Bert Gunter writes:
> However, if I understand correctly, using pls or anything else to try
> to fit (some combination of) 501 variables to 16 data points -- and
> then crossvalidate with 6 data points -- is utter nonsense. You just
> have a fancy random number generator!
That is incorrect. PLS
I think this wants a statistical discussion, which is OT here.
stats.stackexchange.com would be a better place to post for that.
However, if I understand correctly, using pls or anything else to try
to fit (some combination of) 501 variables to 16 data points -- and
then crossvalidate with 6 data
Hi,
I'm trying to fit PLSR model in R with 'pls' package with 22 samples (16
train, 6 test). I know that basic for considering of number of component is
cross-validation (in my case 'LOO') and then I should choose number of
component with minimum of RMSEP (or first minimum). But problem i
Hi All,
I am using the cppls function in the pls package, and I want to use cross
validation to determine the best number of components. Since Hastie et al
recommended a "one standard error rule", i.e., choose the most parsimonious
model whose error is no more than one standard error above
the err
Wolfgang Obermeier writes:
> how is it possible that the loadings of the second or even third component of
> a PLS-Analysis show higher values than the first component? Somebody got an
> idea??
The loadings of a PLS regression are simply the coefficients that are
multiplied with the X variables
Dear Subscribers,
how is it possible that the loadings of the second or even third
component of a PLS-Analysis show higher values than the first component?
Somebody got an idea??
Thanks in advance,
Wolfgang
--
Dipl. Geogr. Wolfgang Obermeier
Faculty of Geography
Philipps-University of Marbu
On 2011-06-27 03:02, 新鼎-智慧製造事業處-鄭紹文 wrote:
Dear sir,
If I have a vector as:
y<- c(1:4)
and a matrix as:
x<- matrix (5:12, nrow=4, ncol=2)
Then I create a data frame as:
t<- data.frame(y, x)
R will generate a data frame of 3 columns
names(t)
"y" "X1" "X2"
is.vector(t$X1) returns TRUE
is
hi,
because of my ignorance I can't figure out why would you need such a structure.
However the following trick works:
> a <- matrix(1:6,3)
> b <- data.frame(1:3)
> b$a <- a
> b
X1.3 a.1 a.2
11 1 4
22 2 5
33 3 6
> names(b)
[1] "X1.3" "a"
> class(b$a)
[1] "matrix"
> is.mat
Dear sir,
If I have a vector as:
>y <- c(1:4)
and a matrix as:
>x <- matrix (5:12, nrow=4, ncol=2)
Then I create a data frame as:
>t <- data.frame(y, x)
R will generate a data frame of 3 columns
>names(t)
> "y" "X1" "X2"
>is.vector(t$X1) returns TRUE
> is.vector(t$X2) returns TRUE
R splits the
Thank you so very much. Yes, a statistician friend expressed his
certainty that NAs cannot be handled by such algorithms, and you just
answered the R specific questions.
I will prune the data more and feed them into pls as matrices.
I am, in fact, following the examples your provide with the
Payam Minoofar writes:
> I have managed to format my data into a single datframe consisting of two
> AsIs response and predictor dataframes in order to supply the plsr command of
> the pls package for principal components analysis.
>
> When I execute the command, however, I get this error:
>> f
Hi,
I have managed to format my data into a single datframe consisting of two AsIs
response and predictor dataframes in order to supply the plsr command of the
pls package for principal components analysis.
When I execute the command, however, I get this error:
> fiber1 <- plsr(respmat ~ predma
Hi Damien,
How did you install the package? Usually this error pops up when people
simply download the zip file and then unzip into their library directory.
If you use the package installation functions in R, you shouldn't have
this problem:
install.packages("pls")
Best,
Jim
mienad wro
Hi,
I am using R 2.8.1 version on Windows with RGui. I have loaded pls package
lattest version (2.1-0). When I try to load this package in R using
library(pls) command, the following error message appear:
Erreur dans library(pls) :
'pls' n'est pas un package valide -- a-t-il été installé < 2
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