That is very helpful--I've run your code, and it works perfectly with the example data. However,
I'm having some problems using my actual data--probably because my Time variable isn't actually a
regular series (as it was in the example: Time=1:100). When run, it's producing "estim"
and "error" vectors of lengths much greater than N. Here's a subset of my actual data:
dput(example)
structure(list(Time = c(3075L, 3168L, 3318L, 3410L, 3534L, 3715L,
3776L, 3926L, 3987L, 4110L, 4232L, 4291L, 4413L, 4505L, 4536L,
4656L, 4782L, 4886L, 5018L, 5138L, 5187L, 5253L, 5384L, 5540L,
5669L, 5740L, 5796L, 5887L, 5963L, 6042L, 6126L, 6197L, 6280L,
6405L, 6464L, 6553L, 6659L, 6755L, 6847L, 6917L, 7001L, 7120L,
7216L), Price = c(2.08, 3.55, 5.75, 5.69, 4.47, 5.11, 2.74, 3.04,
3.87, 4.7, 6.61, 3.95, 4.63, 7.11, 3.08, 4.476628726, 7.472854559,
8.775893276, 6.34, 5.79, 3.988888889, 4.019166667, 3.69, 4.603636364,
5.242094366, 6.854871699, 5.163700257, 9.154206814, 8.712059541,
10.60635248, 10.58180221, 10.55396909, 10.67812007, 9.985298266,
10.57385693, 9.644945417, 11.62, 12.615, 13.61, 10.833333, 8.38,
12.7, 8.94)), .Names = c("Time", "Price"), row.names = c(NA,
-43L), class = "data.frame")
Is this as simple as replacing the expression:
for (i in Time) {
with
for (i in 1:length(Time)) {
or somesuch?
Mendi
________________________________
De : Rui Barradas <ruipbarra...@sapo.pt>
À : "triskell4-umbre...@yahoo.fr" <triskell4-umbre...@yahoo.fr>
Cc : "r-help@r-project.org" <r-help@r-project.org>
Envoyé le : Mardi 2 avril 2013 11h51
Objet : Re: [R] Iterative regression through a series
Hello,
The error comes from NAs where you would expect coefficients. Try the
following.
set.seed(7511) # Make the example reproducible
N <- 100
Time <-1:N
Price <- rnorm(N, 8, 2)
estim <- numeric(N) # It's better to allocate the results
error <- numeric(N) # vectors in advance
for (i in Time) {
regr <- lm(Price[1:i] ~ Time[1:i])
estim[i] <- coef(regr)[2]
if(is.na(coef(regr)[2]))
error[i] <- NA
else
error[i] <- summary(regr)$coefficients[2,2]
}
Hope this helps,
Rui Barradas
Em 02-04-2013 17:44, triskell4-umbre...@yahoo.fr escreveu:
Hello,
Some context: let's say I have a data series (let's call it PRICE, for
simplicity), sample size N. I have a desire to regress that on TIME, and then
to use the TIME and intercept coefficients to predict the price in the next
period and to use the standard error to calculate a confidence interval. This
is all very easy.
However, what I need help for is to calculate a confidence interval for each
point in time: imagining that at the end of the 10th period I have 10 data
points, and wish to regress them on the 10 periods to create a confidence
interval for the next 'predicted' price. And so on from TIME[10:100]. So the
first regression would be of PRICE[1:10] on TIME[1:10], the second of
PRICE[1:11] on TIME[1:11], the third of PRICE[1:11] on TIME[1:11], and so on to
PRICE[1:N] and TIME[1:N]. I'd like to be able to vary the starting point (so
it would need to be an argument in the function, in this case it would be 10).
The ultimate output of the code would be to save the TIME coefficients and
standard errors it generates to two vectors, say TIME.coef and TIME.SE.
I'm not sure if lapply() can be bent to my will, or if a for loop would be too
inefficient, or what. I'm not new to R, but I'm fairly new to this kind of
programming.
This is a bungled mess of a narrative, and I apologize. Please feel free to
use TIME=1:100 and PRICE=rnorm(100,8,2).
Here's an attempt, which has failed for reasons I can only imagine. Any help
getting this to work would be greatly appreciated. Any help doing this without
loops would be even better.
Time=1:100
Price=rnorm(100,8,2)
estim=0 #I'm hoping this will be the Time coefficient
error=0 #I'm hoping this will be the standard error of the Time coefficient
for (i in Time) {
+ regr=lm(Price[1:i]~Time[1:i])
+ estim=c(estim,coef(summary(regr))[2,1])
+ error=c(error,coef(summary(regr))[2,1])
+ }
Error: subscript out of bounds
Many, many thanks in advance.
Mendi
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