Subject: Re: [R] User error in calling predict/model.frame
On Sun, Jan 30, 2011 at 5:59 PM, Russell Pierce wrote:
> R-help list and interested parties,
>
> On Cross Validated mpiktas correctly noted that both the I() and rms
> Glm/Predict solution produce incorrect results
You proba
On Sun, Jan 30, 2011 at 5:59 PM, Russell Pierce wrote:
> R-help list and interested parties,
>
> On Cross Validated mpiktas correctly noted that both the I() and rms
> Glm/Predict solution produce incorrect results
You probably meant it this way anyways, but I would say it produces
undesired resu
R-help list and interested parties,
On Cross Validated mpiktas correctly noted that both the I() and rms
Glm/Predict solution produce incorrect results
(http://stats.stackexchange.com/questions/6684/how-can-one-use-the-predict-function-on-a-lm-object-where-the-ivs-have-been-dynam/6718#6718).
As f
On Sat, Jan 29, 2011 at 1:23 PM, Joshua Wiley wrote:
>> xxAB <- dat$xxA * dat$xxB
>> meanAB <- meanA * meanB
>> sdAB <- sqrt(sdA^2 + sdB^2 + meanA^2 * sdB^2 + meanB^2 * sdA^2)
don't know why I did that the hard way (besides I think I messed up
the formula slightly), but the point is there's still
On Sat, Jan 29, 2011 at 1:05 PM, David Winsemius wrote:
>
> On Jan 29, 2011, at 3:49 PM, Joshua Wiley wrote:
>
>> On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
>> wrote:
>>>
>>> Huh?. With the same model and data, they should be the same:
>>
>> I must be missing something, because that is wha
On Jan 29, 2011, at 3:49 PM, Joshua Wiley wrote:
On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
wrote:
Huh?. With the same model and data, they should be the same:
I must be missing something, because that is what I would have
expected too, but it is not what I get (at least when I run t
On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
wrote:
> Huh?. With the same model and data, they should be the same:
I must be missing something, because that is what I would have
expected too, but it is not what I get (at least when I run the code
as shown below).
>> lm.mod2 <- lm(out ~ scal
On Jan 29, 2011, at 2:36 PM, Joshua Wiley wrote:
On Sat, Jan 29, 2011 at 11:10 AM, David Winsemius
wrote:
On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using
David,
Thanks for your feedback. However, I think if you look at the mean of
your output (which, as a minor point, doesn't provide the same data to
predict as expand.grid(X1=c(-1,0,1),X2=c(-1,0,1)), you'll see it is
very different from the mean of dat$out. Like Ista's previous answer,
I think in
Another option is
set.seed(10)
dat <- data.frame(xxA = rnorm(20,10), xxB = rnorm(20,20))
dat$out <- with(dat,xxA+xxB+xxA*xxB+rnorm(20,20))
lm.mod <- lm(out ~ I(scale(xxA))*I(scale(xxB)), data=dat)
newdata <- data.frame(xxA=c(-1,0,1),xxB=c(-1,0,1))
preds <- predict(lm.mod, newdata)
Best,
Ista
On Sat, Jan 29, 2011 at 11:10 AM, David Winsemius
wrote:
>
> On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
>
>> Just in case someone else stumbles onto this thread and is facing a
>> similar issue: The quick solution for me turned out to be using Glm
>> and Predict in the rms package. Than
On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using Glm
and Predict in the rms package. Thanks go to Joshua and Ista for
helping me out with this issue. Double
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using Glm
and Predict in the rms package. Thanks go to Joshua and Ista for
helping me out with this issue. Double thanks go to Joshua for
suggesting I take a closer look
Dear Russell,
On Fri, Jan 28, 2011 at 5:59 PM, Russell Pierce wrote:
> Ista & r-help list,
>
> I guess I left out the most important part of any question, my reason
> for doing this. I am interested in the predicted y value at the mean,
> 1 SD above of the mean, and 1 SD below the mean for each
Ista & r-help list,
I guess I left out the most important part of any question, my reason
for doing this. I am interested in the predicted y value at the mean,
1 SD above of the mean, and 1 SD below the mean for each predictor.
Since I conducted my analysis with my IVs in the scale of Z (i.e. my
On Fri, Jan 28, 2011 at 6:26 PM, Russell Pierce wrote:
> Thanks for your response Ista,
> I'm looking at the results in newdata following your command. I agree that
> predict ran, but I don't think it did what I expect it to do. I may be
> mistaken, but shouldn't the mean of dat$out be close to t
Thanks for your response Ista,
I'm looking at the results in newdata following your command. I agree that
predict ran, but I don't think it did what I expect it to do. I may be
mistaken, but shouldn't the mean of dat$out be close to the mean of
newdata$Y? Shouldn't the values in newdata$Y (assum
Hi Russell,
There may be some subtleties that I'm not picking up on, but the
obvious problem is that the names of the predictors in newdata do not
match the names of the predictors in dat.
names(newdata) <- names(dat)[1:2]
newdata$Y <- predict(lm.obj,newdata)
does work on my machine.
Best,
Is
I want to predict values from an existing lm (linear model, e.g.
lm.obj) result in R using a new set of predictor variables (e.g.
newdata). However, it seems that because my linear models was made by
calling scale() on the target predictor that predict exits with an
error, "Error in scale(xxA, cen
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