Displaying 20 results from an estimated 8000 matches similar to: "operating on arrays of unknown dimensionality"
2009 Jul 21
1
subscript into matrix discards row/column names
Dear R-help,
When the result of a matrix subscription degenerates to a scalar the names implied by the dimnames are discarded.
> x <- matrix(0, 1, 1, dimnames=list('a', 'x'))
## below I expected result to have names='x', it's not
> x[1,]
[1] 0
## below I expected result to have names='a', it's not
> x[,1]
[1] 0
This is probably a side effect
2004 Nov 24
2
LDA with previous PCA for dimensionality reduction
Dear all, not really a R question but:
If I want to check for the classification accuracy of a LDA with
previous PCA for dimensionality reduction by means of the LOOCV method:
Is it ok to do the PCA on the WHOLE dataset ONCE and then run the LDA
with the CV option set to TRUE (runs LOOCV)
-- OR--
do I need
- to compute for each 'test-bag' (the n-1 observations) a PCA
2010 Mar 01
0
Is package "dr" appropriate for reducing the dimensionality of molecules conformational space ?
I anticipate lacking of prior experience with dimensionality reduction problems.
Some scientists concerned with drug discovery performed several steered Molecular Dynamics simulations of the
alanine-dipeptide molecule dragged by a radial force from an equilibrium conformation to another different equilibrium conformation.
They sampled at regular intervals 7 dihedral angles, 5 bending angles, and
2018 Jul 07
2
Testing for vectors
On Sat, Jul 7, 2018 at 12:54 PM, Duncan Murdoch
<murdoch.duncan at gmail.com> wrote:
> On 07/07/2018 1:20 PM, Hadley Wickham wrote:
>>
>> Hi all,
>>
>> Is there are base function that I've missed that tests if an object is
>> a vector in the dimensionality sense, rather than the data structure
>> sense? i.e. something that checks is.null(dim(x)) ?
2005 May 08
3
Light-weight data.frame class: was: how to add method to .Primitive function
Hi,
Encouraged by a tip from Simon Urbanek I tried to use the S3 machinery
to write a faster version of the data.frame class.
This quickly hits a snag: the "[.default"(x, i) for some reason cares
about the dimensionality of x.
In the end there is a full transcript of my R session. It includes the
motivation for writing the class and the problems I have encountered.
As a result I see
2010 Feb 14
1
Dimensional reduction package
Is there any R package which implements non-linear dimensionality reduction (LLE, ISOMAP, GTM, and so on) and/or intrinsic dimensionality estimation ?
Thank you,
Maura
tutti i telefonini TIM!
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2008 Jul 23
2
shQuote and cat
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right?
I am not sure cat() correctly prints strings which are generated by triple-shQuote():
> shQuote(shQuote("a"))
[1] "\"\\\"a\\\"\""
> cat(shQuote(shQuote(shQuote("a"))), '\n')
2008 Jul 23
2
shQuote and cat
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right?
I am not sure cat() correctly prints strings which are generated by triple-shQuote():
> shQuote(shQuote("a"))
[1] "\"\\\"a\\\"\""
> cat(shQuote(shQuote(shQuote("a"))), '\n')
2010 Jun 26
3
converting result of substitute to 'ordidnary' expression
Dear R users,
As substitute() help page points out:
Substituting and quoting often causes confusion when the argument
is 'expression(...)'. The result is a call to the 'expression'
constructor function and needs to be evaluated with 'eval' to give
the actual expression object.
And indeed I am confused. Consider:
> dat <- data.frame(x=1:10,
2010 Jun 26
3
converting result of substitute to 'ordidnary' expression
Dear R users,
As substitute() help page points out:
Substituting and quoting often causes confusion when the argument
is 'expression(...)'. The result is a call to the 'expression'
constructor function and needs to be evaluated with 'eval' to give
the actual expression object.
And indeed I am confused. Consider:
> dat <- data.frame(x=1:10,
2010 Jun 26
2
subset arg in subset(). was: converting result of substitute to 'ordidnary' expression
Dear R users,
Please disregard my previous post "converting result of substitute to 'ordidnary' expression". The problem I have has nothing to do with substitute.
Consider:
> dat <- data.frame(x=1:10, y=1:10)
> subsetexp <- expression(5<x)
> ## this does work
> subset(dat, eval(subsetexp))
x y
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
> ##
2010 Feb 19
0
"tsne" package for t-SNE dimensionality reduction
I'd like to announce the availability of the "tsne" package.
It provides the T distributed Stocastic Neighbor Embedding algorithm by van
der Maaten and Hinton:
1. http://ict.ewi.tudelft.nl/~lvandermaaten/t-SNE.html
2.
L.J.P. van der Maaten and G.E. Hinton. Visualizing High-Dimensional Data
Using t-SNE. Journal of Machine Learning Research 9(Nov):2579-2605, 2008.
[
2010 Feb 19
0
"tsne" package for t-SNE dimensionality reduction
I'd like to announce the availability of the "tsne" package.
It provides the T distributed Stocastic Neighbor Embedding algorithm by van
der Maaten and Hinton:
1. http://ict.ewi.tudelft.nl/~lvandermaaten/t-SNE.html
2.
L.J.P. van der Maaten and G.E. Hinton. Visualizing High-Dimensional Data
Using t-SNE. Journal of Machine Learning Research 9(Nov):2579-2605, 2008.
[
2006 May 30
1
when dimensionality is larger than the number of observations?
Hi, there:
Can anyone here kindly point some good reference or links on this topic?
Esp. some solutions from BioConductor or R, when dealing with
microarray-like, "fat" data?
thanks,
--
Weiwei Shi, Ph.D
"Did you always know?"
"No, I did not. But I believed..."
---Matrix III
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2008 Oct 03
2
computing on expressions
Dear R-users,
Suppose I have an expression:
expr = expression(a>0)
and now I want to modify it to expression(a>0 & b>0). The following doesn't work:
expr = expression(expr & b>0)
What would be a good way of doing this?
Thanks,
Vadim
________________________________
Note: This email is for the confidential use of the named addressee(s) only and may contain
2010 Oct 15
1
calling browser on error
Dear R-developers,
I am trying to figure out a way to call browser() when an error occur, and naturally I want the browser() to be called in the environment of the error.
I tried something simple in vain:
> f <- function() { x <- 1; stop('ok') }
> tryCatch(f(), error=browser())
Called from: tryCatch(f(), error = browser())
## if browser() was called in the local environment
2008 Aug 15
1
stopifnot message mutation
Dear R-users,
Could someone please explain why the message printed by function stopifnot2, see below, is different from that of stopifnot itself?
Thank you for your help,
Vadim
> stopifnot2 <- function(...) stopifnot(...)
> stopifnot(F)
Error: F is not TRUE
> stopifnot2(F)
Error: ..1 is not TRUE
> version
_
platform i386-pc-mingw32
arch i386
os
2009 Mar 09
1
detecting NULL in recursive lists
Dear R-users,
How can I detect a NULL in a recursive list?
For a regular list I could use lapply:
> lapply(list(x=NULL), is.null)
$x
[1] TRUE
However that doesn't work for structures like list(list(x=NULL)). I tried rapply but it treats NULL as a list and discards them:
> rapply(list(a=1, b=list(x=NULL)), is.null)
a
FALSE
Any suggestion?
Thank you for your help,
Vadim
Note:
2008 Nov 28
1
names generated in list indexing
Dear R-devel,
When a character vector is used to subscript a list and when some of the subscripts are not present in the list names R returns NULL for those subscripts and generate NA names for each of them:
> list(b=1)[c('a','b')]
$<NA> <<-- generated name
NULL
$b
[1] 1
Wouldn't it be more intuitive to use the subscript name rather than to generate an NA?
2009 Jul 21
1
bug in approx crashes R
Dear R-devel,
The following line crashes R
> approx(1, 1, 0, method='const', rule=2, f=0, yleft=NULL, ties='ordered')$y
Process R:2 exited abnormally with code 5 at Tue Jul 21 14:18:09 2009
> version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 2
minor 9.1
year