# st: Re: RE: Re: eliminate negative values and their positive counterpart

 From "Sergiy Radyakin" To Subject st: Re: RE: Re: eliminate negative values and their positive counterpart Date Tue, 6 Mar 2007 22:53:10 +0100

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-

Sincerely yours,

----- Original Message ----- From: "Nick Cox" <n.j.cox@durham.ac.uk>
To: <statalist@hsphsun2.harvard.edu>
Sent: Tuesday, March 06, 2007 10:20 PM
Subject: st: RE: Re: eliminate negative values and their positive counterpart

```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

Nick
n.j.cox@durham.ac.uk

```
```you will get a dozen of ways to do this more efficiently in
several minutes,
but here is a straightforward way:

+------------+
| 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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