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From |
"Martin Weiss" <martin.weiss1@gmx.de> |

To |
<statalist@hsphsun2.harvard.edu> |

Subject |
st: Re: comparing across observations and variables |

Date |
Thu, 9 Apr 2009 22:50:20 +0200 |

<>

**** clear* inp str2 firm str2 parent str2 ultimate_parent A A1 A2 A1 A1 A1 A3 A4 A end levelsof ultimate_parent, local(mylevels) stack firm parent ultimate_parent, into(countvar) clear foreach lev of local mylevels{ qui cou if countv=="`lev'" di in red "`lev' " `r(N)' } **** HTH Martin _______________________

To: <statalist@hsphsun2.harvard.edu> Sent: Thursday, April 09, 2009 10:33 PM Subject: st: comparing across observations and variables

hi,I have two sets of data: data about control relationships between firms(firm, parent firm, ultimate parent firm) and alliances between firms. Ineed a way to figure out who is at the top of the control pyramid usingthe first set of data, and then, in the alliance data, replace all firmslower down on the control pyramid with the firm at the top of the controlpyramid.1) For instance, my data about control relations between firms is shownbelow. I need a way to automatically figure out that A2 is at the top ofthis control pyramid.firm, parent, ultimate parent A, A1, A2 A1, A1, A1 A3, A4, A 2) My data about alliances between firms is shown below: Participant 1, Participant2 A, B A3, C A1, D3) I need to incorporate the data about control relationships (A2 is atthe top of the control pyramid involving A, A1 A3, A4) in the data aboutalliances so that my alliance file looks like this:A2, B A2, C A2, D I will be very grateful for any ideas you can suggest to do the above. thanks so much dalhia * * 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 * http://www.ats.ucla.edu/stat/stata/

**References**:**st: comparing across observations and variables***From:*Dalhia <ggs_da@yahoo.com>

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