Remove everything except what is in parentheses?

[pacbloyd]

How would I convert something like 'US (United States)' to just 'United States' ? Have to do this for multiple countries

[Thableaus]

Hi @pacbloyd could the row have more than one "string between parenthesis" occurence?

[pacbloyd]

Hello @Thableaus - I don't think so. I just need to do this for many countries where there's an abbreviation before the full country name in parentheses.

What about a solution which simply trims off the first 4 characters (i.e. the 2 letter abbreviation, space, and opening parentheses, and then another formula to trim off the last character (i.e. the closing parentheses)?

[pacbloyd]

I think I figured it out using a right and left function:

right([Country],Length([Country])-4)

left([Country],Length([Country])-1)

However, if there is a better formula, or a formula that extracts from parentheses, let me know

[Thableaus]

@pacbloyd 

REGEX in tokenize mode to split only what you need:

Expression: \([^()]+\)

\( - open parenthesis

[^()]+ capture everything that is not an open/closing parenthesis

\) - closing parenthesis

I think that might work.

[Thableaus · Accepted solution]

I'm sorry, if you don't wanna catch the parenthesis as well, you could use "unmarked groups".

(?:\()([^()]+)(?:\))

add ?: before open/closing parenthesis character.

[Thableaus]

The nice thing about using Tokenize that it's dynamic enough to grab multiple occurences.

That's for example if I had 3 instances of the same pattern. And if you want to make even more dynamic you could use Tokenize and Split to Rows, so you dont have to specify a number of columns.

[binuacs · Accepted solution]

@pacbloyd another option using RegExReplace formula

[pacbloyd]

Thank you, @binuacs !

Bonus question, how would I do the same, but for brackets instead of parentheses?

[binuacs]

@pacbloyd the below RegEx formula remove both ( and [ from the country name

REGEX_Replace([Country], '(.+?)[\(\[](.+?)[\)\]]', '$2')