That's what I am looking for. Thanks for being so helpful, I appreciate it
very much.
On Sun, Feb 23, 2014 at 10:16 PM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> On Sun, Feb 23, 2014 at 9:06 PM, C W wrote:
> > Is there a way to rbind this? Do I have to use use a package like 'zoo
On Sun, Feb 23, 2014 at 9:06 PM, C W wrote:
> Is there a way to rbind this? Do I have to use use a package like 'zoo' and
> merge()?
>
> p <- predict(model, data.frame(t = 1, q = factor(1, 1:4))
> rbind(tsdat, p)
>
If you want to append this to the end of the series then try this:
ts(c(tsdat,
Is there a way to rbind this? Do I have to use use a package like 'zoo'
and merge()?
p <- predict(model, data.frame(t = 1, q = factor(1, 1:4))
rbind(tsdat, p)
On Sun, Feb 23, 2014 at 8:43 PM, C W wrote:
> I guess this is for anyone in the future.
>
> > predict(model, data.frame(t = 1, q = fac
I guess this is for anyone in the future.
> predict(model, data.frame(t = 1, q = factor(1, 1:4))
This would be the answer. Thanks again, Gabor!
Mike
On Sun, Feb 23, 2014 at 8:23 PM, Gabor Grothendieck wrote:
> On Sun, Feb 23, 2014 at 7:40 PM, C W wrote:
> > Gabor,
> > Your response worked p
On Sun, Feb 23, 2014 at 7:40 PM, C W wrote:
> Gabor,
> Your response worked perfectly, list(t = 1, q = factor(1, 1:4) was what I
> couldn't figure out. Thank you.
>
> In predict.lm(model, newdata), 2nd argument MUST be data.frame(). Why does
> list() also work?
>
Yes, it would be better to use
Gabor,
Your response worked perfectly, list(t = 1, q = factor(1, 1:4) was what I
couldn't figure out. Thank you.
In predict.lm(model, newdata), 2nd argument MUST be data.frame(). Why does
list() also work?
Mike
On Sun, Feb 23, 2014 at 7:32 PM, Gabor Grothendieck wrote:
> On Sun, Feb 23, 20
On Sun, Feb 23, 2014 at 7:25 PM, C W wrote:
> Gabor,
> Let me change newdata since it's confusing.
>
> Suppose I want to predict, year 1990, and quarter 2.
>> newdata <- data.frame(c(1990, 1, 0, 0)
>
> Since Q1 is a baseline, we will only see Q2, Q3, Q4. So, 4 parameters in
> total.
>
The formul
Gabor,
Let me change newdata since it's confusing.
Suppose I want to predict, year 1990, and quarter 2.
> newdata <- data.frame(c(1990, 1, 0, 0)
Since Q1 is a baseline, we will only see Q2, Q3, Q4. So, 4 parameters in
total.
Mike
On Sun, Feb 23, 2014 at 7:13 PM, Gabor Grothendieck wrote:
>
On Sun, Feb 23, 2014 at 6:56 PM, C W wrote:
> Gabor,
> I want the new data to be this,
> newdata <- data.frame(c(1, 0, 0, 0))
>
Its not clear what this means. There are two input variables so we
must specify two inputs.
For example, this would get the prediction for t=1 and for level 1 of
q whic
Gabor,
I want the new data to be this,
newdata <- data.frame(c(1, 0, 0, 0))
Here the code with the new data,
> dat <- rnorm(20)
> tsdat <- ts(dat, start=c(1900, 1), freq=4)
> q <- as.factor(rep(1:4, 5))
> t <- 1:20
> lm(tsdat~t+q)
Call:
lm(formula = tsdat ~ t + q)
Coefficients:
(Intercept)
On Sun, Feb 23, 2014 at 6:34 PM, C W wrote:
> Hello,
> I don't know how to use predict.lm() for ts object.
>
> Here's the time series regression.
> y = t + Q1 + Q2 + Q3 + Q4
>
> Here's my R code,
>
>> dat <- rnorm(20)
>> tsdat <- ts(dat, start=c(1900, 1), freq=4)
>> q <- as.factor(rep(1:4, 5))
>>
Hello list,
Does predict.lm() take ts object. I am aware that it requires data to come
in data.frame
Here's the time series regression.
y = t + Q1 + Q2 + Q3 + Q4
Here's my R code,
> dat <- rnorm(20)
> tsdat <- ts(dat, start=c(1900, 1), freq=4)
> q <- as.factor(rep(1:4, 5))
> t <- 1:20
> lm(ts
Hello,
I don't know how to use predict.lm() for ts object.
Here's the time series regression.
y = t + Q1 + Q2 + Q3 + Q4
Here's my R code,
> dat <- rnorm(20)
> tsdat <- ts(dat, start=c(1900, 1), freq=4)
> q <- as.factor(rep(1:4, 5))
> t <- 1:20
> lm(tsdat~t+q)
Call:
lm(formula = tsdat ~ t + q)
Michael Haenlein wrote
Dear all,
I would like to use predict.lm to obtain a set of predicted values
based on
a regression model I estimated.
When I apply predict.lm to two vectors that have the same values, the
predicted values will be identical. I know that my regress
> I would like to use predict.lm to obtain a set of predicted
> values based on a regression model I estimated.
>
Do you want predictions - which will always be the same - or randomly
distributed data that is based on the model prediction plus random error?
*
Dear all,
I would like to use predict.lm to obtain a set of predicted values based on
a regression model I estimated.
When I apply predict.lm to two vectors that have the same values, the
predicted values will be identical. I know that my regression model is not
perfect and I would like to take a
The most common case that I see that error is when someone fits their
model using syntax like:
fit <- lm( mydata$y ~ mydata$x )
instead of the preferred method:
fit <- lm( y ~ x, data=mydata )
The fix (if this is what you did and why you are getting the error) is
to not use the first way and in
e, because only data.frame forces all of
its components to have the same length.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of frauke
> Sent: Wed
> Of course I can extend the new dataframe with a few thousands
> NAs, but is there a more elegant solution?
That should not be necessary: predict.lm should work on any number of newdata
rows, whether longer or shorter than the original data set.
However, the help page for predict.lm says (among
Hi everybody,
recently a member of the community pointed me to the useful predict.lm()
comment. While I was toying with it, I stumbled across the following
problem.
I do the regression with data from five years. But I want to do a prediction
with predict.lm for only one year. Thus my dataframe f
.
Proposed:
newdata An optional data frame in which to
look for new values of terms with
which to predict. If omitted, the
fitted values are used.
-John Thaden, Ph.D.
College Station, TX
--- On Sun, 9/2/12, peter dalgaard wrote:
From: peter dalgaard
h to predict. If omitted, the
fitted values are used.
-John Thaden, Ph.D.
College Station, TX
--- On Sun, 9/2/12, peter dalgaard wrote:
> From: peter dalgaard
> Subject: Re: [R] predict.lm(...,type="terms") question
> To: "David Winsemius"
> Cc: "R
On Sep 2, 2012, at 03:38 , David Winsemius wrote:
>
> Why should predict not complain when it is offered a newdata argument that
> does no contain a vector of values for "x"? The whole point of the terms
> method of prediction is to offer estimates for specific values of items on
> the RHS of
On Sep 1, 2012, at 4:33 AM, Rui Barradas wrote:
Hello,
John is right, there seems to be an error in predict.lm. It can be
made to work but if the model is fitted with lm(...,data) then it
messes things up. Apparently predict.lm disregards the data argument
and uses whatever it finds in t
Hello,
John is right, there seems to be an error in predict.lm. It can be made
to work but if the model is fitted with lm(...,data) then it messes
things up. Apparently predict.lm disregards the data argument and uses
whatever it finds in the global environment. In the examples below,
after r
On Aug 31, 2012, at 3:48 PM, jjthaden wrote:
On Aug 30, 2012, at 4:35 PM, David Windemius wrote:
David said my newdata data frame 'new' must have a column named
'area'.
It did. Nonetheless predict.lm throws an error with type = "terms"
and
newdata = new. I see nothing in the predict.lm
On Aug 31, 2012, at 5:27 PM, jjthaden wrote:
The error is generated in the last line of code shown here from
predict.lm
The code you offered had an extra comma in the 'area' vector. Removing
allowed the fitting and prediction to proceed without error, I do not
see why you are producing c
On Aug 30, 2012, at 4:35 PM, David Windemius wrote:
>> David said my newdata data frame 'new' must have a column named 'area'.
>> It did. Nonetheless predict.lm throws an error with type = "terms" and
>> newdata = new. I see nothing in the predict.lm documentation that
>> bars this usage. Is the
The error is generated in the last line of code shown here from predict.lm
> predict.lm
function (object, newdata, se.fit = FALSE, scale = NULL, df = Inf,
interval = c("none", "confidence", "prediction"), level = 0.95,
type = c("response", "terms"), terms = NULL, na.action = na.pass,
Dave said my newdata data frame 'new' must have a column named 'area'.
It did. Nonetheless predict.lm throws an error with type = "terms" and
newdata = new. I see nothing in the predict.lm documentation that
bars this usage. Is there a bug?
To illustrate an OLS behavior, I had cited Ludbrook '12.
On Aug 30, 2012, at 2:29 PM, John Thaden wrote:
> Dave said my newdata data frame 'new' must have a column named 'area'.
> It did. Nonetheless predict.lm throws an error with type = "terms" and
> newdata = new. I see nothing in the predict.lm documentation that
> bars this usage. Is there a bug?
Hello,
Instead of reversing the regression, that, like you say, may have
problems, it's very easy to wrap the formula
x' <- (y' - beta0)/beta1
in a function and use the direct regression to get new 'x' values from
new 'y' ones.
This function assumes a first order ols model.
invpredict <- f
On Aug 29, 2012, at 4:08 PM, John Thaden wrote:
Draper & Smith sections (3.2, 9.6) address prediction interval
issues, but
I'm also concerned with the linear fit itself. The Model II regression
literature
Citations?
makes it abundantly clear that OLS regression of x on y
frequently yields
Draper & Smith sections (3.2, 9.6) address prediction interval issues, but
I'm also concerned with the linear fit itself. The Model II regression
literature makes it abundantly clear that OLS regression of x on y
frequently yields a different line than of y on x. The example below is not
so extreme
Hello,
Ok, got it, thanks.
Apparently I was right, the estimator is still x.pred = (y.new - b0)/b1
which can be obtained from predict.lm with the inverse regression model
formula. According to the litterature, the main differences are in the
confidence intervals for the true value of x.pred,
On Aug 29, 2012, at 8:06 AM, John Thaden wrote:
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted?
The new data argument MUST contain a column with the name "area". If
it does not hten the original data is used.
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted? If so,
what numbers would I fill it with? Or should my newdata object include
the original data, too? Is there another mailing list I can ask?
Thanks,
-John
On Wed, Aug
I think that what the OP is looking for comes under the heading of
"inverse regression" or the "calibration" problem. One reference
with a simple explanation including confidence intervals is "Applied
regression analysis" by Draper and Smith. (It's in section 3.2 in
my 3rd edition).
Peter Ehlers
I think I may be misreading the help pages, too, but misreading how?
I agree that inverting the fitted model is simpler, but I worry that I'm
misusing ordinary least squares regression by treating my response, with
its error distribution, as a predictor with no such error. In practice,
with my rea
Hello,
Inline.
Em 29-08-2012 16:06, John Thaden escreveu:
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted? If so,
what numbers would I fill it with? Or should my newdata object include
the original data, too? Is the
Hello,
predict(..., type = "terms") returns the predicted regressors multiplied
by the respective beta centered. The centering constant is also returned:
model <- lm(area ~ concn, data) # Already run
predict(model, type = "terms")
concn
1 -11542.321
2 -8244.515
3 1648.903
4 18137.9
Hello,
You seem to be misreading the help pages for lm and predict.lm, argument
'terms'.
A much simpler way of solving your problem should be to invert the
fitted model using lm():
model <- lm(area ~ concn, data) # Your original model
inv.model <- lm(concn ~ area, data = data) # Your probl
Hello all,
How do I actually use the output of predict.lm(..., type="terms") to
predict new term values from new response values?
I'm a chromatographer trying to use R (2.15.1) for one of the most
common calculations in that business:
- Given several chromatographic peak areas measured for
On Mar 23, 2011, at 10:05 AM, agent dunham wrote:
Dear all,
I've fitted a lm using 61 data (training data), and I'left 10 as
test data.
Training data and test data are stored in an excell.
training <- read.xls("C:/./training.xls") , the same for test.
That is:
v1
v2
...
v15
When I
Dear all,
I've fitted a lm using 61 data (training data), and I'left 10 as test data.
Training data and test data are stored in an excell.
training <- read.xls("C:/./training.xls") , the same for test. That is:
v1
v2
...
v15
When I type str(training) and str(test), both sets have the sam
Thanks very much Dennis and Michael. That makes sense. I had some very
sloppy code which did the equivalent of
y<-rnorm(100)
x<-list()
x[[1]]<-rnorm(100)
x[[2]]<-rnorm(100)
lm.yx<-lm(y~x[[1]]+x[[2]])
To predict from lm.yx, I was trying to feed predict.lm a data.frame
with columns called x[[1]] an
Hi:
On Wed, Dec 15, 2010 at 10:05 PM, Anirban Mukherjee wrote:
> Hi all,
>
> Suppose:
>
> y<-rnorm(100)
> x1<-rnorm(100)
> lm.yx<-lm(y~x1)
>
> To predict from a new data source, one can use:
>
> # works as expected
> dum<-data.frame(x1=rnorm(200))
> predict(lm.yx, newdata=dum)
>
Yup.
>
> Suppos
Hi all,
Suppose:
y<-rnorm(100)
x1<-rnorm(100)
lm.yx<-lm(y~x1)
To predict from a new data source, one can use:
# works as expected
dum<-data.frame(x1=rnorm(200))
predict(lm.yx, newdata=dum)
Suppose lm.yx has been run and we have the lm object. And we have a
dataframe that has columns that don't
Hi Anirban,
You can do it like this...
lm.foobar <- lm(foo ~ bar)
some.data <- rnorm(200)
predict(lm.foobar, newdata=list(bar=some.data))
Hope that helps.
Michael
On 16 December 2010 17:05, Anirban Mukherjee wrote:
> Hi all,
>
> Suppose:
>
> y<-rnorm(100)
> x1<-rnorm(100)
> lm.yx<-lm(y~x1)
>
Hi all,
Suppose:
y<-rnorm(100)
x1<-rnorm(100)
lm.yx<-lm(y~x1)
To predict from a new data source, one can use:
# works as expected
dum<-data.frame(x1=rnorm(200))
predict(lm.yx, newdata=dum)
Suppose lm.yx has been run and we have the lm object. And we have a
dataframe that has columns that don't
))
Dave
From:
"Thaler, Thorn, LAUSANNE, Applied Mathematics"
To:
Date:
12/13/2010 12:16 PM
Subject:
[R] predict.lm[e] with formula passed as a variable
Sent by:
r-help-boun...@r-project.org
Dear all,
In a function I paste a string and convert it to a formula which I pass
to lm[e]. T
Dear all,
In a function I paste a string and convert it to a formula which I pass
to lm[e]. The idea is to write a function which takes the name of the
response variable and the explanatory variable and the data frame as an
argument and calculates an lm[e]. (see example below)
This works fine, bu
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Stephan Kolassa
> Sent: Tuesday, August 17, 2010 5:25 AM
> To: R-help@r-project.org
> Subject: [R] predict.lm, matrix in formula and newdata
>
> Dear a
Of Stephan Kolassa
Sent: 17 August 2010 14:25
To: R-help@r-project.org
Subject: [R] predict.lm, matrix in formula and newdata
Dear all,
I am stumped at what should be a painfully easy task: predicting from an lm
object. A toy example would be this:
XX <- matrix(runif(8),ncol=2)
yy <- runif(4
try it better this way:
XX <- matrix(runif(8), ncol = 2)
DF <- as.data.frame(XX)
DF$yy <- runif(4)
model <- lm(yy ~ ., DF)
XX.pred <- as.data.frame(matrix(runif(6), ncol = 2))
predict(model, XX.pred)
I hope it helps.
Best,
Dimitris
On 8/17/2010 2:24 PM, Stephan Kolassa wrote:
Dear all,
I
Dear all,
I am stumped at what should be a painfully easy task: predicting from an lm
object. A toy example would be this:
XX <- matrix(runif(8),ncol=2)
yy <- runif(4)
model <- lm(yy~XX)
XX.pred <- data.frame(matrix(runif(6),ncol=2))
colnames(XX.pred) <- c("XX1","XX2")
predict(model,newdata=XX.p
You can use predict() by specifying a complete data.frame() for prediction to
the argument newdata=. Look:
da <- expand.grid(x1=LETTERS[1:4], x2=1:9)
da$y <- rnorm(da$x1)
da$y[sample(length(da$y), 5)] <- NA
m0 <- lm(y~x1+x2, data=da)
predict(m0) # NA not predicted
predict(m0, newdata=da) # NA pr
see ?na.exclude
you can set na.action='na.exclude' when fit the model.
On 15 April 2010 09:06, Martin Batholdy wrote:
> Hi,
>
> I wanted to use the predict.lm() function to compare the empirical data with
> the predicted values.
> The problem is that I have NAs in my data.
>
> I wanted to cbin
Hi,
I wanted to use the predict.lm() function to compare the empirical data with
the predicted values.
The problem is that I have NAs in my data.
I wanted to cbind my data.frame with the empirical values with the vector I get
from predict.lm.
But they don't have the same length because predict.
Hi Felipe,
Make sure "predictors" data.frame has the EXACT same column names as the
data.frame you used for creating ModeloLineal.
Cheers,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www
Hello I am trying to use predict.lm, but I am having trouble getting out of
sample predictions. I am getting the same output if I use the following
three commands:
predict(ModeloLineal,predictors[721:768,])
predict(ModeloLineal,predictors[1:768,])
predict(ModeloLineal)
where ModeloLineal is the o
Hello!
I'm still working on my problem, which also occurs with the predict.lm()
function. - Providing newdata, which is a data.frame with all variables
being "numeric", as str() shows, R tells me the following:
ar1.xpred.test.pred <- predict(ar1.xpred.fitted, regdata.test, se.fit =
FALSE)
Feh
You called lm() with a predictor named ``D$X'' and called predict.lm()
with a predictor name ``X''.
Simplest remedy: Use
fit <- lm(Y ~ X, data = D)
Remark: Not a good idea to use ``D'' as the name of your data frame
(``D'' is
the name of a function --- derivative). Likewise don't u
On 07/04/2008 5:57 PM, Chip Barnaby wrote:
> Dear R-people ...
>
> I'm a new user. I can't get predict.lm() to produce predictions for
> new independent data. There are some messages in archived help about
> this problem, but I still don't see my error after reviewing
> those. I understand t
Dear R-people ...
I'm a new user. I can't get predict.lm() to produce predictions for
new independent data. There are some messages in archived help about
this problem, but I still don't see my error after reviewing
those. I understand that the new independent data must have the same
name(s
Marlin Keith Cox wrote:
> Thank you in advance for helping me with this.
> I included a part of the actual data. The result of pred.est and
> pred.est1are different, but they should be identical. For
> pred.est, I entered the slope and y intercept and received a value for each
> individual numb
Thank you in advance for helping me with this.
I included a part of the actual data. The result of pred.est and
pred.est1are different, but they should be identical. For
pred.est, I entered the slope and y intercept and received a value for each
individual number in the matrix (Z). For pred.est1
Marlin Keith Cox wrote:
> Z is a matrix and when I run the following line, it creates a prediction
> estimate using each column, how can I get it an estimate for each individual
> number. I have tried changing Z to a data.frame, but this does not do it
> either.
>
> model.lm<-lm(w~x)
>
> pred.
Z is a matrix and when I run the following line, it creates a prediction
estimate using each column, how can I get it an estimate for each individual
number. I have tried changing Z to a data.frame, but this does not do it
either.
model.lm<-lm(w~x)
pred.est <- predict.lm(model.lm, data.frame(x=Z
This is an inconsistency in drop:
> x <- matrix(1:4, 4,1, dimnames=list(letters[1:4], NULL))
> x
[,1]
a1
b2
c3
d4
> drop(x)
a b c d
1 2 3 4
> drop(x[1,,drop=FALSE])
[1] 1
S does not do that, and I don't think R should, given its documentation.
(Note that x[1,] also drops names
predict.lm keeps row names when working from several rows in
newdata, but always removes rowname from a single row.
The rownames are removed by the line in predict.lm
predictor <- drop(X[, piv, drop = FALSE] %*% beta[piv])
What is the reason for that decision? I usually want to
retain the row
71 matches
Mail list logo
