No interpolation [NCA / SHAM]

posted by Helmut Homepage – Vienna, Austria, 2013-05-23 01:25 (3962 d 21:15 ago) – Posting: # 10623
Views: 17,475

Hi Yung-jin,

❝ I will change the way of AUC calculation with bear later.


Relax. Would make only sense if the user has a choice. Some people still get crazy if one tries anything else than the linear trapezoidal.
Don’t use my code without improving it! In its current form it will not work with NAs and/or isolated BQLs within the profile.

❝ I remembered that I read something about method comparisons for trapezoidal rule in one textbook, but just cannot remember which book.


Yes. I think I have posted even references somewhere. To lazy to search now.

❝ ❝ ...

❝ ❝   B1.1   <- C0.1*k01.1/(k01.1-K)*exp(k10*tlag1)

❝ ❝   B2.1   <- C0.1*k01.1/(k01.1-K)*exp(k01.1*tlag1)

❝ Here I got Error: object 'K' not found and some error messages after this.


Sorry, I made a copy/paste error. Should work now (use k10 instead of K).

❝ One more question, in your above presentation (slide#17[edited]; missing data occurred in R at 12h), it was a lin-up/log-down plot.


Nope. 16/17 are linear plots. In slide 16 I connected all points by straight lines (what the linear trap. would do), and in slide 17 by straight lines if ascending and by an exponential if descending.

❝ As you said that no need to do data imputation if using lin-up/log-down. Do you do any line smoothing with your data first?


No.

❝ Otherwise, the line should not be connected in that way. It should be connected as the line that the arrow points.


Why?

❝ Do I miss anything here?


Yes, you do. ;-) If the 12 h sample is missing the linear trapezoidal works as if we would impute a value by linear interpolation. The lin-up/log-down works as if we would impute a value by logarithmic interpolation. But the trick is that nothing has to be imputed in the lin-up/log-down at all!
  t    C
  8  33.17
 12  miss.
 16   7.86


Interpolated values for imputation:
linear: C12*=33.17+|(12-8)/(16-8)|(7.86-33.17)=20.52
linlog: C12*=exp(log(33.17)+|(12-8)/(16-8)|log(7.86/33.17))=16.15


Calculation based on imputed values:
linear/lin imp.: pAUC8-16=0.5(12- 8)(33.17+20.52)+
                          0.5(16-12)(20.52+ 7.86)=164.1
linear/log imp.: pAUC8-16=0.5(12- 8)(33.17+20.52)+
                          0.5(16-12)(20.52+ 7.86)=146.7
linlog/log imp.: pAUC8-16=(12- 8)(16.15-33.17)/log(16.15/33.17)+
                          (16-12)( 7.86-16.15)/log( 7.86/16.15)=140.6


Direct calculation without imputation:
linear: pAUC8-16=0.5(16-8)(33.17+7.86)=164.1
linlog: pAUC8-16=(16-8)(7.86-33.17)/log(7.86/33.17)=140.6


BTW, the theoretical pAUC8-16 based on the model is 140.62…

To make a long story short: Only if one has to use the linear trapezoidal (for whatever wacky reasons) I would impute an estimate based on the log/linear interpolation (second variant above). Otherwise the bias is substantial. Q.E.D.

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