No pity from me on that score.
by t abs_charge:egen npos = count(charge>0)
and
by t abs_charge:egen npos = total(charge>0)
are both legal. The second is often useful.
Nick
[email protected]
Sergiy Radyakin
> Sorry, (my mistake) this line should be simply omitted. Both
> npos and nneg
> are defined later on.
> I guess the solution proposed by Mr Blasnik will exclude
> observations if the
> sum
> over a group is zero, but should it be a necessary condition?
>
> 30
> -30
> 30
>
> (with all other variables being same) -- e.g. bought apples
> for 30, than
> returned them to the cashier, than decided to buy them again :)
> This should simplify to 30, but the sum over the group is not 0.
>
> it is a pity that I can't do as I have written, i.e. -count
> if- after -egen-
Nick Cox
> >I stopped at this line. There is a typo there
> > somewhere, as the code is illegal:
> >
> > by t abs_charge:egen npos=count if charge>0
Sergiy Radyakin
> >> you will get a dozen of ways to do this more efficiently in
> >> several minutes,
> >> but here is a straightforward way:
> >>
> >> Assume your data is:
> >> +------------+
> >> | t charge |
> >> |------------|
> >> 1. | 1 10 |
> >> 2. | 2 20 |
> >> 3. | 2 20 |
> >> 4. | 3 30 |
> >> 5. | 3 20 |
> >> |------------|
> >> 6. | 3 -30 |
> >> 7. | 3 -30 |
> >> 8. | 4 10 |
> >> 9. | 4 20 |
> >> 10. | 4 30 |
> >> |------------|
> >> 11. | 4 40 |
> >> 12. | 5 10 |
> >> 13. | 5 20 |
> >> 14. | 5 -10 |
> >> 15. | 5 -10 |
> >> |------------|
> >> 16. | 5 10 |
> >> 17. | 5 10 |
> >> 18. | 5 20 |
> >> 19. | 5 10 |
> >> 20. | 5 10 |
> >> |------------|
> >> 21. | 5 10 |
> >> 22. | 5 -10 |
> >> 23. | 5 20 |
> >> 24. | 5 -20 |
> >> +------------+
> >>
> >>
> >> Then
> >>
> >>
> >>
> >> gen abs_charge=abs(charge)
> >> sort t abs_charge
> >> by t abs_charge:egen npos=count if charge>0
> >>
> >> gen posit=1 if charge>0
> >> gen negat=1 if charge<0
> >>
> >> by t abs_charge:egen npos=count(posit)
> >> by t abs_charge:egen nneg=count(negat)
> >>
> >> by t abs_charge:gen ndel=npos*(npos<nneg)+nneg*(nneg<npos)
> >>
> >> by t abs_charge:drop if _n<ndel | _n>_N-ndel
> >>
> >>
> >> Or something similar? change t as appropriate to define a group.
> >
> > sara borelli
> >
> >> > This is an exctract of my data:
> >> >
> >> > personid charge proc date
> >> > 1000124 +13 80048 6/6/2001
> >> > 1000124 +13 80076 6/6/2001
> >> > ...
> >> > 1000124 +13 80048 6/7/2001
> >> > 1000124 +13 80076 6/7/2001
> >> > ...
> >> > 1000124 -13 80048 6/7/2001
> >> > 1000124 -13 80076 6/7/2001
> >> > ...
> >> > 1000124 +13 80048 6/8/2001
> >> > ...
> >> > 1000124 +13 80048 6/9/2001
> >> > ...
> >> > 1000124 +13 81001 6/11/2001
> >> > ...
> >> > 1000124 +13 80048 6/11/2001
> >> > 1000124 +13 80048 6/12/2001
> >> >
> >> > where the dots indicate that other values for the same
> >> > personid are in between. I need to eliminate the
> >> > negative charges AND their positive counterpart with
> >> > the same proc and date. Thus, for example I need to
> >> > eliminate the negative
> >> > -13 80048 6/7/2001
> >> > AND its positive counterpart with the same proc and
> >> > date:
> >> > +13 80048 6/7/2001,
> >> > and so on.
> >> > I should find a way to construct an algorithm that
> >> > identifies and eliminates the negatives AND their
> >> > poistive counterpart with he same date and procedure,
> >> > but I cannot figure that out.
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