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Re: st: Re: string functions quotation marks


From   Nick Cox <njcoxstata@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Re: string functions quotation marks
Date   Wed, 28 Sep 2011 11:02:25 +0100

This possibility was addressed in
http://www.stata.com/statalist/archive/2011-09/msg01101.html

from which a solution for removing first and last characters that are
quotation marks is (with your variable name)

 replace country = substr(country, 2, length(country) - 2)

Eric or Ric didn't reply to the list, so we presume that he is content somehow.

Nick

On Wed, Sep 28, 2011 at 10:25 AM, Oliver Jones
<ojones@wiwi.uni-bielefeld.de> wrote:
> Hi, can you get rid of the first quotation mark by
>
> replace country = substr(country, 2, .)
>
> where I assume that the variable holding your country names is called
> country?
>
> If yes, than you can figure out how to get rid of the last one by first
> finding out the length of the string...
>
> If this just cuts of the first letter of the country name then the string
> doesn't actually contain any quotation marks. They are probably just
> displayed because of thats the way Stata displays string variables.
>
> Best
> Oliver
>
>
> Am 24.09.2011 17:06, schrieb Eric Uslaner:
>>
>> Greetings,
>>
>> I have a dataset (obviously not my own) in which the names of countries
>> are all in quotation marks.  I need to get rid of the quotation marks to
>> merge this dataset with another.  However, the situation is somewhat
>> complicated.  Country names may have one word (e.g., "Canada" or "Yemen"),
>> two words ("North Korea" is one example), three words ("United States of
>> America" or "Yemen Arab Republic"), or more ("Democratic Republic of the
>> Congo").  Is there a way to get rid of these quotation marks?  I have read
>> the manual and looked at help from within Stata and can't find a solution.
>>
>> Thanks for any help,

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