Hope it helps.

That does not seem to find the substring. P. Match(data) returns nothing.

– foosion Oct 16 at 0:16.

This : result = re. Sub("(?si)(.*? )*>.

**>(.*)", r"\1 \2", subject) Applied to : the currency Produces : the currency Although I would strongly suggest against using regex with xml/html/xhtml. Ts generic regex will remove all elements and capture any text before / after to groups 1,2.

The reason the html seems reversed is that you're seeing the end of one paragraph and the beginning of another. That regex seems very complex – foosion Oct 16 at 0:20 @foosion Well this regex will work regardless of any elements you throw in it. – FailedDev Oct 16 at 0:26.

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