Code:
* Example generated by -dataex-. For more info, type help dataex clear input str12 stkcd int year byte multi "000002" 2009 0 "000002" 2010 0 "000002" 2011 0 "000002" 2012 0 "000002" 2013 0 "000002" 2014 1 "000002" 2015 1 "000002" 2016 1 "000002" 2017 1 "000002" 2018 1 "000002" 2019 1 "000004" 2011 0 "000004" 2012 0 "000004" 2013 0 "000004" 2014 0 "000004" 2015 1 "000004" 2016 1 "000004" 2017 0 "000004" 2018 0 "000004" 2019 0 "000005" 2009 0 "000005" 2010 0 "000005" 2011 0 "000005" 2012 0 "000005" 2013 0 "000005" 2014 0 "000005" 2015 0 "000005" 2016 0 "000005" 2017 0 "000005" 2018 0 "000005" 2019 0 "000006" 2009 0 "000006" 2010 1 "000006" 2011 1 "000006" 2012 1 "000006" 2013 1 "000006" 2014 0 "000006" 2015 0 "000006" 2016 0 "000006" 2017 0 "000006" 2018 0 "000006" 2019 0 "000007" 2012 0 "000007" 2013 0 "000007" 2014 0 "000007" 2015 0 "000007" 2016 1 "000008" 2013 0 "000008" 2014 0 "000008" 2015 0 "000008" 2016 0 "000008" 2017 0 "000008" 2018 1 "000008" 2019 1 "000009" 2011 0 "000009" 2012 0 "000009" 2013 0 "000009" 2014 0 "000009" 2015 0 "000009" 2016 0 "000009" 2017 0 "000009" 2018 0 "000009" 2019 0 "000010" 2013 1 "000010" 2014 1 "000010" 2015 0 "000010" 2016 0 "000010" 2017 0 "000010" 2018 0 "000011" 2009 0 "000011" 2010 0 "000011" 2011 0 "000011" 2012 0 "000011" 2013 0 "000011" 2014 0 "000011" 2015 0 "000011" 2016 0 "000011" 2017 0 "000011" 2018 0 "000011" 2019 0 "000012" 2015 0 "000012" 2016 0 "000012" 2017 0 "000012" 2018 0 "000012" 2019 0 "000014" 2009 0 "000014" 2010 0 "000014" 2011 0 "000014" 2012 0 "000014" 2013 0 "000014" 2014 0 "000014" 2015 1 "000014" 2016 1 "000014" 2017 0 "000014" 2018 0 "000014" 2019 0 "000016" 2009 0 "000016" 2010 0 "000016" 2011 0 "000016" 2012 0 end
- For each `stkcd', if a value 0 of `multi' is (imediately) followed by 1, then v1=1 for this `stkcd', 0 otherwise.
- In contrast, for each `stkcd', if a value 1 of `multi' is (imediately) followed by 0, then v2=1 for this `stkcd', 0 otherwise.
0 Response to identify data pattern.
Post a Comment