Following the above, you could use year(), month(), day() to extract
the date components if that is your aim.
On Sun, Aug 24, 2008 at 10:43 AM, Salah Mahmud <[email protected]> wrote:
> Hi Michael,
>
> Have you tried the date function?
>
> input str12 date
> "12/12/08"
> "12/02/08"
> "4/14/08"
> "4/04/08"
> end
> gen d=date(date, "MDY", 2010)
> format d %d
> list
>
> Tested using Stata 10.1
>
> /salah
>
>
>
> On Sat, Aug 23, 2008 at 9:14 PM, Michael McCulloch <[email protected]> wrote:
>> Hello,
>> I have dates in an Excel *.csv file as:
>>
>> date
>> 12/12/08
>> 12/02/08
>> 04/14/08
>> 04/04/08
>>
>> I wrote the following code to convert to Stata date:
>> gen str date_year = substr(date, -2,.)
>> gen str date_month = substr(date, 1, 2)
>> gen str date_day = substr(date, 4, 2)
>> destring date_year date_month date_day, replace
>> list date date_month date_day date_year
>>
>> However, when I import the *.csv file, the leading zeroes are dropped
>> from months 1-9.
>> Is there a way to solve this in Stata, without having to create
>> separate columns for m, d, and y in Excel?
>>
>>
>> --
>>
>> Best wishes,
>> Michael McCulloch
>>
>>
>>
>> Pine Street Foundation
>> 124 Pine St., San Anselmo, CA 94960-2674
>> Tel: (415) 407-1357
>> Fax: (415) 485-1065
>> [email protected]
>> www.pinestreetfoundation.org
>>
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>>
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
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