Re: [R] Iterative regression through a series
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)$coefficients[2,2]
}
Hope this helps,
Rui Barradas
That is very helpful--I've run your code, and it works perfectly with the
example data. However, I'm having some problems using my actual
data--probably because my Time variable isn't actually a regular series (as
it was in the example: Time=1:100). When run, it's producing estim and
error vectors of lengths much greater than N. Here's a subset of my actual
data:
dput(example)
structure(list(Time = c(3075L, 3168L, 3318L, 3410L, 3534L, 3715L,
3776L, 3926L, 3987L, 4110L, 4232L, 4291L, 4413L, 4505L, 4536L,
4656L, 4782L, 4886L, 5018L, 5138L, 5187L, 5253L, 5384L, 5540L,
5669L, 5740L, 5796L, 5887L, 5963L, 6042L, 6126L, 6197L, 6280L,
6405L, 6464L, 6553L, 6659L, 6755L, 6847L, 6917L, 7001L, 7120L,
7216L), Price = c(2.08, 3.55, 5.75, 5.69, 4.47, 5.11, 2.74, 3.04,
3.87, 4.7, 6.61, 3.95, 4.63, 7.11, 3.08, 4.476628726, 7.472854559,
8.775893276, 6.34, 5.79, 3.98889, 4.01917, 3.69, 4.603636364,
5.242094366, 6.854871699, 5.163700257, 9.154206814, 8.712059541,
10.60635248, 10.58180221, 10.55396909, 10.67812007, 9.985298266,
10.57385693, 9.644945417, 11.62, 12.615, 13.61, 10.83, 8.38,
12.7, 8.94)), .Names = c(Time, Price), row.names = c(NA,
-43L), class = data.frame)
Is this as simple as replacing the expression:
for (i in Time) {
with
for (i in 1:length(Time)) {
or somesuch?
Mendi
De : Rui Barradas ruipbarra...@sapo.pt
Cc : r-help@r-project.org r-help@r-project.org
Envoyé le : Mardi 2 avril 2013 11h51
Objet : Re: [R] Iterative regression through a series
Hello,
The error comes from NAs where you would expect coefficients. Try the
following.
set.seed(7511) # Make the example reproducible
N - 100
Time -1:N
Price - rnorm(N, 8, 2)
estim - numeric(N) # It's better to allocate the results
error - numeric(N) # vectors in advance
for (i in Time) {
regr - lm(Price[1:i] ~ Time[1:i])
estim[i] - coef(regr)[2]
if(is.na(coef(regr)[2]))
error[i] - NA
else
error[i] - summary(regr)$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
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
Em 03-04-2013 01:57, triskell4-umbre...@yahoo.fr escreveu:
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, 3.98889, 4.01917, 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, 10.83, 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
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
[R] Iterative regression through a series
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.
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
[[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.
__
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.
Re: [R] Iterative regression through a series
Also look at zoo's rollapply.
MW
On Apr 2, 2013, at 13:51, Rui Barradas ruipbarra...@sapo.pt wrote:
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
[[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.
__
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.
__
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.
Re: [R] Iterative regression through a series
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, 3.98889, 4.01917, 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, 10.83, 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.
[[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.
Re: [R] Iterative regression through a series
Thank you, Michael--it looks like a fantastic package. I'm fairly new to R,
and not at all familiar with the zoo class, but it seems to have great
potential. I'll need to spend some time sorting out how to use regressions in
it, because the syntax used in the examples is alien to me.
Mendi
De : Michael Weylandt michael.weyla...@gmail.com
À : Rui Barradas ruipbarra...@sapo.pt
@r-project.org r-help@r-project.org
Envoyé le : Mardi 2 avril 2013 11h58
Objet : Re: [R] Iterative regression through a series
Also look at zoo's rollapply.
MW
On Apr 2, 2013, at 13:51, Rui Barradas ruipbarra...@sapo.pt wrote:
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
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Re: [R] Iterative regression through a series
Hi: rollapply is a fine solution but if you frame your problem as a state
space
model where the observation equation is the linear regression you're
interested in,
the state space framework provided by the DLM package will give you the
output
you want at each step t. the gory details are explained in the paper below
but
I only have a hard copy ( somewhere ? ) and it's 14.00 from JSTOR.
http://www.jstor.org/discover/10.2307/2284643?uid=3739832uid=2uid=4uid=3739256sid=21101972960881
On Tue, Apr 2, 2013 at 8:58 PM, David Chertudi david_txert...@yahoo.frwrote:
Thank you, Michael--it looks like a fantastic package. I'm fairly new to
R, and not at all familiar with the zoo class, but it seems to have great
potential. I'll need to spend some time sorting out how to use regressions
in it, because the syntax used in the examples is alien to me.
Mendi
De : Michael Weylandt michael.weyla...@gmail.com
À : Rui Barradas ruipbarra...@sapo.pt
@r-project.org r-help@r-project.org
Envoyé le : Mardi 2 avril 2013 11h58
Objet : Re: [R] Iterative regression through a series
Also look at zoo's rollapply.
MW
On Apr 2, 2013, at 13:51, Rui Barradas ruipbarra...@sapo.pt wrote:
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.
__
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.
[[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.
[[alternative HTML version deleted]]
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