f <- function(x, y, criticalY) {
# return x value of first upcrossing of y past criticalY
stopifnot(length(x) == length(y), length(x)>1, length(criticalY)==1)
i <- seq_along(x)[-1]
w <- which(y[i] >= criticalY & y[i-1] <= criticalY)[1] # pos of first
upcrossing
if (is.na(w)) { # no upcrossing
NA
} else {
denom <- y[w+1] - y[w]
if (denom == 0) { # take left side of flat spot
x[w]
} else { # the nice case
x[w] + (x[w+1] - x[w]) / denom * (criticalY - y[w])
}
}
}
> f(x=c(1,2,3,4,5), y=c(0, 1.5, 3, 1, 4), criticalY=2)
[1] 2.333333
> f(x=c(1,2,3,4,5), y=c(0, 1.5, 3, 1, 4), criticalY=100)
[1] NA
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Andras Farkas
> Sent: Tuesday, February 26, 2013 1:21 PM
> To: r-help@r-project.org
> Subject: [R] Help on a code
>
> Dear List,
>
> I have the following code:
>
> x <-c(0, 13.8333333333333, 38.1666666666667, 62.1666666666667,
> 85.9166666666667,
> 108.916666666667)
> y <-c(1.77, 2.39, 3, 2.65, 2.62, 1.8)
> Interpolated <- approx(x, y,xout=0:tail(x, n=1),method="linear")
> plot(Interpolated)
>
> in this code x is time in hours (cumulative), and y is a biological variable.
> I am using linear
> interpolation assuming getting from y(i) to y(i+1) is a linear path. I would
> like to calculate
> the time in this case that it takes to surpass the y value of 2 for the first
> time. If you look
> at the plot, you will see that the y value of 2 is crossed on 2 occasions:
> between the first
> and second value of y (ie:1.77 to 2.39) and the fifth and sixth value of y
> (ie2.62 and 1.8).
> In my case I really only care about the first "section" (ie: how long it
> takes to get from
> 1.77 to 2.0). Any thoughts on how I could calculate that chunk of the time
> would be
> greatly appreciated.
>
> let me present it in a simpler idea wher I can actually tell you what the
> answer should be
> without having the code:
>
> x <-c(0, 12, 24, 36, 48, 60)
> y <-c(1.5, 2, 3, 2.65, 2.62, 1.8)
> Interpolated <- approx(x, y,xout=0:tail(x, n=1),method="linear")
> plot(Interpolated)
>
> the answer in this case is 12 hours, which is the time spent in getting from
> 1.5 to 2...
>
> hope I explained clearly what I am trying to do
>
> appreciate the help,
>
> Andras
>
> [[alternative HTML version deleted]]
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