Thank you , Rui--it works very well now. I'll have to check out the attributes
and classes of the things you changed to understand what was going wrong
before. Integer versus numeric, perhaps. I'll look into seq_len() as well.
Many, many thanks for your help. It will make this and several future projects
much more streamlined.
Regards,
Mendi
________________________________
De : Rui Barradas <ruipbarra...@sapo.pt>
Cc : "r-help@r-project.org" <r-help@r-project.org>
Envoyé le : Mercredi 3 avril 2013 3h43
Objet : Re: [R] Iterative regression through a series
Hello,
I've made a samll change to the code, and with your example data it's
now working without errors.
N <- nrow(example)
estim <- numeric(N) # It's better to allocate the results
error <- numeric(N) # vectors in advance
for (i in seq_len(N)) {
regr <- lm(example$Price[1:i] ~ example$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
> 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>
> 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
>
>> 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
>> [[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.
>>
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