On May 24, 2014, at 6:58 AM, Jason Stout, M.D. wrote:
> Hi R-users,
>
> I'm trying to simulate the outcome of several diagnostic tests with binary
> outcomes (positive/negative) and different performance characteristics. What
> I would like to generate is a dataframe with the first column representing
> the result of a "perfect" test, and different columns to the right simulating
> tests with different combinations of sensitivity and specificity. Here's the
> code I used to attempt this:
>
> x3<-rmvbin(10000,margprob=0.2) # results of perfect test, disease prevalence
> 20%
> sens1<-0.7 # sensitivity of test 1
> spec1<-0.8 # specificity of test 1
> sens2<-0.8 # sensitivity of test 2
> spec2<-0.7 # specificity of test 2
> funcsensspec<-function(x,sens,spec) {
> result<-ifelse(x==1,ifelse(runif(1,0,1)<=sens,1,0),
> ifelse(runif(1,0,1)<=spec,0,1))
> return(result)
> }
> x4<-funcsensspec(x3,sens1,spec1)
> x5<-funcsensspec(x3,sens2,spec2)
> xx<-cbind(x3,x4,x5)
>
> The problem is that this is not behaving as I expected. Ideally I wanted R
> to randomly reassign values for each row with probability based on test
> characteristics, but what I think it is doing is generating one random value
> and using that to reassign the entire vector. Here is sample output:
I think you need to re-examine your understanding of sensitivity and
specificity. They are mappings from data to a continuous variable in [0,1] for
varying choice of a continuously variable threshold, rather than a sampling
parameter. The typical display of ROC curves is if no help to physicians
because it obscures this relationship by failing to include annotations with
the threshold values.
--
David Winsemius, MD
>
>
>> head(xx,30)
> [,1] [,2] [,3]
> [1,] 0 1 0
> [2,] 0 1 0
> [3,] 0 1 0
> [4,] 0 1 0
> [5,] 0 1 0
> [6,] 0 1 0
> [7,] 1 1 1
> [8,] 0 1 0
> [9,] 1 1 1
> [10,] 0 1 0
> [11,] 1 1 1
> [12,] 0 1 0
> [13,] 0 1 0
> [14,] 1 1 1
> [15,] 0 1 0
> [16,] 1 1 1
> [17,] 0 1 0
> [18,] 0 1 0
> [19,] 0 1 0
> [20,] 0 1 0
> [21,] 0 1 0
> [22,] 0 1 0
> [23,] 0 1 0
> [24,] 1 1 1
> [25,] 0 1 0
> [26,] 0 1 0
> [27,] 0 1 0
> [28,] 0 1 0
> [29,] 0 1 0
> [30,] 0 1 0
>
> I know I could do this task by creating two actual vectors of random numbers
> and using them to assign probabilities, but am wondering if there is a
> simpler and more elegant way to accomplish the task (and this would provide
> some insight into how ifelse is working within this function).
>
> Thanks for any assistance.
>
>
> Jason Stout, MD, MHS
> Box 102359-DUMC
> Durham, NC 27710
> FAX 919-681-7494
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
