Hi:
Here's one way using package plyr and its ddply() function. ddply()
takes a data frame as input and expects to output either a scalar or a
data frame. In this case, we want the latter.
library(plyr)
f <- function(df) {
mn <- min(df$result)
tms <- df$time[df$result == mn]
subdf <- df[max(tms):nrow(df), ]
b1 <- coef(lm(result ~ time, data = subdf))[2]
data.frame(NadirLevel = mn, NadirFirstTime = min(tms),
NadirLastTime = max(tms), Slope = b1)
}
This function takes a data frame df as input - in practice, it will be
a sub-data frame associated with a level of ID. We find the minimum of
result and assign it to mn, and then find the times that match the
minimum.
Next, we construct the subdata on which to run the simple linear
regression line. Finally, an output data frame is created. ddply()
will add in the ID variable. Calling your example data frame d,
> ddply(d, 'ID', f)
ID NadirLevel NadirFirstTime NadirLastTime Slope
1 A 1 3 5 1
2 B 2 2 2 2
HTH,
Dennis
On Mon, Jun 6, 2011 at 10:04 PM, Williams Scott
<Scott.Williams at petermac.org> wrote:> I have a large dataset similar to this:
>
> ID ? ? ?time ? ?result
> A ? ? ? 1 ? ? ? 5
> A ? ? ? 2 ? ? ? 2
> A ? ? ? 3 ? ? ? 1
> A ? ? ? 4 ? ? ? 1
> A ? ? ? 5 ? ? ? 1
> A ? ? ? 6 ? ? ? 2
> A ? ? ? 7 ? ? ? 3
> A ? ? ? 8 ? ? ? 4
> B ? ? ? 1 ? ? ? 3
> B ? ? ? 2 ? ? ? 2
> B ? ? ? 3 ? ? ? 4
> B ? ? ? 4 ? ? ? 6
> B ? ? ? 5 ? ? ? 8
>
> I need to extract a number of features for each individual in it
(identified by "ID"). These are:
> * The lowest result (the nadir)
> * The time of the nadir - but if the nadir level is present at >1 time
point, I need the minimum and maximum time of nadir
> * For the time period from maximum time of nadir to the last result, I need
the coefficient from a lm(result~time)
>
> The result would be a table looking like:
>
> ID ? ? ?NadirLevel ? ? ?NadirFirstTime ?NadirLastTime ? Slope
> A ? ? ? 1 ? ? ? ? ? ? ? 3 ? ? ? ? ? ? ? ? ? ? ? 5 ? ? ? ? ? ? ? ? ? ? ? 1
> B ? ? ? 2 ? ? ? ? ? ? ? 2 ? ? ? ? ? ? ? ? ? ? ? 2 ? ? ? ? ? ? ? ? ? ? ? 2
>
> I can manage to extract all the required elements in a very cumbersome
loop, but I am sure an elegant method using apply() or the like could be devised
but I cant presently understand the necessary syntax. An suggestions
appreciated.
>
> Thanks
> Scott
> _____________________________
>
> Dr. Scott Williams
> Peter MacCallum Cancer Centre
> Melbourne, Australia
> ph +61 3 9656 1111
> fax +61 3 9656 1424
> scott.williams at petermac.org
>
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