How to replace the newline/nextline character in a column with space character

[Pratap Pundir]

Hello!

I understand that subinstr can be used to replace a substring in a column
https://www.stata.com/statalist/arch.../msg00386.html

gen newvar = subinstr(oldvar,"dis","reg",.)

I would like to replace all appearances of newline, i.e. cr/lf - char(10) or char(13) - in a column with the space character. I used

gen newvar = subinstr(oldvar,"`=char(10)'"," ",.)
gen newvar2 = subinstr(newvar,"`=char(13)'"," ",.)

Is this the best way of doing this? Is there a more efficient, single line command?

Also when thinking about next line character, are carriage return and line feed the only two I need to worry about, or are there other characters I am missing?

[George Ford]

You can imbed subinstr within subinstr

https://www.statalist.org/forums/forum/general-stata-discussion/general/1455172-substr-within-a-subinstr

]

[George Ford]

try this.

 
 gen newvar2 = subinstr(subinstr(oldvar,"`=char(10)'"," ",.),"`=char(13)'"," ",.)

[Pratap Pundir]

Nice, thanks, and chat(10) and char(13) are the only 2 I should worry about, correct?

[George Ford]

I have no idea what you are up to aside from string replacement.

[Bjarte Aagnes]

gen v2 = ustrregexra(v1, "\p{Space_Separator}", char(32))

[Pratap Pundir]

gen v2 = ustrregexra(v1, "\p{Space_Separator}", char(32))

Nice! I should do that...