Dear NMusers,
I currently building an indirect response model
C1=A(1)/V1
DADT(2)=KIN-KOUT*INH*A(2)
, where the KOUT is inhibited by INH=1-(SLOPE*C1). SLOPE should be estimated
with variability. How can I make sure that INH stays between 0 and 1?
I wrote it like this
IF ((SLOPE*C1).GT.1) THEN
INH=0
ENDIF
IF ((SLOPE*C1).LT.0) THEN
INH=1
ENDIF
IF((SLOPE*C1).LE.1.AND.(SLOPE*C1).GE.0) THEN
INH=1-SLOPE*C1
ENDIF
, but I think it is not optimal to write it in conditions. Does anyone know a
better way to write this?
Thank you.
-----------------------------
Hauke Rühs
Apotheker
Pharmazeutisches Institut
- Klinische Pharmazie -
An der Immenburg 4
53121 Bonn
Tel: + 49-(0)228 73-5781
Fax: + 49-(0)228 73-9757
http://www.klinische-pharmazie.info/ www.klinische-pharmazie.info
linear relation
Dear Hauke,
I never tried this myself, but I think the following transformation should
give you what you described
INH=1-(exp(slope*C1)-1)/exp(slope*C1)
Jun Shen
Seventh Wave Labs
Quoted reply history
On Wed, Oct 13, 2010 at 10:30 AM, Hauke Rühs <[email protected]> wrote:
> Dear NMusers,
>
>
>
> I currently building an indirect response model
>
>
>
> C1=A(1)/V1
>
> DADT(2)=KIN-KOUT*INH*A(2)
>
>
>
> , where the KOUT is inhibited by INH=1-(SLOPE*C1). SLOPE should be
> estimated with variability. How can I make sure that INH stays between 0 and
> 1?
>
>
>
> I wrote it like this
>
>
>
> IF ((SLOPE*C1).GT.1) THEN
>
> INH=0
>
> ENDIF
>
>
>
> IF ((SLOPE*C1).LT.0) THEN
>
> INH=1
>
> ENDIF
>
>
>
> IF((SLOPE*C1).LE.1.AND.(SLOPE*C1).GE.0) THEN
>
> INH=1-SLOPE*C1
>
> ENDIF
>
>
>
> , but I think it is not optimal to write it in conditions. Does anyone know
> a better way to write this?
>
>
>
> Thank you.
>
>
>
>
>
>
>
> -----------------------------
>
> Hauke Rühs
>
> Apotheker
>
> Pharmazeutisches Institut
>
> - Klinische Pharmazie -
>
> An der Immenburg 4
>
> 53121 Bonn
>
>
>
> Tel: + 49-(0)228 73-5781
>
> Fax: + 49-(0)228 73-9757
>
>
>
> www.klinische-pharmazie.info
>
>
>
>
>