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From |
"Nick Cox" <n.j.cox@durham.ac.uk> |

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

Subject |
RE: Re: st: making sure on the graph |

Date |
Fri, 4 Sep 2009 14:29:06 +0100 |

Maarten's solution is good, but there is no need for the sorting within -yearofbirth-. . bys yearofbirth: gen byte first = _n == 1 . egen meanwage = mean(wage), by(yearofbirth) . scatter meanwage yearofbirth if first It doesn't matter which observation you pick up within each group of -yearofbirth-, as -meanwage- will be the same for all within the same group. Note that this is also equivalent to what -egen, tag()- does. You could go egen tag = tag(yearofbirth) egen meanwage = mean(wage), by(yearofbirth) scatter meanwage yearofbirth if tag The -egen- route is strictly inefficient. You make Stata do quite a lot of work, yet in this case there is an equivalent one-liner that gets you there directly. Also, getting familiar with -by:- tricks is a good long-term idea. Nick n.j.cox@durham.ac.uk Maarten buis An intermediate solution that will get the small graph while keeping the data intact is: . bys yearofbirth (wage): gen byte first = _n == 1 . egen meanwage = mean(wage), by(yearofbirth) . scatter meanwage yearofbirth if first --- Ulrich Kohler wrote: > . egen meanwage = mean(wage), by(yearofbirth) > . scatter meanwage yearofbirth > > is one possibility. The advantage of this solution is that it keeps your > data as it is. The disadvantage is that the file size of the graph > becomes arbitrary large if you have many observations. > > A solution that destroys the data in memory but produces smaller graphs > (in terms of bandwidth) is > > . collapse (mean) wage, by(yearofbirth) > . scatter wage yearofbirth * * 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**:**Re: st: making sure on the graph***From:*Maarten buis <maartenbuis@yahoo.co.uk>

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