On Feb 3, 2010, at 3:36 PM, stephen sefick wrote:
This is a subset of a much larger dataframe. I would like to be
able
to automate finding the pair of x, y coordinates where the line
crosses zero agian
x <- (structure(list(bankfull_depths_m = c(0, 0.17, 0.38, 0.37,
0.36,
0.39, 0.47, 0.48, 0.19, 0.05, -0.05, -0.09), measurment_num_m =
c(0.2,
0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4)), .Names =
c("bankfull_depths_m",
"measurment_num_m"), class = "data.frame", row.names = 97:108))
qplot(measurment_num_m, bankfull_depths_m, data=x)
in this case it is 2.1, 0
I cannot quite get there with a naive application of approxfun using
"reversed arguments", since the inverse function that would be
created is
not a legitimate function. I can do it in segments, though:
y.x <- approxfun(x=x$bankfull_depths_m[9:12], y=x
$measurment_num_m[9:12])
y.x(0)
[1] 2.1 # so you can do it piecewise in regions where the x-y
function is
monotonic
y.x <- approxfun(x=x$bankfull_depths_m, y=x$measurment_num_m)
y.x(0)
[1] 0.2 # an almost trivial result at the LHS of the range.
You could also try to do a Newtonian walk (at least that is my
guess for the
underlayment of uniroot() ) along the result of the the unreversed
arguments.
x.y <- approxfun(y=x$bankfull_depths_m, x=x$measurment_num_m)
uniroot(x.y, range(x$measurment_num_m) )
$root
[1] 0.2
$f.root
[1] 0
$iter
[1] 0
$estim.prec
[1] 0
# That was the first root and this is the second.
uniroot(x.y, c(1, 2.3) )
$root
[1] 2.1
$f.root
[1] 0
$iter
[1] 3
$estim.prec
[1] 0.01162791
So you still need to apply some guidance to the functions.
--
david
I need a way to process a whole bunch of data points, and I am at a
loss. Thanks for any help.
regards,
Whole bunch??? Can you be any more vague, please?
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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