Re: [R] Array analogue of row and col
slice.index() in base On 4/2/2013 6:36 AM, Enrico Bibbona wrote: Is there any function that extends to multidimentional arrays the functionalities of row and col which are just defined for matrices? Thanks, Enrico Bibbona [[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] Array analogue of row and col
slice.index() in base
On 4/2/2013 9:53 AM, Enrico Bibbona wrote:
Great!
Thanks a lot, Enrico
2013/4/2 Duncan Murdoch murdoch.dun...@gmail.com
On 02/04/2013 6:36 AM, Enrico Bibbona wrote:
Is there any function that extends to multidimentional arrays the
functionalities of row and col which are just defined for matrices?
Thanks, Enrico Bibbona
Not as far as I know, but there are a lot of functions in packages.
You could write your own something like this. I've skipped any error
checking; you'll want to add that.
indices - function(a, which) {
d - dim(a)
prod_before - prod(d[seq_len(which-1)])
result - rep(seq_len(d[which]), each=prod_before, length.out = prod(d))
dim(result) - d
result
}
Then indices(a, 1) gives an array of the same shape as a where entry i, j,
k, ... is i, indices(a, 2) has entry i,j,k, ... equal to j, etc.
Duncan Murdoch
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Re: [R] list of matrices -- array
abind() (from package 'abind') can take a list of arrays as its first argument, so in general, no need for do.call() with abind(). As another poster pointed out, simplify2array() can also be used; while abind() gives more options regarding which dimension is created and how dimension names are constructed. x - list(A=cbind(X=c(a=1,b=2,c=3,d=4),Y=5:8,Z=9:12), B=cbind(X=c(a=13,b=14,c=15,d=16),Y=17:20,Z=21:24)) $A X Y Z a 1 5 9 b 2 6 10 c 3 7 11 d 4 8 12 $B X Y Z a 13 17 21 b 14 18 22 c 15 19 23 d 16 20 24 dim(abind(x, along=3)) [1] 4 3 2 dim(abind(x, along=1.5)) [1] 4 2 3 dim(abind(x, along=0.5)) [1] 2 4 3 dim(abind(x, along=1, hier.names=T)) # construct rownames in a hierarchical manner A.a, A.b, etc [1] 8 3 dim(abind(x, along=2, hier.names=T)) # construct colnames in a hierarchical manner [1] 4 6 abind(x, along=2, hier.names=T) A.X A.Y A.Z B.X B.Y B.Z a 1 5 9 13 17 21 b 2 6 10 14 18 22 c 3 7 11 15 19 23 d 4 8 12 16 20 24 On 2/14/2013 3:53 AM, Rolf Turner wrote: require(abind) do.call(abind,c(my_list,list(along=0))) # Gives 2 x 4 x 5 do.call(abind,c(my_list,list(along=3))) # Gives 4 x 5 x 2 The latter seems more natural to me. cheers, Rolf Turner On 02/14/2013 07:03 PM, Murat Tasan wrote: i'm somehow embarrassed to even ask this, but is there any built-in method for doing this: my_list - list() my_list[[1]] - matrix(1:20, ncol = 5) my_list[[2]] - matrix(20:1, ncol = 5) now, knowing that these matrices are identical in dimension, i'd like to unfold the list to a 2x4x5 (or some other permutation of the dim sizes) array. i know i can initialize the array, then loop through my_list to fill the array, but somehow this seems inelegant. i also know i can vectorize the matrices and unlist the list, then build the array from that single vector, but this also seems inelegant (and an easy place to introduce errors/bugs). i can't seem to find any built-in that handles this already... but maybe i just haven't looked hard enough :-/ __ 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] Problem in installing and starting Rattle
On Mon, Mar 7, 2011 at 6:57 AM, nerice neil.r...@plymouth.ac.uk wrote: CHECK FOR CONFLICTS IN YOUR PATH !!! I had a related problem when trying to use library RGtk2 for the first time. My problem was that when loading the library R was looking for the file zlib1.dll but couldn't find the procedure to launch RGtk2. I was getting an Entry Point not found error from Rgui.exe. The reason was because I had another package in my PATH environment variable (C:\program files\Intel\WiFi\bin) which had a CONFLICTING version of the zlib1.dll - and it was looking in this file and not the zlib1.dll which came with GTK+. Removed this conflict from the PATH and all was ok. I also had a related problem trying to use libary RGtk2 under Windows XP. I kept getting the message unable to load shared object C:\R\site-library\RGtk2\libs\i386\RGtk2.dll Uninstalling and reinstalling various versions of GTK2, both through R and outside R, many times, did not help. With the help of Neil Rice's comment, I found that there was another version of zlib1.dll in a directory on my PATH. Removing that directory from the path (inside R) fixed the problem and RGtk2 runs fine now. Here's an R expression to look for other copies on zlib1.dll on the path: with(list(x=file.path(strsplit(Sys.getenv(PATH), ;)$PATH, zlib1.dll)), x[file.exists(x)]) [1] C:\\R\\GTK2-Runtime\\bin/zlib1.dll There are many ways to modify the PATH. To set PATH inside R: Sys.setenv(PATH=paste(grep(UNWANTEDPATH, strsplit(Sys.getenv(PATH), ;)[[1]], value=T, invert=T), collapse=;)) (substitute some pattern that matches the unwanted directory for UNWANTEDPATH, remembering to use four backslashes to match one). -- Tony Plate __ 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] mvbutils and trackObjs
trackObjs will not work together with mvbutils -- both use active bindings to store objects on disk, and I would expect that trying to use both together would cause all sort of nasty conflicts. I would think it would be possible to fold the creation/modification time recording from trackObjs into mvbutils, but it would probably be a significant amount of work. -- Tony Plate On 05/31/2010 09:56 PM, Day, Roger S wrote: Hello Colleagues, I've recently become a fan of Mark Bravington's mvbutils package for organizing analysis projects in a tree. Using cd(), Save(), fixr(), mlazy() etcetera solves nicely some of the nuisances that have worried or annoyed me and sometimes caused big problems over the years. Well thought out. Now one feature that would be fabulous would be automatic time-stamping of objects. The trackObjs package from provides this, among other services. The question is, can the two packages work together peacefully? One curiosity: they have opposite ideas of the word cache. In mvbutils, a cached object is one that is stored on disk, only retrieved into memory when needed. In trackObjs, a cached object is one that is stored in memory as well as on disk. More than a curiosity, also potentially a bad sign for compatibility between the two packages. Anyone have experience with the two together, and care to share it? Thanks! Roger Day University of Pittsburgh Departments of Biomedical Informatics and Biostatistics University of Pittsburgh Cancer Institute University of Pittsburgh Molecular Medicine Institute - Room 310, Suite 301 Cancer Pavilion (CNPAV) 5150 Centre Ave. Pittsburgh, PA 15232 e-mail: da...@pitt.edu cell phone 412-609-3918 assistant: Lucy Cafeo: (412) 623-2952 __ 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] Variable variables using R ... e.g., looping over data frames with a numeric separator
On 05/17/2010 03:51 PM, Monte Shaffer wrote:
for(i in 1:L-1)
{
dataStr = gsub(' ','',paste(fData.,i));
dataVar = eval(dataStr);
## GOAL is to grab data frame 'fData.1' and do stuff with it, then in next
loop grab data frame 'fData.2' and do stuff with it
}
As Dan Davison said, the more standard R way would be to put all your
data frames in a list, then iterate over the list.
However, if do want to have your data frames in separate variables, and
then get each data frame in a loop similar to the code fragment above,
try something like this:
for (i in 1:(L-1)) {
dataName - paste(fData., i, sep=)
df - get(dataName)
... do something with data frame df ...
}
You can also give additional arguments to get() to tell it where to look
(pos=,envir=), and whether to look in parent environments
(inherits=TRUE/FALSE).
-- Tony Plate
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Re: [R] 2D array of strings
matrix(str, ncol=1) Francesco Napolitano wrote: Sorry for the dumb question, but I couldn't figure this out myself. Consider the following: str - c(abc,def) array(str, c(2,1)) [,1] [1,] abc [2,] def How can i obtain the outcome of the second instruction without specifying the number of rows? Thank you in advance, Francesco. __ 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] How to change the default Date format for write.csvfunction?
You can use a numeric value for the quote= argument to write.table to specify
which columns should have quotes.
d - data.frame(ticker=c(IBM, IBM), date = as.Date(c(2009-12-03,
2009-12-04)), price=c(120.00, 123.00))
d1 - as.data.frame(lapply(d, function(x) if (is(x, Date)) format(x,
%m/%d/%Y) else x))
write.table(d1, quote=which(sapply(d, function(x) !is.numeric(x) !is(x,
Date
ticker date price
1 IBM 12/03/2009 120
2 IBM 12/04/2009 123
-- Tony Plate
William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of
george@bnymellon.com
Sent: Monday, December 28, 2009 8:18 AM
To: r-help@r-project.org
Subject: Re: [R] How to change the default Date format for
write.csvfunction?
Hi,
This problem might be a little harder than it appears.
I receive a few emails all suggesting that convert the Date field to
character by calling format(date, %m/%d/%Y) in one way or
another. Well,
this is not the solution I'm looking for and it doesn't work
for me. All
the date fields are generated with quotes around them, which will be
treated by other software as string instead of date. Please
note, the
write.csv() function doesn't put quotes around date. All I
need is to
change the format behavior of Date without adding any quotes. So the
output of CSV I'm looking for should be:
ticker,date,price
IBM,12/03/2009,120
IBM,12/04/2009,123
Not this:
ticker,date,price
IBM,12/03/2009,120
IBM,12/04/2009,123
Write a function that adds double quotes
for the columns with classes that you want
quoted and call write.csv with quote=FALSE.
E.g., the following function f puts double
quotes around character and factor columns:
f - function (dataframe)
{
doubleQuoteNoFancy - function(x) paste(\, x, \, sep = )
for (i in seq_along(dataframe)) {
if (is(dataframe[[i]], character))
dataframe[[i]] - doubleQuoteNoFancy(dataframe[[i]])
else if (is(dataframe[[i]], factor))
levels(dataframe[[i]]) -
doubleQuoteNoFancy(levels(dataframe[[i]]))
else if (is(dataframe[[i]], Date))
dataframe[[i]] - format(dataframe[[i]], %m/%d/%Y)
}
colnames(dataframe) - doubleQuoteNoFancy(colnames(dataframe))
dataframe
}
Use it as:
write.csv(f(d), file=stdout(), quote=FALSE, row.names=FALSE)
ticker,date,price
IBM,12/03/2009,120
IBM,12/04/2009,123
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
Thanks for trying though.
George
From:
george@bnymellon.com
To:
r-help@r-project.org
Date:
12/28/2009 10:20 AM
Subject:
[R] How to change the default Date format for write.csv function?
Sent by:
r-help-boun...@r-project.org
Hi,
I have a data.frame containing a Date column. When using write.csv()
function to generate a CSV file, I always get the Date column
formatted as
-MM-DD. I would like to have it formatted as
MM/DD/, but
could not find an easy way to do it.Here is the test code:
d - data.frame(ticker=c(IBM, IBM), date =
as.Date(c(2009-12-03,
2009-12-04)), price=c(120.00, 123.00))
write.csv(d, file=C:/temp/test.csv, row.names=FALSE)
The test.csv generated looks like this:
ticker,date,price
IBM,2009-12-03,120
IBM,2009-12-04,123
I would like to have the date fields in the CSV formatted as
MM/DD/.
Is there any easy way to do this?
Thanks in advance.
George Zou
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Re: [R] [Rd] Error in namespaceExport(ns, exports) :
Try looking in the NAMESPACE file (in the same directory as the DESCRIPTION file). -- Tony Plate David Scherrer wrote: Dear all, I get the error Error in namespaceExport(ns, exports) : undefined exports function1 , function2 when compiling or even when I roxygen my package. The two function I once had in my package but I deleted them including their .Rd files. I also can't find them in any other function or help file. So does anybody know where these functions are still listed that causes this error? Many thanks, David [[alternative HTML version deleted]] __ r-de...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel __ 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] Help with complicated regular expression
One of these should be a start. If there can be no extra text at the beginning or end, start with ^ and end with $. x - c(WORD ( 123), WORD(1 ), WORD\t ( 21\t), WORD \t ( 1 \t ), decoy((2)), more words in front(2)) grep([[:alpha:]]+[ \t]*\\([ \t]*[0-9]+[ \t]*\\), x) [1] 1 2 3 4 6 grep(^[[:alpha:]]+[ \t]*\\([ \t]*[0-9]+[ \t]*\\), x) [1] 1 2 3 4 -- Tony Plate Dennis Fisher wrote: Colleagues, I am using R (2.9.2, all platforms) to search for a complicated text string using regular expressions. I would appreciate any help you can provide. The string consists of the following elements: SOMEWORDWITHNOSPACES any number of spaces and/or tabs ( any number of spaces and/or tabs integer any number of spaces and/or tabs ) Examples include: WORD ( 123) WORD(1 ) WORD\t ( 21\t) WORD \t ( 1 \t ) etc. I don't need to substitute anything, only to identify if such a string exists. Any help with regular expressions would be appreciated. Thanks. Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com __ 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] Loadings and scores from fastICA?
ICA and PCA both model the data as a product of two matrices (usually
called something like components or loadings weights or scores). It's
how those matrices are constructed that differs. PCA is often a first
step in doing ICA. I'd suggest reading the ICA tutorial by *Aapo
Hyvärinen and Erkki Oja *
(http://www.cis.hut.fi/aapo/papers/IJCNN99_tutorialweb/) -- it's an
excellent introduction.
-- Tony Plate
Joel Fürstenberg-Hägg wrote:
Ok, so then the S gives the individual components, good. Thanks Tony!
But what about the principal components from the PCA plot, how are
they calculated?
And are the linear mixing matrix A really the same as the
loadings/weights? There must be different loadings for the PCA and ICA
right?
Best regards,
Joel
Date: Wed, 11 Nov 2009 14:29:06 -0700
From: tpl...@acm.org
To: joel_furstenberg_h...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R] Loadings and scores from fastICA?
The help for fastICA says:
The data matrix X is considered to be a linear combination of
non-Gaussian (independent) components i.e. X = SA where columns of
S contain the independent components and A is a linear mixing
matrix.
The value of fastICA is a list with components S (the estimated
source matrix) and A (the estimated mixing matrix). Are these what
you want?
-- Tony Plate
Joel Fürstenberg-Hägg wrote:
Hi all,
Does anyone know how to get the independent components and
loadings from an Independent Component Analysis (ICA), as well as
principal components and loadings from a Pricipal Component analysis
(PCA) using the fastICA package? Or perhaps if there's another way to
do ICAs in R?
Below is an example from the fastICA manual
(http://cran.r-project.org/web/packages/fastICA/fastICA.pdf)
if(require(MASS))
{
x - mvrnorm(n = 1000, mu = c(0, 0), Sigma = matrix(c(10, 3, 3,
1), 2, 2))
x1 - mvrnorm(n = 1000, mu = c(-1, 2), Sigma = matrix(c(10, 3, 3,
1), 2, 2))
X - rbind(x, x1)
a - fastICA(X, 2, alg.typ = deflation, fun = logcosh, alpha =
1, method = R, row.norm = FALSE, maxit = 200, tol = 0.0001, verbose
= TRUE)
par(mfrow = c(1, 3))
plot(a$X, main = Pre-processed data)
plot(a$X%*%a$K, main = PCA components)
plot(a$S, main = ICA components)
}
Best regards,
Joel
_
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Re: [R] Comparison of vectors in a matrix
This is a tricky data entry problem. The right technique will depend on the fine details of the
data, and it's not clear what those are. E.g., when you say In my first column, for example,
I have henry , it's unclear to me whether or not the double quotes are part of
the data or not - which is why it's nice to provide reproducible examples.
But, if you do have quoted strings in your data fields as they exist in an R
matrix, you can do something like the following:
# each element of the matrix x contains one or more quoted strings, separated
by commas
x - matrix(c('a, b', 'c', 'b', 'd'), ncol=2, dimnames=list(c(row1, row2),
c(X,Y)))
x
X Y
row1 \a\, \b\ \b\
row2 \c\\d\
# use R's parsing and evaluation to turn 'a, b' into c(a, b), and turn
that
# into a matrix containing character vectors of various lengths.
matrix(lapply(parse(text=paste(c(, x, ))), eval), ncol=ncol(x),
dimnames=dimnames(x))
X Y
row1 Character,2 b
row2 c d
- Tony Plate
esterhazy wrote:
Yes, thanks for this, this is exactly what I want to do.
However, I have a remaining problem which is how to get R to understand that
each entry in my matrix is a vector of names.
I have been trying to import my text file with the names in each vector of
names enclosed in quotes and separated by commas, or separated by spaces, or
without quotes, etc, with no luck.
Everytime, R seems to consider the vector of names as just one long name.
In my first colum, for example, I have henry, in the second, mary,
ruth, and in the third mary, joseph, and I have no idea how to get R
to see that mary, ruth, for example, is composed of two strings of text,
rather than just one.
Thanks for any further help!
http://old.nabble.com/file/p26305756/ffoexample.txt ffoexample.txt
Tony Plate wrote:
Nice problem!
If I understand you correctly, here's how to do it (with list-based
matrices):
set.seed(1)
(x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10],
size=k)), ncol=2, dimnames=list(1:5,c(A,B
A B
1 Character,2 Character,5
2 Character,2 Character,5
3 Character,3 Character,3
4 Character,5 Character,3
5 Character,2 i
x[1,1]
[[1]]
[1] c b
x[1,2]
[[1]]
[1] c d a j f
(y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]],
ab[[2]]
A B A-B
1 Character,2 Character,5 b
2 Character,2 Character,5 g
3 Character,3 Character,3 Character,3
4 Character,5 Character,3 Character,2
5 Character,2 i Character,2
y[1,3]
[[1]]
[1] b
-- Tony Plate
esterhazy wrote:
Hi,
I have a matrix with two columns, and the elements of the matrix are
vectors.
So for example, in line 3 of column 1 I have a vector v31=(marc,
robert,
marie).
What I need to do is to compare all vectors in column 1 and 2, so as to
get,
for example setdiff(v31,v32) into a new column.
Is there a way to do this in R?
Thanks!
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Re: [R] partial cumsum
William Dunlap wrote:
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 smu
Sent: Wednesday, November 11, 2009 7:58 AM
To: r-help@r-project.org
Subject: [R] partial cumsum
Hello,
I am searching for a function to calculate partial cumsums.
For example it should calculate the cumulative sums until a
NA appears,
and restart the cumsum calculation after the NA.
this:
x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10)
should become this:
1 3 6 NA 5 11 18 26 35 45
Perhaps
ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum)
[1] 1 3 6 NA 5 11 18 26 35 45
Nice simple function! Here's a different approach I use that's faster for long
vectors with many NA values. Note however, that this approach can suffer from
catastrophic round-off error because it does a cumsum over the whole vector
(after replacing NA's with zeros) and then subtracting out the cumsum at the
most recent NA values. Most of the body of this function is devoted to
allowing (an unreasonable degree of) flexibility in specification of where to
reset.
cumsum.reset - function(x, reset.at=which(is.na(x)), na.rm=F) {
# compute the cumsum of x, resetting the cumsum to 0 at each element indexed
by reset.at
if (is.logical(reset.at)) {
if (length(reset.at)length(x)) {
if ((length(reset.at) %% length(x))!=0)
stop(length of reset.at must be a multiple of length of x)
x - rep(x, len=length(reset.at))
} else if (length(reset.at)length(x)) {
if ((length(x) %% length(reset.at))!=0)
stop(length of x must be a multiple of length of reset.at)
reset.at - rep(reset.at, len=length(x))
}
reset.at - which(reset.at)
} else if (!is.numeric(reset.at)) {
stop(reset.at must be logical or numeric)
}
if (length(i - which(reset.at=1)))
reset.at - reset.at[-i]
if (any(i - is.na(x[reset.at])))
x[reset.at[i]] - 0
if (na.rm any(i - which(is.na(x
x[i] - 0
y - cumsum(x)
d - diff(c(0, y[reset.at-1]))
r - rep(0, length(x))
r[reset.at] - d
return(y - cumsum(r))
}
x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10)
cumsum.reset(x)
[1] 1 3 6 0 5 11 18 26 35 45
ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum)
[1] 1 3 6 NA 5 11 18 26 35 45
The speedup from not breaking the input vector into smaller vectors is (to me)
surprisingly small -- only a factor of 3:
x - replace(rnorm(1e6), sample(1e6, 1), NA)
all.equal(replace(ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum), is.na(x), 0),
cumsum.reset(x))
[1] TRUE
system.time(cumsum.reset(x))
user system elapsed
0.310.030.35
system.time(ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum))
user system elapsed
0.990.051.15
So, I'd go with the ave() approach unless this type of cumsum is the core of a
long computationally intensive job. And if that's the case, it would make
sense to code it in C.
-- Tony Plate
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
any ideas?
thank you and best regards,
stefan
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Re: [R] Loadings and scores from fastICA?
The help for fastICA says:
The data matrix X is considered to be a linear combination of
non-Gaussian (independent) components i.e. X = SA where columns of
S contain the independent components and A is a linear mixing
matrix.
The value of fastICA is a list with components S (the estimated source matrix) and
A (the estimated mixing matrix). Are these what you want?
-- Tony Plate
Joel Fürstenberg-Hägg wrote:
Hi all,
Does anyone know how to get the independent components and loadings from an
Independent Component Analysis (ICA), as well as principal components and
loadings from a Pricipal Component analysis (PCA) using the fastICA package? Or
perhaps if there's another way to do ICAs in R?
Below is an example from the fastICA manual
(http://cran.r-project.org/web/packages/fastICA/fastICA.pdf)
if(require(MASS))
{
x - mvrnorm(n = 1000, mu = c(0, 0), Sigma = matrix(c(10, 3, 3, 1), 2, 2))
x1 - mvrnorm(n = 1000, mu = c(-1, 2), Sigma = matrix(c(10, 3, 3, 1), 2,
2))
X - rbind(x, x1)
a - fastICA(X, 2, alg.typ = deflation, fun = logcosh, alpha = 1, method =
R, row.norm = FALSE, maxit = 200, tol = 0.0001, verbose = TRUE)
par(mfrow = c(1, 3))
plot(a$X, main = Pre-processed data)
plot(a$X%*%a$K, main = PCA components)
plot(a$S, main = ICA components)
}
Best regards,
Joel
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Re: [R] Comparison of vectors in a matrix
Nice problem! If I understand you correctly, here's how to do it (with list-based matrices): set.seed(1) (x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10], size=k)), ncol=2, dimnames=list(1:5,c(A,B A B 1 Character,2 Character,5 2 Character,2 Character,5 3 Character,3 Character,3 4 Character,5 Character,3 5 Character,2 i x[1,1] [[1]] [1] c b x[1,2] [[1]] [1] c d a j f (y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]], ab[[2]] A B A-B 1 Character,2 Character,5 b 2 Character,2 Character,5 g 3 Character,3 Character,3 Character,3 4 Character,5 Character,3 Character,2 5 Character,2 i Character,2 y[1,3] [[1]] [1] b -- Tony Plate esterhazy wrote: Hi, I have a matrix with two columns, and the elements of the matrix are vectors. So for example, in line 3 of column 1 I have a vector v31=(marc, robert, marie). What I need to do is to compare all vectors in column 1 and 2, so as to get, for example setdiff(v31,v32) into a new column. Is there a way to do this in R? Thanks! __ 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] prcomp - principal components in R
The output of summary prcomp displays the cumulative amount of variance explained relative to the total variance explained by the principal components PRESENT in the object. So, it is always guaranteed to be at 100% for the last principal component present. You can see this from the code in summary.prcomp() (see this code with getAnywhere(summary.prcomp)). Here's how to get the output you want (the last line in the transcript below): set.seed(1) summary(pc1 - prcomp(x)) Importance of components: PC1 PC2 PC3 PC4 PC5 Standard deviation 1.175 1.058 0.976 0.916 0.850 Proportion of Variance 0.275 0.223 0.190 0.167 0.144 Cumulative Proportion 0.275 0.498 0.688 0.856 1.000 summary(pc2 - prcomp(x, tol=0.8)) Importance of components: PC1 PC2 PC3 Standard deviation 1.17 1.058 0.976 Proportion of Variance 0.40 0.324 0.276 Cumulative Proportion 0.40 0.724 1.000 pc2$sdev [1] 1.1749061 1.0581362 0.9759016 pc1$sdev [1] 1.1749061 1.0581362 0.9759016 0.9164905 0.8503122 svd(scale(x, center=T, scale=F))$d / sqrt(nrow(x)-1) [1] 1.1749061 1.0581362 0.9759016 0.9164905 0.8503122 cumsum(pc1$sdev^2) / sum((svd(scale(x, center=T, scale=F))$d / sqrt(nrow(x)-1))^2) [1] 0.2752317 0.4984734 0.6883643 0.8558386 1.000 # output in terms of the cumulative % of the total variance cumsum(pc2$sdev^2) / sum((svd(scale(x, center=T, scale=F))$d / sqrt(nrow(x)-1))^2) [1] 0.2752317 0.4984734 0.6883643 It's probably better to get prcomp to compute all the components in the first place, because the SVD is the bulk of the computation anyway (so doing it again will be slower for large matrices.) Then just look at the most important principal components. However, there may be a shortcut for computing the values of D in the SVD of a matrix -- you could look for that if you have demanding computations (e.g., the sqrts of the eigen values of the covariance matrix of scaled x: sqrt(eigen(var(scale(x, center=T, scale=F)), only.values=T)$values)). -- Tony Plate zubin wrote: Hello, not understanding the output of prcomp, I reduce the number of components and the output continues to show cumulative 100% of the variance explained, which can't be the case dropping from 8 components to 3. How do i get the output in terms of the cumulative % of the total variance, so when i go from total solution of 8 (8 variables in the data set), to a reduced number of components, i can evaluate % of variance explained, or am I missing something?? 8 variables in the data set princ = prcomp(df[,-1],rotate=varimax,scale=TRUE) summary(princ) Importance of components: PC1 PC2 PC3 PC4 PC5 PC6PC7PC8 Standard deviation 1.381 1.247 1.211 0.994 0.927 0.764 0.6708 0.4366 Proportion of Variance 0.238 0.194 0.183 0.124 0.107 0.073 0.0562 0.0238 Cumulative Proportion 0.238 0.433 0.616 0.740 0.847 0.920 0.9762 *1.* princ = prcomp(df[,-1],rotate=varimax,scale=TRUE,tol=.75) summary(princ) Importance of components: PC1 PC2 PC3 Standard deviation 1.381 1.247 1.211 Proportion of Variance 0.387 0.316 0.297 Cumulative Proportion 0.387 0.703 *1.000* [[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] rm(list-ls()) error
- and = are not universally interchangable. args(rm) function (..., list = character(0L), pos = -1, envir = as.environment(pos), inherits = FALSE) The call rm(list - ls()) assigns the result of ls() to the variable 'list' and passes that value as an anonymous argument to rm() (Probably more than you want to know: or it would if rm() didn't have non-standard evaluation rules -- as it happens, list - ls() is recognized as an invalid argument before it is evaluated.) The call rm(list=ls()) calls rm() with the 'list' argument having the value of ls() Here's an example that doesn't confuse things by having non-standard evaluation rules: f - function(a=1, b=2) cat(a=, a, b=, b, \n) b Error: object 'b' not found f(b - 33) a= 33 b= 2 b [1] 33 f(b=33) a= 1 b= 33 -- Tony Plate Feng Li wrote: Dear R, Why rm(list-ls()) gives an error but rm(list=ls()) not? I remember the operator ‘-’ can be used anywhere... Thanks! Feng __ 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] Create Artificial Binary Matrix based on probability
x - matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30) table(x) x 01 1131 69 x - matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30) table(x) x 01 1151 49 bikemike42 wrote: Dear All, I am trying to create an artificial binary matrix such that each cell has a probability of 0.048 of having a 1. So far the closest I've come is us by using a random poisson distribution with a mean of 0.048, but I can't figure out how to limit the max value to 1. Otherwise that would work fine it seems. Any suggestions? The main code I've got to create said matrix so far is: a-replicate(26,rpois(57,0.048)) Thanks in Advance, Mike __ 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] column names of a correlation matrix
Here's a simple example that might help you get what you want:
set.seed(1)
x - matrix(rnorm(30), ncol=3, dimnames=list(NULL, letters[1:3]))
(xc - cor(x))
a b c
a 1.000 -0.3767034 -0.7158385
b -0.3767034 1.000 0.6040273
c -0.7158385 0.6040273 1.000
(cd - data.frame(elt=outer(colnames(xc), colnames(xc), paste,
sep=:)[upper.tri(xc)], row=row(xc)[upper.tri(xc)], col=col(xc)[upper.tri(xc)],
cor=xc[upper.tri(xc)]))
elt row colcor
1 a:b 1 2 -0.3767034
2 a:c 1 3 -0.7158385
3 b:c 2 3 0.6040273
cd[order(-cd$cor),]
elt row colcor
3 b:c 2 3 0.6040273
1 a:b 1 2 -0.3767034
2 a:c 1 3 -0.7158385
If you need something more efficient, try using which(..., arr.ind) to pick out
matrix style indices, e.g.:
(ii - which(xc -0.4 upper.tri(xc), arr.ind=T))
row col
a 1 2
b 2 3
cbind(ii, cor=xc[ii])
row colcor
a 1 2 -0.3767034
b 2 3 0.6040273
-- Tony Plate
Lee William wrote:
Hi! All,
I am working on a correlation matrix of 4217x4217 named 'cor_expN'. I wish
to obtain pairs with highest correlation values. So, I did this
b=matrix(data=NA,nrow=4217,ncol=1)
rownames(b)=rownames(cor_expN)
for(i in 1:4217){b[i,]=max(cor_expN[i,])}
head(b)
[,1]
aaeA_b3241_14 0.7181912
aaeB_b3240_15 0.7513084
aaeR_b3243_15 0.7681684
aaeX_b3242_12 0.5230587
aas_b2836_14 0.6615927
aat_b0885_140.6344144
Now I want the corresponding columns for the above values. For that I tried
this
c=matrix(data=NA,nrow=4217,ncol=1)
for(i in 1:4217){b[i,]=colnames(max(cor_expN[i,]))}
And got the following error:
Error in b[i, ] = colnames(max(cor_expN[i, ])) : number of items to replace
is not a multiple of replacement length
Any thoughts?
Lee
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Re: [R] LDA Precdict - Seems to be predicting on the Training Data
Maybe you're getting strange results because you're not supplying a data object to lda() when you build your fit. When I do it the standard way, predict.lda() uses the new data and produces a result of length 6 as expected: myDat - read.csv(clipboard, sep=\t) fit - lda(c1 ~ v1 + v2 + v3, data=myDat[1:10,]) predict(fit, myDat[11:16,]) $class [1] c c c b c a Levels: a b c ... -- Tony Plate BostonR wrote: When I import a simple dataset, run LDA, and then try to use the model to forecast out of sample data, I get a forecast for the training set not the out of sample set. Others have posted this question, but I do not see the answers to their posts. Here is some sample data: DateNames v1 v2 v3 c1 1/31/2009 Name1 0.714472361 0.902552278 0.783353694 a 1/31/2009 Name2 0.512158919 0.770451596 0.111853346 a 1/31/2009 Name3 0.470693282 0.129200065 0.800973877 a 1/31/2009 Name4 0.24236898 0.472219638 0.486599763 b 1/31/2009 Name5 0.785619735 0.628511593 0.106868172 b 1/31/2009 Name6 0.718718387 0.697257275 0.690326648 b 1/31/2009 Name7 0.327331186 0.01715109 0.861421706 c 1/31/2009 Name8 0.632011743 0.599040196 0.320741634 c 1/31/2009 Name9 0.302804404 0.475166304 0.907143632 c 1/31/2009 Name10 0.545284813 0.967196462 0.945163717 a 1/31/2009 Name11 0.563720418 0.024862018 0.970685281 a 1/31/2009 Name12 0.357614427 0.417490445 0.415162276 a 1/31/2009 Name13 0.154971203 0.425227967 0.856866993 b 1/31/2009 Name14 0.935080173 0.488659307 0.194967973 a 1/31/2009 Name15 0.363069339 0.334206603 0.639795596 b 1/31/2009 Name16 0.862889297 0.821752532 0.549552875 a Attached is the code: myDat -read.csv(file=f:\\Systematiq\\data\\TestData.csv, header=TRUE,sep=,) myData - data.frame(myDat) length(myDat[,1]) train - myDat[1:10,] outOfSample - myDat[11:16,] outOfSample - (cbind(outOfSample$v1,outOfSample$v2,outOfSample$v3)) outOfSample -data.frame(outOfSample) length(train[,1]) length(outOfSample[,1]) fit - lda(train$c1~train$v1+train$v2+train$v3) forecast - predict(fit,outOfSample)$class length(forecast)# I am expecting this to be same as lengthoutOfSample[,1]), which is 6 Output: length(forecast)# I am expecting this to be same as lengthoutOfSample[,1]), which is 6 [1] 10 __ 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] rbind to array members
Unfortunately, I can't read your examples. Can you repost without the formatting characters which are confusing when rendered in plain text? One thing to keep in mind is that a R array is a regular object -- e.g., if you have a 2 x 3 x 4 array, then if you want to add a row to a slice, you must add a row to all the slices. When I read to add a row in place to a single table of the 3 dimensional array it sounds like you might be trying to do something that's not possible with R arrays. However, if I could see your examples, then I probably give more help. -- Tony Plate Another Oneforyou wrote: 4adbca02.8020...@temple.edu Content-Type: text/plain; charset=Windows-1252 Content-Transfer-Encoding: quoted-printable MIME-Version: 1.0 =20 library(abind) ## array binding I've looked into using abind()=2C but it seems I might not understand it pr= operly. I can build my 2 table array=2C and insert a row into each table using: =A0=A0=A0=A0=A0=A0=A0 x - array(0=2Cc(1=2C3=2C2)) =A0=A0=A0=A0=A0=A0=A0 x[=2C=2C1] - c(1=2C2=2C3) =A0=A0=A0=A0=A0=A0=A0 x[=2C=2C2] - c(7=2C8=2C9) And I can use rbind() or abind() to add a row to the first table and assign= it to a separate object. =A0=A0=A0=A0=A0=A0=A0 y - rbind(x[=2C=2C1]=2C c(4=2C5=2C6)) =A0=A0=A0=A0=A0=A0=A0 z - abind(x[=2C=2C1]=2C c(4=2C5=2C6)=2Calong=3D0) But I can't determine how to add a row in place to a single table of the 3 = dimensional array. For example=2C this does not work: =A0=A0=A0=A0=A0=A0=A0 x[=2C=2C1] - abind(x[=2C=2C1]=2C c(4=2C5=2C6)=2Calon= g=3D0) Error in x[=2C =2C 1] - abind(x[=2C =2C 1]=2C c(4=2C 5=2C 6)=2C along =3D = 0) :=20 =A0 number of items to replace is not a multiple of replacement length I would like 'x' to be: x =2C =2C 1 =A0=A0=A0=A0 [=2C1] [=2C2] [=2C3] [1=2C]=A0=A0=A0 1=A0=A0=A0 2=A0=A0=A0 3 [2=2C]=A0=A0=A0 4=A0=A0=A0 5=A0=A0=A0 6 =2C =2C 2 =A0=A0=A0=A0 [=2C1] [=2C2] [=2C3] [1=2C]=A0=A0=A0 7=A0=A0=A0 8=A0=A0=A0 9 Thanks for any help. =20 _ Hotmail: Trusted email with Microsoft=92s powerful SPAM protection. __ 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] populating an array
R doesn't access arrays like C, use [i,j] to access a 2-d array, e.g.:
my_array - array(0,dim=c(2,2))
for(i in seq(1,2,by=1)){
+ for(j in seq(1,2,by=1)){
+ my_array[i,j] = i+j
+ }
+ }
my_array
[,1] [,2]
[1,]23
[2,]34
tdm wrote:
Hi,
Can someone please give me a pointer as to how I can set values of an array?
Why does the code below not work?
my_array - array(dim=c(2,2))
my_array[][] = 0
my_array
[,1] [,2]
[1,]00
[2,]00
for(i in seq(1,2,by=1)){
for(j in seq(1,2,by=1)){
my_array[i][j] = 5
}
}
Warning messages:
1: In my_array[i][j] = 5 :
number of items to replace is not a multiple of replacement length
2: In my_array[i][j] = 5 :
number of items to replace is not a multiple of replacement length
my_array
[,1] [,2]
[1,]50
[2,]50
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Re: [R] help with the use of mtext to create main title over multiple plots
Try playing around with the oma setting in par() -- it sets the outer
margins, which by default are zero.
The following shows the mtext label for me, using the windows device:
par(mfrow=c(2,2))
par(oma)
[1] 0 0 0 0
par(oma=c(0,0,2,0))
for (i in 1:4) plot(0:1,0:1)
mtext(text = my test plots, side = 3, outer = TRUE)
Mark Kimpel wrote:
I'm trying to use mtext to create a main title over multiple plots. Below is
a simple self-contained example and my sessionInfo (I should note I've also
tried this with R-2.8.1 with the same results). When I execute the code
chunk below, I get the plots, but no title. I've tried this using the screen
driver, pdf, and postscript. I've used different sizes of paper. I suspect I
am making an elementary error but searching the help files and help archives
hasn't provided me an answer.
Thanks for any help, Mark
#
setwd(~/Desktop)
pdf(my.test.plots.pdf, paper = letter)
par(mfrow=c(2,2))
for (i in 1:4){
plot(1:6, 1:6)
}
mtext(text = my test plots, side = 3, outer = TRUE)
dev.off()
#
R version 2.10.0 Under development (unstable) (2009-09-21 r49771)
x86_64-unknown-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] car_1.2-15
loaded via a namespace (and not attached):
[1] tools_2.10.0
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, Mobile VoiceMail
(317) 399-1219 Skype No Voicemail please
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Re: [R] Re use objects from within a custom made function
test$x doesn't evaluate the function, you want something like test(1,2)$x, e.g.:
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(list(x=x,y=y,z=z))
}
test(1,2)$x
[1] 1 2
test(1,2)$y
[1] 2
test(1,2)$z
[1] 0.5
Or if you want to avoid evaluating your function multiple times:
res - test(1,2)
res$x
[1] 1 2
res$y
[1] 2
res$z
[1] 0.5
res
$x
[1] 1 2
$y
[1] 2
$z
[1] 0.5
Stropharia wrote:
Hi everyone,
i'm having a problem extracting objects out of functions i've created, so i
can use them for further analysis. Here's a small example:
# ---
test - function(i, j){
x - i:j
y - i*j
z - i/j
return(x,y,z)
}
# ---
This returns the 3 objects as $x, $y and $z. I cannot, however, access these
objects individually by typing for example:
test$x
I know i can do this by adding an extra arrow head to the assignment arrow
(-), but I am sure that is not how it is done in some of the established R
functions (like when calling lm$coef out of the lm function). Is there a
simple command i've omitted from the function that allows access to objects
inside it?
Thanks in advance,
Steve
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Re: [R] how to have 'match' ignore no-matches
x - data.frame(d=letters[1:3], e=letters[3:5])
lookuptable - c(a=aa, c=cc, e=ee)
match.or.keep - function(x, lookuptable) {if (is.factor(x)) x - as.character(x);
m - match(x, names(lookuptable)); ifelse(is.na(m), x, lookuptable[m])}
# to return a matrix
apply(x, 2, match.or.keep, lookuptable=lookuptable)
de
[1,] aa cc
[2,] b d
[3,] cc ee
# to return a data frame
as.data.frame(lapply(x, match.or.keep, lookuptable=lookuptable))
d e
1 aa cc
2 b d
3 cc ee
Jill Hollenbach wrote:
Let me clarify:
I'm using this--
dfnew- sapply(df, function(df) lookuptable[match(df, lookuptable [ ,1]),
2])
lookup
0101 01:01
0201 02:01
0301 03:01
0401 04:01
df
0101 0301
0201 0401
0101 0502
dfnew
01:01 03:01
02:01 04:01
01:01 NA
but what I want is:
dfnew2
01:01 03:01
02:01 04:01
01:01 0502
thanks again,
Jill
Jill Hollenbach wrote:
Hi all,
I think this is a very basic question, but I'm new to this so please bear
with me.
I'm using match to translate elements of a data frame using a lookup
table. If the content of a particular cell is not found in the lookup
table, the function returns NA. I'm wondering how I can just ignore those
cells, and return the original contents if no match is found in the lookup
table.
Many thanks in advance, this site has been extremely helpful for me so
far,
Jill
Jill Hollenbach, PhD, MPH
Assistant Staff Scientist
Center for Genetics
Children's Hospital Oakland Research Institute
jhollenb...@chori.org
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Re: [R] grep or other complex string matching approach to capture necessary information...
You could use grep, but it's probably easier to use %in% (see also is.element()), e.g.: house_info[ house_info[,1] %in% c(Water damage, water pipes damaged, leaking water), ] water_evaluation.water_evaluation_selection. house_number 6 water pipes damaged 489 8 water pipes damaged 512 11 water pipes damaged 597 19 Water damage 478 21 water pipes damaged 373 23 Water damage 465 house_info[ house_info[,1] %in% c(Water damage, water pipes damaged, leaking water), 2] [1] 489 512 597 478 373 465 337 362 234 535 551 351 415 495 220 216 317 443 346 577 585 268 463 441 225 200 304 486 390 476 485 247 [33] 399 504 262 551 575 359 538 sort(unique(house_info[ house_info[,1] %in% c(Water damage, water pipes damaged, leaking water), 2])) [1] 200 216 220 225 234 247 262 268 304 317 337 346 351 359 362 373 390 399 415 441 443 463 465 476 478 485 486 489 495 504 512 535 [33] 538 551 575 577 585 597 Also, an easier way to generated random integers is sample(), e.g. sample(1:3, size=5, rep=T) [1] 3 1 2 1 1 (This is more straightforward, and more easily avoids possibly unintended errors such as floor(runif(100, 1,6) never generating a 6, but do be careful of the gotcha that sample(2:3, ...) will generate a selection of 2's and 3's, while sample(3,...) will generate samples from 1, 2, and 3.) -- Tony Plate Jason Rupert wrote: Say I have the following data: house_number-floor(runif(100, 200, 600)) water_evaluation-c(No water damage, Water damage, Water On, Water off, water pipes damaged, leaking water) water_evaluation_selection-floor(runif(100, 1,6)) house_info-data.frame(water_evaluation[water_evaluation_selection], house_number) And, that I only want to pull out the ones with negative water evaluations, i.e. Water damage, water pipes damaged, and leaking water. Should/could I use grep in order to pull the house numbers out of house_info with those negative water evaluations? I guess I want to know the house numbers from house_info where the water evaluation is negative. Is there a way to use grep or another R function in order to acquire that information? Thank you again in advance for any insights. __ 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] Downloading currency data from from Yahoo
you simply had the wrong symbol. a search on yahoo.com suggests the symbol might be
GBP=x,
getSymbols('GBP=x',src='yahoo')
str(get(GBP=X))
An ‘xts’ object from 2007-01-03 to 2009-09-23 containing:
Data: num [1:707, 1:6] 0.51 0.51 0.52 0.52 0.52 0.52 0.51 0.51 0.51 0.51 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:6] GBP=X.Open GBP=X.High GBP=X.Low GBP=X.Close ...
Indexed by objects of class: [Date] TZ: GMT
xts Attributes:
List of 2
$ src: chr yahoo
$ updated: POSIXct[1:1], format: 2009-09-24 19:58:01
-- Tony Plate
Bogaso wrote:
Hi,
I wanted to download some currency data using quantmod package, however
got following error :
getSymbols('USD/GBP',src='yahoo')
Error in download.file(paste(yahoo.URL, s=, Symbols.name, a=, from.m,
:
cannot open URL
'http://chart.yahoo.com/table.csv?s=USD/GBPa=0b=01c=2007d=8e=24f=2009g=dq=qy=0z=USD/GBPx=.csv'
In addition: Warning message:
In download.file(paste(yahoo.URL, s=, Symbols.name, a=, from.m, :
cannot open: HTTP status was '404 Not Found'
Can anyone please tell me how to get rid of that? Your help will be highly
appreciated.
Thanks
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Re: [R] How can I get predict.lm results with manual calculations ? (a floating point problem)
Results can be slightly different when matrix algebra routines are called.
Here's your example again. When the prediction is computed directly using
matrix multiplication, the result is the same as 'predict' produces (at least
in this case.)
set.seed(1)
n - 100
x - rnorm(n)
y - rnorm(n)
aa - lm(y ~ x)
all.equal(as.numeric(predict(aa, new)), as.numeric(aa$coef[1] + aa$coef[2] *
new$x), tol=0)
[1] Mean relative difference: 1.840916e-16
all.equal(as.numeric(predict(aa, new)), as.numeric(cbind(1, new$x) %*%
aa$coef), tol=0)
[1] TRUE
These types of small differences are often not indicative of lower precision in
one method, but rather just random floating-point inaccuracies that can depend
on things like the order numbers are summed in (e.g., ((bigNegNum + bigPosNum)
+ smallPosNum) will often be slightly different to ((bigPosNum + smallPosNum) +
bigNegNum). They can also depend on whether intermediate results are kept in
CPU registers, which sometimes have higher precision than 64 bits. Usually,
they're nothing to worry about, which is one of the major reasons that
all.equal() has a non-zero default for the tol= argument.
-- Tony Plate
Tal Galili wrote:
Hello dear r-help group
I am turning for you for help with FAQ number 7.31: Why doesn't R think
these numbers are equal?
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
*My story* is this:
I wish to run many lm predictions and need to have them run fast.
Using predict.lm is relatively slow, so I tried having it run faster by
doing the prediction calculations manually.
But doing that gave me problematic results (I won't go into the details of
how I found that out).
I then discovered that the problem was that the manual calculations I used
for the lm predictions yielded different results than that of predict.lm,
*here is an example*:
predict.lm.diff.from.manual.compute - function(sample.size = 100)
{
x - rnorm(sample.size)
y - x + rnorm(sample.size)
new - data.frame(x = seq(-3, 3, length.out = sample.size))
aa - lm(y ~ x)
predict.lm.result - sum(predict(aa, new, se.fit = F))
manual.lm.compute.result - sum(aa$coeff[1]+ new * aa$coeff[2])
# manual.lm.compute.result == predict.lm.result
return(all.equal(manual.lm.compute.result , predict.lm.result, tol=0))
}
# and here are the results of running the code several times:
predict.lm.diff.from.manual.compute(100)
[1] Mean relative difference: 1.046407e-15
predict.lm.diff.from.manual.compute(1000)
[1] Mean relative difference: 4.113951e-16
predict.lm.diff.from.manual.compute(1)
[1] Mean relative difference: 2.047455e-14
predict.lm.diff.from.manual.compute(10)
[1] Mean relative difference: 1.294251e-14
predict.lm.diff.from.manual.compute(100)
[1] Mean relative difference: 5.508314e-13
And that leaves me with *the question*:
Can I reproduce more accurate results from the manual calculations (as the
ones I might have gotten from predict.lm) ?
Maybe some parameter to increase the precision of the computation ?
Many thanks,
Tal
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Re: [R] un run run...
You could try setting options(error=function() NULL). This should cause R in batch mode to continue running after an error (the same way it does in interactive mode.) -- Tony Plate Nir Shachaf wrote: Hi All, I am running an Rscript with a bunch of algorithms that are UNSTABLE under some parameter settings. At a certain point one of them sends error massage and my whole run STOPS! What I would like is to save the error massage in some file or variable and carry on to the next command line without stopping this run... Any help or ideas would be welcome, please, with a concrete example (not just - have you thought of using 'tryCatch' etc.). Thanks all! Nir __ 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] A question on operation on list
Here's one way: set.seed(1) x1 - lapply(1:5, function(i) rnorm(2)) x2 - lapply(x1, function(x) outer(x, x)) Reduce(+, x2, 0) [,1] [,2] [1,] 1.768406 -1.534413 [2,] -1.534413 3.890200 and another way using the abind package: library(abind) dim(abind(along=0, x2)) [1] 5 2 2 colSums(abind(along=0, x2)) [,1] [,2] [1,] 1.768406 -1.534413 [2,] -1.534413 3.890200 -- Tony Plate megh wrote: Hi, I have created a list object like that : x = vector(list) for (i in 1:5) x[[i]] = rnorm(2) x Now I want to do two things : 1. for each i, I want to do following matrix calculation : t(x[[i]]) %*% x[[i]] i.e. for each i, I want to get a 2x2 matrix 2. Next I want to get x[[1]] + x[[2]] + I did following : res=vector(list); res = sapply(x, function(i) t(x[[i]]) %*% x[[i]]) However above syntax is not giving desired result. Any suggestion please? __ 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] Checking a (new) package - examples require other package functions
Did you try putting library(your-package-name, char=TRUE) at the start of the examples? -- Tony Plate Rebecca Sela wrote: I am creating an R package. I ran R CMD check on the package, and everything passed until it tried to run the examples. Then, the result was: * checking examples ... ERROR Running examples in REEMtree-Ex.R failed. The error most likely occurred in: ### * AutoCorrelationLRtest flush(stderr()); flush(stdout()) ### Name: AutoCorrelationLRtest ### Title: Test for autocorrelation in the residuals of a RE-EM tree ### Aliases: AutoCorrelationLRtest ### Keywords: htest tree models ### ** Examples # Estimation without autocorrelation simpleEMresult-RandomEffectsTree(Y~D+t+X, data=simpleREEMdata, random=~1|ID, simpleREEMdata$ID) Error: couldn't find function RandomEffectsTree Execution halted The function RandomEffectsTree is defined in the R code for the package. How can I refer to other functions from the package in examples? (I have the Writing R-extensions PDF, so it would be enough to point me to the right page, if the answer is in there and I just missed it.) Thanks! Rebecca __ 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] Finding rows common to two datasets
I think merge() can do what's wanted, but you do have to be careful that values match exactly. Here's an example where two data frames print the same in a row for columns 'a' and 'b', but are not exactly same. merge() returns zero rows. This problem can be fixed in this case by rounding, but that's not a good general solution because very close numbers can round to different numbers, e.g., 1.499 and 1.501. Here are examples: x - data.frame(a=c(1.001,2), b=c(3,4), c=LETTERS[1:2]) y - data.frame(a=c(1,2), b=c(3,5), c=LETTERS[3:4]) x a b c 1 1 3 A 2 2 4 B y a b c 1 1 3 C 2 2 5 D # x[1,a] and y[1,a] look the same, but are very slightly different merge(x, y, by=c(a, b)) [1] a b c.x c.y 0 rows (or 0-length row.names) # make x1 a version of x where the values are rounded to whole numbers x1 - x x1$a - round(x1$a) merge(x1, y, by=c(a, b)) a b c.x c.y 1 1 3 A C # intersect() returns columns that are the same in each dataframe, not rows intersect(x, y) c 1 C 2 D intersect(x1, y) a c 1 1 C 2 2 D -- Tony Plate jim holtman wrote: You are missing a comma: common - intersect(data_frame_x[,c(Latitude, Longitude)], data_frame_y[,c(Latitude,Longitude)]) On Tue, Apr 28, 2009 at 5:49 AM, Steve Murray smurray...@hotmail.com wrote: Thanks for the reply, however, when I do the following command, I receive the message: 'data frame with 0 columns and 0 rows'. I've checked again though, and there should be several thousand rows where the Latitude and Longitude pairs are the same. common - intersect(data_frame_x[c(Latitude, Longitude)], data_frame_y[c(Latitude,Longitude)]) common data frame with 0 columns and 0 rows Is there an obvious solution to this? Should I be using 'unique' instead, and if so, how would I get the above to correspond to this command? Thanks, Steve Date: Tue, 28 Apr 2009 13:36:51 +0530 Subject: Re: [R] Finding rows common to two datasets From: umesh.sriniva...@gmail.com To: smurray...@hotmail.com CC: r-help@r-project.org Dear Steve, Try ? intersect and see if that might help. Cheers, Umesh On Tue, Apr 28, 2009 at 1:29 PM, Steve Murray wrote: Dear all, I have 2 data frames, both with 14 columns of data and differing numbers of rows. The first two columns are 'Latitude' and 'Longitude'. I want to find the pairs of Latitude and Longitude coordinates which are common to both datasets, and output a new data frame which is composed of these coincident rows. I tried using the 'unique' command, but had difficulties interpreting the help file. Many thanks for any help offered, Steve __ 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] Puzzled by an error with apply()
One simple explanation for the error message you received is that you have a typo: 'lapply' instead of 'apply': x - matrix(1:6, 2) apply(x, 1, sum) [1] 9 12 lapply(x, 1, sum) Error in match.fun(FUN) : '1' is not a function, character or symbol However, it's difficult to diagnose without at least seeing some cut'n'pasted transcripts. The output of traceback() would also be useful. -- Tony Plate Gang Chen wrote: I've written a function, myFunc, that works fine with myFunc(data, ...), but when I use apply() to run it with an array of data apply(myArray, 1, myFunc, ...) I get a strange error: Error in match.fun(FUN) : '1' is not a function, character or symbol which really puzzles me because '1' is meant to be the margin of the array I want to apply over, but how come does apply() treat it as a function? I have been successfully using apply() for a while, so I must have made a stupid mistake this time. Hopefully somebody can point out something obviously wrong without me providing any details of the function. TIA, Gang __ 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] sorting by creation time in ls()
R doesn't keep track of when objects were created, so that's not possible. If this is important to you, you could look at the 'trackObjs' package, which does this and also stores individual objects in individual files (and writes them to the file when they are changed in R). -- Tony Plate Alexy Khrabrov wrote: When trying to remember what did I do in the session, especially after coming back to it after a few days, I'd like to mimic Unix's ls -ltrh -- does R retain the timing a certain variable is created? If not, would it make a useful addition, to have ls with an option to sort by creation time? Cheers, Alexy __ 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] Strange behaviour of ISOdatetime
Have you checked that that time exists in the time zone you are using? From ?ISOdatetime: Note ... Remember that in most timezones some times do not occur and some occur twice because of transitions to/from summer time. What happens in those cases is OS-specific. You could try working out what your system is using as the transition to/from summer time. (If you need to generate times that are 2 hours after midnight, try using ISOdatetime to generate the midnight times and add 2 hours). On my system, all this works fine: ISOdatetime(1995,03,26,2,0,0) [1] 1995-03-26 02:00:00 MST ISOdatetime(1995,03,26,0,0,0) + 2 * 60 * 60 [1] 1995-03-26 02:00:00 MST sessionInfo() R version 2.8.1 (2008-12-22) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] TimeWarp_0.7abind_1.2-0 trackObjs_0.8-0 tap.misc_1.0 [5] bmc.misc_1.0RtTests_0.1-5 -- Tony Plate Pedro de Barros wrote: Hi All, I am watching a strange behaviour of ISOdatetime. In my work computer, I get NA when I try to do ISOdatetime(1995,03,26,2,0,0) [1] NA But on other dates and/or times (hour) works OK ISOdatetime(1995,03,25,2,0,0) [1] 1995-03-25 02:00:00 GMT In my home computer, I do not have this problem. I am running the same version of R (2.8.1 patched) on both machines, the same version of Gnu Emacs (22.3.1) and the same version of ESS (5.3.10). Both are running Windows XP. Has anyone experienced this before? Pedro __ 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] abind
It looks like you are trying to construct a ragged array, where the extent of
the dimensions varies.
However, in R, ordinary arrays have a regular structure, e.g., the rows of a
matrix always have the same number of columns. This is the kind of object
abind() constructs.
So, to bind your two matrices together, abind() requires that their dimensions
match
a - matrix(1:6, ncol=3, byrow=T)
b - matrix(7:10, ncol=2, byrow=T)
a
[,1] [,2] [,3]
[1,]123
[2,]456
b
[,1] [,2]
[1,]78
[2,]9 10
library(abind)
abind(a, b, along=3)
Error in abind(a, b, along = 3) :
arg 'X2' has dims=2, 2, 1; but need dims=2, 3, X
The only way to get abind to bind these together on the third dimension is to
pad out the smaller matrix with NA's, e.g.:
abind(a, cbind(b, NA), along=3)
, , 1
[,1] [,2] [,3]
[1,]123
[2,]456
, , 2
[,1] [,2] [,3]
[1,]78 NA
[2,]9 10 NA
('a' and 'b' do match on the number of rows, so you can bind them together as
columns as does cbind(), e.g.:
abind(a, b, along=2)
[,1] [,2] [,3] [,4] [,5]
[1,]12378
[2,]4569 10
)
If none of this is what you want, you could consider storing the matrices in a
list, as another poster suggested.
-- Tony Plate
Suyan Tian wrote:
I am trying to combine two arrays with different dimensions into one.
For example
The first one is
1 2 3
4 5 6
The second one is
7 8
9 10
The resulted one would be like
, , 1
1 2 3
4 5 6
, , 2
7 8
9 10
I used abind to do this, but failed. Could somebody please let me know
how to do this in R? Thanks so many.
Suyan
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Re: [R] name returned by lapply
If you return the value as named list, you get your answer
using unlist(res, recursive=F):
res - lapply(1:2, function(i) {val - list(i);
names(val) - paste(Hugo, i, sep=_); return(val)})
unlist(res, rec=F)
$Hugo_1
[1] 1
$Hugo_2
[1] 2
Antje wrote:
Oh true, this would solve the problem too :-)
Thanks a lot for the suggestions!
Antje
Martin Morgan schrieb:
Antje [EMAIL PROTECTED] writes:
Thanks a lot for your help!
I know that I cannot directly access the list created, I just was not
sure if there is any format of the return value which could provide
additionally a name for the returned list.
I tried to return the values as list with the appropriate name but
then I end up with a list entry as list entry...
Okay, then I'll solve it with a loop and thanks for the hint with the
article
maybe this:
res - lapply(1:5, function(i) list(key=paste(Hugo, i, sep=_),
val=i))
val - lapply(res, [[, val)
names(val) - lapply(res, [[, key)
val
$Hugo_1
[1] 1
$Hugo_2
[1] 2
$Hugo_3
[1] 3
$Hugo_4
[1] 4
$Hugo_5
[1] 5
Martin
Ciao,
Antje
Gavin Simpson schrieb:
On Fri, 2008-07-18 at 14:19 +0200, Antje wrote:
Hi Gavin,
thanks a lot for your answer.
Maybe I did not explain very well what I want to do and probably
chose a bad example. I don't mind spaces or names starting with a
number. I could even name it:
Hugo1, Hugo2, ...
My biggest problem is, that not only the values are
calculated/estimated within my function but also the names (Yes, in
reality my funtion is more complicated).
Maybe it's easier to explain like this. the parameter x can be a
coordinate position of mountains on earth. Within the funtion the
height of the mountain is estimated and it's name.
In the end, I'd like to get a list, where the entry is named like
the mountain and it contains its height (or other measurements...)
## now that we have a list, we change the names to what you want
names(ret) - paste(1:10, info_within_function)
so this would not work, because I don't have the information
anymore about the naming...
OK, so you can't do what you want to do in the manner you tried, via
lapply as you don't have control of how the list is produced once the
loop over 1:10 has been performed. At the stage that 'test' is being
applied, all it knows about is 'x' and it doesn;t have access to the
list being built up by lapply().
The *apply family of functions help us to *not* write out formal
loops
in R, but here this is causing you a problem. So we can specify an
explicit loop and fill in information as and when we want from within
the loop
## create list to hold results
n - 10
ret - vector(mode = list, length = n)
## initialise loop
for(i in seq_len(n)) {
## do whatever you need to do here, but this line just
## replicates what 'test' did earlier
ret[[i]] - c(1,2,3,4,5)
## now add the name in
names(ret)[i] - paste(Mountain, i, sep = )
}
ret
Alternatively, collect a vector of names during the loop and then
once
the loop is finished do a single call to names(ret) to replace all the
names at once:
n - 10
ret - vector(mode = list, length = n)
## new vector to hold vector of names
name.vec - character(n)
for(i in seq_len(n)) {
ret[[i]] - c(1,2,3,4,5)
## now we just fill in this vector as we go
name.vec[i] - paste(Mountain, i, sep = )
}
## now replace all the names at once
names(ret) - name.vec
ret
This latter version is likely to more efficient if n is big so you
don't
incur the overhead of the repeated calls to names()
The moral of the story is to not jump to using *apply all the time
to
avoid loops. Loops in R are just fine, so use the tool that helps
you do
the job most efficiently *and* most transparently.
Take a look at the R Help Desk article by Uwe Ligges and John Fox in
the
current issue of RNews:
http://www.r-project.org/doc/Rnews/Rnews_2008-1.pdf
Which goes into this in much more detail
HTH
G
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Re: [R] automation of R? running an R script at a certain time each night?
Felipe Carrillo wrote: Hi: Edward, were you able to automate the process? if so, do you mind giving me a hint on how you did it? I am facing the same problem. I created a batch file which it runs fine using the task scheduler but only opens Tinn-R and R but it doesn't execute my script. My task scheduler executes everyday at 8:00 am This is my batch file: @echo off Start C:\Program Files\R\R-2.7.0\bin\Rgui.exe start C:\Documents and settings\Desktop\Software\MyScript.r What am I missing? Looks to me that the batch file above starts up the R GUI, and opens an editor, so it sounds like it is doing exactly what you're asking it to. Here's how I can get R to run as a scheduled task in Windows XP: Create a new Scheduled Task with the following in the Run: box: C:\Rbuild\R-2.6.2\bin\Rcmd.exe BATCH c:/Temp/Rtest.R Then schedule it how I want it to run. Here's the contents of my .R file and the output: $ cat c:/Temp/Rtest.R cat(file=c:/Temp/Rtest.txt, date(), \n, append=TRUE) $ cat c:/Temp/Rtest.txt Fri May 23 10:34:22 2008 Fri May 23 10:34:50 2008 Fri May 23 10:40:01 2008 (I'm sure there are dozens of other ways of doing this, but the above worked for me.) -- Tony Plate --- Edward Wijaya [EMAIL PROTECTED] wrote: You might try cron job under Windows. http://drupal.org/node/31506 HTH. - Edward On Thu, May 22, 2008 at 8:51 AM, Thomas Pujol [EMAIL PROTECTED] wrote: I am using R in a Windows environment. I store my data in a Microsoft SQL database that gets updated automatically nightly. Once my SQL db is updated, I wish to automatically run an R script Any tips on good ways to approach this task? Is there an easy way to launch an R script using the Windows or other scheduler? Can I have an R-script run every hour, and within the script check to determine if my database is updated, and proceed only if it is updated? [[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. Felipe D. Carrillo Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ 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] R is a virus, spyware or malware (gasp!)
And the installed Rgui.exe in 2.6.2 is 10240 bytes, which is their most common file size. They also say (fill in your own comments...): RGUI.EXE has been seen to perform the following behavior(s): * Can communicate with other computer systems using HTTP protocols * This Process Creates Other Processes On Disk * This Process Deletes Other Processes From Disk * Executes a Process * The Process is packed and/or encrypted using a software packing process * Accesses the MS Outlook Address Book and What you should do about RGUI.EXE: Check Your PC Now The most common objects with the name of RGUI.EXE have yet to be classified as safe by our research department. So, I suspect that they don't actually classify it as a virus, they just haven't classified it as safe... (I couldn't see any email contact address on their web page.) -- Tony Plate Peter Dalgaard wrote: Ioannis Dimakos wrote: Of course it's a virus. Once you catch the virus, you feel this rush, this fever to abandon all other statistical packages. On a more serious note, though, the size of the RGUI.exe file as reported in the webpage is not even near close to the actual size of the R distribution exe pack. Hmm, the installed Rgui.exe in 2.7.0rc (which is all I have, under Wine on a Fedora machine) appears to be 27648 bytes, which is one of the cited numbers. Presumably something needs to be done... I On Sun, May 18, 2008 17:30, Philippe Grosjean wrote: After a search session in Google, I found this page: http://www.prevx.com/filenames/X1993788672854780728-0/RGUI.EXE.html which classifies Rgui.exe (clearly stated as R for Windows GUI front-end) in a database of virus, spyware and malware! No comments! Philippe Grosjean __ 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] Accessing items in a list of lists
Try this:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
sapply(comb, [[, a)
data1 data2 data3
1 4 3
# Also, this can be useful:
comb[[c(data2, b)]]
[1] 5
[EMAIL PROTECTED] wrote:
Using R 2.6.2, say I have the following list of lists, comb:
data1 - list(a = 1, b = 2, c = 3)
data2 - list(a = 4, b = 5, c = 6)
data3 - list(a = 3, b = 6, c = 9)
comb - list(data1 = data1, data2 = data2, data3 = data3)
So that all names for the lowest level list are common. How can I most
efficiently access all of the sublist items a indexed by the outer
list names? For example, I can loop through comb[[i]], unlisting as I
go, and then look up the field a, as below, but there has got to be a
cleaner way.
finaldata - double(0)
for(i in 1:length(names(comb))) {
test - unlist(comb[[i]])
finaldata - c(finaldata, test[which(names(test) == a)])
}
data.frame(names(comb), finaldata)
Gives what I want:
names.comb. finaldata
1 data1 1
2 data2 4
3 data3 3
Any help you can give would be greatly appreciated. Thanks.
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Re: [R] Max consecutive increase in sequence
If the increases or decreases could be any size,
rle(sign(diff(x))) could do it:
x - c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1)
r - rle(sign(diff(x)))
r
Run Length Encoding
lengths: int [1:5] 3 2 2 5 4
values : num [1:5] 1 0 1 -1 0
i1 - which(r$lengths==max(r$lengths[r$values==1])
r$values==1)[1]
i2 - which(r$lengths==max(r$lengths[r$values==-1])
r$values==-1)[1]
i1
[1] 1
i2
[1] 4
rbind(up=c(start=cumsum(c(1, r$lengths))[i1],
len=r$lengths[i1]), down=c(start=cumsum(c(1,
r$lengths))[i2], len=r$lengths[i2]))
start len
up 1 3
down 8 5
Ingmar Visser wrote:
rle(diff(sq)) could be helpful here,
best, Ingmar
On May 13, 2008, at 11:19 PM, Marko Milicic wrote:
Hi all R helpers,
I'm trying to comeup with nice and elegant way of detecting consecutive
increases/decreases in the sequence of numbers. I'm trying with
combination
of which() and diff() functions but unsuccesifuly.
For example:
sq - c(1, 2, 3, 4, 4, 4, 5, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1);
I'd like to find way to calculate
a) maximum consecutive increase = 3 (from 1 to 4)
b) maximum consecutive decrease = 5 (from 6 to 1)
All ideas are highly welcomed!
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Re: [R] test
You probably should check this section in your R-help subscription options (via https://stat.ethz.ch/mailman/options/r-help/, I think): Receive your own posts to the list? Ordinarily, you will get a copy of every message you post to the list. If you don't want to receive this copy, set this option to No. I see 5 identical posts with the subject Several questions about MCMClogit on R-help recently. -- Tony Plate j t wrote: Sorry to bother your. I am trying to post my question for more than 10 times, but I still didn't see it. It drives my crazy!!! It is a test for posting some simple pure text. Chao __ 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] Replicating Rows
Another way is using straightforward indexing: x - cbind(trips=c(1,3,2), y=1:3, z=4:6) x trips y z [1,] 1 1 4 [2,] 3 2 5 [3,] 2 3 6 # generate row indices with the appropriate # number of repeats ii - rep(seq(len=nrow(x)), x[,1]) [1] 1 2 2 2 3 3 # use these indices to select data rows x[ii, -1] y z [1,] 1 4 [2,] 2 5 [3,] 2 5 [4,] 2 5 [5,] 3 6 [6,] 3 6 Jorge Ivan Velez wrote: Hi Marion, Try this: set.seed(123) mydf=data.frame(trips=rpois(10,5), matrix(rnorm(10*5),ncol=5)) mydf sapply(mydf[,-1],rep,mydf[,1]) HTH, Jorge On Wed, May 7, 2008 at 11:41 PM, [EMAIL PROTECTED] wrote: Hi, I have a data matrix in which there are 1000 rows by 30 columns. The first column of each row is a numeric indicating the number of trips taken to a particular location with location attributes in the following column entries for that row. I want to repeat each row based on the number of trips taken (as indicated by the number in the first column)...i.e., if 1,1 indicates 4 trips, I want to replicate row 1 four times, and do this for each entry of column 1. I have played with rep command with little luck. Can anyone help? This is probably very simple. Thank you, mw Marion Wittmann, Ph.D. candidate Environmental Science and Management University of California Santa Barbara, CA 93106-5131 __ 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. __ 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] How to stop buffering of cat
If you're using Rgui under Windows, see FAQ 7.1: 7.1 When using Rgui the output to the console seems to be delayed. This is deliberate: the console output is buffered ... (the FAQ says how to turn it off -- it's a menu item). Vidhu Choudhary wrote: Hi All, My R code takes very long time to finish the processing. I want to see at what stage the script is running. So I wrote some output messages using cat. But instead of displaying the cat messages at different stages they are buffered and displayed in the end when entire processing is done. Can you please suggest how to stop this buffering or some alternative way to display messages Thank you Vidhu [[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] Applying user function over a large matrix
It's quite possible that much of the time spent in loess() is setting up the data (i.e., the formula, terms, model.frame, etc.), and that much of that is repeated identically for each call to loess(). I would suggest looking at the code of loess() and work out what arguments it is calling simpleLoess() with, and then try calling stats:::simpleLoess() directly. (Of course you have to be careful with this because this is not using the published API). -- Tony Plate Sudipta Sarkar wrote: Respected R experts, I am trying to apply a user function that basically calls and applies the R loess function from stat package over each time series. I have a large matrix of size 21 X 900 and I need to apply the loess for each column and hence I have implemented this separate user function that applies loess over each column and I am calling this function foo as follows: xc-apply(t,2,foo) where t is my 21 X 900 matrix and loess. This is turning out to be a very slow process and I need to repeat this step for 25-30 such large matrix chunks. Is there any trick I can use to make this work faster? Any help will be deeply appreciated. Regards Sudipta Sarkar PhD Senior Analyst/Scientist Lanworth Inc. (Formerly Forest One Inc.) 300 Park Blvd., Ste 425 Itasca, IL Ph: 630-250-0468 __ 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] What objects will save.image saves ? And how to specify objects to be saved..
save.image() is a wrapper for save() (Type 'save.image' (without quotes) at the prompt to see the code for the function.) Try something like this (not tested): load(oldvars.RData) old.vars - ls(all=TRUE) ... computation ... save(list=setdiff(ls(all=TRUE), old.vars), file=newvars.RData) You can play with the args (pattern= all=) of ls() to select which vars you are interested in. If your computations might change some of the existing variables, and you want to save those at the end too, you'll probably have to keep track of those manually. AFAIK, it's not possible to add some new vars to an existing .RData file -- that's why I wrote the above to save them in a .RData file that's different to the one that contained the old variables. -- Tony Plate Ng Stanley wrote: Read and reread, can't make out. Will try an experiment later On Wed, Apr 16, 2008 at 7:51 PM, Henrique Dallazuanna [EMAIL PROTECTED] wrote: See ?save On Wed, Apr 16, 2008 at 8:46 AM, Ng Stanley [EMAIL PROTECTED] wrote: Hi, I have a R script that loads an image R.data, does some operations, then save to the R.data again. Suppose I have done some computation before loading the R script, will all the objects before the R script execution be saved to R.data ? If yes, how can I specify save.image to save only those objects created in the R script ? Thanks Stanley [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] Use of ellipses ... in argument list of optim(), integrate(), etc.
Ravi Varadhan wrote: Hi, I have noticed that there is a change in the use of ellipses or . in R versions 2.6.1 and later. In versions 2.5.1 and earlier, the . were always at the end of the argument list, but in 2.6.1 they are placed after the main arguments and before method control arguments. This results in the user having to specify the exact (complete) names of the control arguments, i.e. partial matching is not allowed. An example with integrate() : integrate(function(x) exp(-x^2), lower=-Inf, upper=L, subdiv=1000) Error in f(x, ...) : unused argument(s) (subdiv = 1000) integrate(function(x) exp(-x^2), lower=-Inf, upper=L, subdivisions=1000) 1.633051 with absolute error 1.6e-06 Here is an example with optim(): res - optim(50, fw, meth=BFGS, control=list(maxit=2, temp=20, parscale=20)) Error in fn(par, ...) : unused argument(s) (meth = BFGS) FYI, I am using R version 2.6.1 on Windows XP. May I ask what the rationale behind this change is and also about the pros and cons of the two different ways of specifying (.)? Putting optim() arguments after the ... disallows the use of abbreviated actual arguments for optim(). This is generally a good thing, because prior to this change, it was impossible to supply, via the '...' arguments of optim(), an argument to fn() whose name was a prefix of one of the arguments of optim(). E.g., if your function had a argument named 'm', you could not previously supply it via the '...' argument of optim(), because if you did something like optim(x, fun, m=240), intending 'm' to be passed to 'fun', the 'm' would instead match the 'method' argument of optim(). The cons of the new argument structure are that abbreviations for names of arguments of optim() can't be used (a minor and debatable con), and that previous code that used abbreviations might break, but it will likely break quickly and noisily, so it's not too bad (the only case where it wouldn't break is when fn has a '...' argument itself, and it ignores unrecognized components, or where the are other argument name collisions). -- Tony Plate Thank you very much. Best, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html [[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] Data frame with 0 rows.
Rolf Turner wrote:
For reasons best known only to myself ( :-) ) I wish to create a data
frame with 0 rows and 9 columns.
The best I've been able to come up with is:
junk - as.data.frame(matrix(0,nrow=0,ncol=9))
Is there a sexier way?
I'm unsure of their virtue or seediness, but here are some alternatives:
data.frame(a=numeric(0), b=numeric(0)) # include 9 arguments if you like
[1] a b
0 rows (or 0-length row.names)
as.data.frame(rep(list(a=numeric(0)), 9))
[1] a a.1 a.2 a.3 a.4 a.5 a.6 a.7 a.8
0 rows (or 0-length row.names)
-- Tony Plate
cheers,
Rolf
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Re: [R] Odp: How to make t.test handle NA and essentially constant values ?
Petr PIKAL wrote:
Hi
[EMAIL PROTECTED] napsal dne 12.02.2008 09:09:23:
Hi,
First problem:
test - matrix(c(1,1,2,1), 2,2)
apply(test, 1, function(x) { t.test(x) $p.value })
Error in t.test.default(x) : data are essentially constant
make your data not constant
Second problem:
test - matrix(c(1,0,NA,1), 2,2)
apply(test, 1, function(x) { t.test(x) $p.value })
Error in t.test.default(x) : not enough 'x' observations
increase number of observations
How to make t-test ignores this errors ?
Well, the procedure is complaining that you do not give it correct data.
You shall be gratefull for a great software which prevent you from making
silly things as try to compute t.test when data have zero variantion or
number of observations is 1.
It's nice that the software recognizes situations in which a sensible
answer can't be computed. At that point, there are two possible actions:
(1) stop with an informative error, and (2) silently return NA. Option (1)
is wonderful for interactive use, but option (2) is easier to handle in
programs where one is making many calls to the function as part of some
automated procedure (e.g., as part of a bootstrap procedure).
Speaking from personal experience, it can be quite a drag when one has set
up and mostly-debugged a long computation only to have it stop with an
error like data are essentially constant right near the end because of
some condition for which the function author thought it better to stop with
an error rather than return NA (or some other indication that there was no
sensible answer) (didn't happen with t.test, but I've experienced it with a
few other functions.)
So, I don't think it's at all unreasonable for the OP to request a way to
make t.test() return NA instead of stopping with an error.
Looking at the code for t.test, it doesn't look like there's any argument
to specify such behavior, so the options are to write one's own version of
t.test, or use try() as other posters have suggested. Here's an example
using try():
my.t.test.p.value - function(...) {
+obj-try(t.test(...), silent=TRUE)
+if (is(obj, try-error)) return(NA) else return(obj$p.value)
+ }
my.t.test.p.value(numeric(0))
[1] NA
my.t.test.p.value(1:10)
[1] 0.000278196
my.t.test.p.value(1)
[1] NA
my.t.test.p.value(c(1,1,1))
[1] NA
my.t.test.p.value(c(1,2,NA))
[1] 0.2048328
my.t.test.p.value(c(1,2))
[1] 0.2048328
hope this helps,
Tony Plate
Regards
Petr
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[R] [R-pkgs] new version of trackObjs
The trackObjs package stores objects in files on disk so that files are automatically rewritten when objects are changed, and so that objects are accessible but do not occupy memory until they are accessed. Also tracks times when objects are created and modified, and caches some basic characteristics of objects to allow for fast summaries of objects. This version trackObjs_0.8-0 fixes some bugs: o Fixed faulty detection of conflicting existing objects when starting to track to an existing directory. o Replaced environment on function that is in the active binding for a tracked object. Previously, that function could, if constructed via track(obj - value), have a copy of the tracked object in its environment, which would stay present taking up memory even if the object was flushed out of the tracking environment. o Fixed bug that stopped track.stop(all=TRUE) from working -- Tony Plate ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] assigning and saving datasets in a loop, with names changing with i
Marie Pierre Sylvestre wrote:
Dear R users,
I am analysing a very large data set and I need to perform several data
manipulations. The dataset is so big that the only way I can play with it
without having memory problems (E.g. cannot allocate vectors of size...)
is to write a batch script to:
1. cut the data into pieces
2. save the pieces in seperate .RData files
3. Remove everything from the environment
4. load one of the piece
5. perform the manipulations on it
6. save it and remove it from the environment
7. Redo 4-6 for every piece
8. Merge everything together at the end
It works if coded line by line but since I'll have to perform these tasks
on other data sets, I am trying to automate this as much as I can.
The trackObjs package is designed to make it easy to work in approximately
this manner -- it saves objects automatically to disk but they are still
accessible as normal.
Here's how you could do the above - this example works with 10 8Mb objects
in a R session with a limit of 40Mb.
# allow R only 40Mb of vector memory
mem.limits(vsize=40e6)
mem.limits()/1e6
library(trackObjs)
# start tracking to store data objects in the directory 'data'
# each object is 8Mb, and we store 10 of them
track.start(data)
n - 10
m - 1e6
constructObject - function(i) i+rnorm(m)
# steps 1, 2 3:
for (i in 1:n) {
xname - paste(x, i, sep=)
cat(, xname)
assign(xname, constructObject(i))
# store in a file, accessible by name:
track(list=xname)
}
cat(\n)
gc(TRUE)
# accessing object by name
object.size(x1)/2^20 # In Mb
mean(x1)
mean(x2)
gc(TRUE)
# steps 4:6
# accessing object through a constructed name
result - sapply(1:n, function(i) mean(get(paste(x, i, sep=
result
# remove the data objects
track.remove(list=paste(x, 1:n, sep=))
track.stop()
Here's the a full transcript of the above - note how whenever gc() is
called there is hardly any vector memory in use.
# allow R only 40Mb of vector memory
mem.limits(vsize=40e6)
nsizevsize
NA 4000
mem.limits()/1e6
nsize vsize
NA40
library(trackObjs)
# start tracking to store data objects in the directory 'data'
# each object is 8Mb, and we store 10 of them
track.start(data)
n - 10
m - 1e6
constructObject - function(i) i+rnorm(m)
# steps 1, 2 3:
for (i in 1:n) {
+xname - paste(x, i, sep=)
+cat(, xname)
+assign(xname, constructObject(i))
+# store in a file, accessible by name:
+track(list=xname)
+ }
x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 cat(\n)
gc(TRUE)
Garbage collection 19 = 6+0+13 (level 2) ...
4.0 Mbytes of cons cells used (42%)
0.7 Mbytes of vectors used (5%)
used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
Ncells 148362 4.0 35 9.4 NA 35 9.4
Vcells 89973 0.71950935 14.9 38.2 2112735 16.2
# accessing object by name
object.size(x1)/2^20 # In Mb
[1] 7.629417
mean(x1)
[1] 0.998635
mean(x2)
[1] 1.999656
gc(TRUE)
Garbage collection 22 = 7+1+14 (level 2) ...
4.0 Mbytes of cons cells used (43%)
0.7 Mbytes of vectors used (6%)
used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
Ncells 149264 4.0 35 9.4 NA 35 9.4
Vcells 90160 0.71560747 12.0 38.2 2112735 16.2
# steps 4:6
result - sapply(1:n, function(i) mean(get(paste(x, i, sep=
result
[1] 0.998635 1.999656 2.997368 4.000197 5.000159 6.001216 6.999552
[8] 7.999743 8.82 10.001355
# remove the data objects
track.remove(list=paste(x, 1:n, sep=))
[1] x1 x2 x3 x4 x5 x6 x7 x8 x9 x10
track.stop()
I am using a loop in which I used 'assign' and 'get' (pseudo code below).
My problem is when I use 'get', it prints the whole object on the screen.
I am wondering whether there is a more efficient way to do what I need to
do. Any help would be appreciated. Please keep in mind that the whole
process is quite computer-intensive, so I can't keep everything in the
environment while R performs calculations.
Say I have 1 big dataframe called data. I use 'split' to divide it into a
list of 12 dataframes (call this list my.list)
my.fun is a function that takes a dataframe, performs several
manipulations on it and returns a dataframe.
for (i in 1:12){
assign( paste( data, i, sep=), my.fun(my.list[i])) # this works
# now I need to save this new object as a RData.
# The following line does not work
save(paste(data, i, sep = ), file = paste( paste(data, i, sep =
), RData, sep=.))
}
# This works but it is a bit convoluted!!!
temp - get(paste(data, i, sep = ))
save(temp, file = lala.RData)
}
I am *sure* there is something more clever to do but I can't find it. Any
help would be appreciated.
best regards,
MP
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Re: [R] Efficient way to find consecutive integers in vector?
Martin Maechler wrote:
MS == Marc Schwartz [EMAIL PROTECTED]
on Thu, 20 Dec 2007 16:33:54 -0600 writes:
MS On Thu, 2007-12-20 at 22:43 +0100, Johannes Graumann wrote:
Hi all,
Does anybody have a magic trick handy to isolate directly consecutive
integers from something like this:
c(1,2,3,4,7,8,9,10,12,13)
The result should be, that groups 1-4, 7-10 and 12-13 are consecutive
integers ...
Thanks for any hints, Joh
MS Not fully tested, but here is one possible approach:
Vec
MS [1] 1 2 3 4 7 8 9 10 12 13
MS Breaks - c(0, which(diff(Vec) != 1), length(Vec))
Breaks
MS [1] 0 4 8 10
sapply(seq(length(Breaks) - 1),
MS function(i) Vec[(Breaks[i] + 1):Breaks[i+1]])
MS [[1]]
MS [1] 1 2 3 4
MS [[2]]
MS [1] 7 8 9 10
MS [[3]]
MS [1] 12 13
MS For a quick test, I tried it on another vector:
MS set.seed(1)
MS Vec - sort(sample(20, 15))
Vec
MS [1] 1 2 3 4 5 6 8 9 10 11 14 15 16 19 20
MS Breaks - c(0, which(diff(Vec) != 1), length(Vec))
Breaks
MS [1] 0 6 10 13 15
sapply(seq(length(Breaks) - 1),
MS function(i) Vec[(Breaks[i] + 1):Breaks[i+1]])
MS [[1]]
MS [1] 1 2 3 4 5 6
MS [[2]]
MS [1] 8 9 10 11
MS [[3]]
MS [1] 14 15 16
MS [[4]]
MS [1] 19 20
Seems ok, but ``only works for increasing sequences''.
More than 12 years ago, I had encountered the same problem and
solved it like this:
In package 'sfsmisc', there has been the function inv.seq(),
named for inversion of seq(),
which does this too, currently returning an expression,
but returning a call in the development version of sfsmisc:
Its definition is currently
inv.seq - function(i) {
## Purpose: 'Inverse seq': Return a short expression for the 'index' `i'
##
## Arguments: i: vector of (usually increasing) integers.
##
## Author: Martin Maechler, Date: 3 Oct 95, 18:08
##
## EXAMPLES: cat(rr - inv.seq(c(3:12, 20:24, 27, 30:33)),\n); eval(rr)
## r2 - inv.seq(c(20:13, 3:12, -1:-4, 27, 30:31)); eval(r2); r2
li - length(i - as.integer(i))
if(li == 0) return(expression(NULL))
else if(li == 1) return(as.expression(i))
##-- now have: length(i) = 2
di1 - abs(diff(i)) == 1#-- those are just simple sequences n1:n2 !
s1 - i[!c(FALSE,di1)] # beginnings
s2 - i[!c(di1,FALSE)] # endings
## using text parse {cheap and dirty} :
mkseq - function(i,j) if(i == j) i else paste(i,:,j, sep=)
parse(text =
paste(c(, paste(mapply(mkseq, s1,s2), collapse = ,), ), sep =
),
srcfile = NULL)[[1]]
}
with example code
v - c(1:10,11,6,5,4,0,1)
(iv - inv.seq(v))
c(1:11, 6:4, 0:1)
stopifnot(identical(eval(iv), as.integer(v)))
iv[[2]]
1:11
str(iv)
language c(1:11, 6:4, 0:1)
str(iv[[2]])
language 1:11
Now, given that this stems from 1995, I should be excused for
using parse(text = *) [see fortune(106) if you don't understand].
However, doing this differently by constructing the resulting
language object directly {using substitute(), as.symbol(),
as.expression() ... etc}
seems not quite trivial.
So here's the Friday afternoon / Christmas break quizz:
What's the most elegant way
to replace the last statements in inv.seq()
## using text parse {cheap and dirty} :
mkseq - function(i,j) if(i == j) i else paste(i,:,j, sep=)
parse(text =
paste(c(, paste(mapply(mkseq, s1,s2), collapse = ,), ), sep =
),
srcfile = NULL)[[1]]
by code that does not use parse (or source() or similar) ???
I don't have an answer yet, at least not at all an elegant one.
And maybe, the solution to the quiz is that there is no elegant
solution.
How about this ? :
i - c(1, 10, 12)
j - c(5, 10, 14)
mkseq - function(i, j) if (i==j) i else call(':', i, j)
as.call(c(list(as.name('c')), mapply(i, j, FUN=mkseq)))
c(1:5, 10, 12:14)
eval(.Last.value)
[1] 1 2 3 4 5 10 12 13 14
-- Tony Plate
Martin
MS HTH,
MS Marc Schwartz
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Re: [R] Function reference
R does this sort of thing easily without any parse/eval acrobatics needed.
E.g., you can do:
stu - function(x) {return( 1 + (2*x*x) - (3*x) )}
(x - 0:3)
[1] 0 1 2 3
stu(x)
[1] 1 0 3 10
metafun - function(FUN, data) FUN(data)
metafun(stu, x)
[1] 1 0 3 10
# if you want to be able use a character-data name for the function:
metafun2 - function(FUN, data) {if (is.character(FUN)) FUN -
getFunction(FUN); FUN(data)}
metafun2(stu, x)
[1] 1 0 3 10
metafun2(stu, x)
[1] 1 0 3 10
What went wrong with your code was that your parse() constructed a list of
4 expressions, and evaluating that returned the value of the last one:
(fun - stu)
[1] stu
paste( fun, (, x, ), sep = )
[1] stu(0) stu(1) stu(2) stu(3)
parse( text = paste( fun, (, x, ), sep = ) )
expression(stu(0), stu(1), stu(2), stu(3))
attr(,srcfile)
text
(Others have observed that in a very large proportion of the situations
where people reach for parse/eval, there's a neater, cleaner more direct
way of doing the job.)
-- Tony Plate
Talbot Katz wrote:
Hi.
I'm looking for an R equivalent to something like function pointers in C/C++.
I have a search procedure that evaluates the fitness of each point it
reaches as it moves along, and decides where to move next based on its
fitness evaluation. I want to be able to pass different fitness functions to
this procedure. I am trying to find a good way to do this. I was thinking
of passing in the name of the function and then using eval. However, I
haven't gotten this to work the way I'd like it to. Consider the following
example:
stu - function(x) {return( 1 + (2*x*x) - (3*x) )} (x=0:3)[1] 0 1 2 3
stu(x)[1] 1 0 3 10 (fun=stu)[1] stu eval( parse( text = paste( fun,
(, x, ), sep = ) ) )[1] 10
Notice that the function I defined called stu will operate on a vector x
and return a vector y = stu(x) such that y[i] equals stu(x[i]). When I tried
to pass stu and x to a procedure that would evaluate stu(x) I only get
stu(x[N]), when N is the last element of x. What am I doing wrong? Is there
a better way to pass function references?
I can get the following to work, but it seems awfully clunky:
sapply( 1:length(x), function(i){ return( eval( parse( text = paste( fun,
(, x[i], ), sep = ) ) ) ) } )[1] 1 0 3 10
Thanks!
-- TMK --212-460-5430 home917-656-5351 cell
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Re: [R] Factor Madness
From ?cbind: Data frame methods The cbind data frame method is just a wrapper for data.frame(..., check.names = FALSE). This means that it will split matrix columns in data frame arguments, and convert character columns to factors unless stringsAsFactors = TRUE is passed. (I'm guessing 'spectrum' is a data.frame before the code fragment you've shown) hope this helps, Tony Plate Johannes Graumann wrote: Why is class(spectrum[[Ion]]) after this factor? spectrum - cbind(spectrum,Ion=rep(, nrow(spectrum)),Deviation.AMU=rep(0.0, nrow(spectrum))) slowly going crazy ... Joh __ 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] Array dimnames
I can't quite understand what you're having difficulty with (is it constructing the array, or coping with the different 'matrices' having different column names, or something else?) However, your sample data looks like it has a mixture of factor (region) and numeric data (Qty), so you're probably storing it in a data frame. AFAIK, there is no 3d object in R that can store mixed-type data like a data frame can. An array object in R has to have the same data type for every column etc. -- Tony Plate dave mitchell wrote: Dear all, Possibly a rudimentary question, however any help is greatly appreciated. I am sorting a large matrix into an array of dim(p(i),q,3). I put each entry into a corresponding matrix (1 of the 3) based on some criteria. I figure this will assist me in condensing code as I can loop through the 3rd dimension of the array instead of generating 3 separate matrices and using the same block of code 3 times. My question is how to get the colnames of the 3 nested matrices in the array to match the colnames of the data matrix. In other words... DATA: Exp region Qty Ct ...q 1 S CB 3.55 2.15 . 2 S TG 4.16 2.18 . 3 C OO 2.36 3.65 . 4 C . . . . . . . . . . . . . . . . . . p ... ARRAY 1 [,1] [,2][,3] [,4]...q 1 SOME DATA WILL FILL THIS . 2 . . .. 3 . . . . 4 .. . . . . . . . . . . .. . . . . . P(1) ... 2 [,1] [,2][,3] [,4]...q 1 SOME DATA WILL FILL THIS . 2 . . .. 3 . . . . 4 .. . . . . . . . . . . .. . . . . . P(2) ... 3 [,1] [,2][,3] [,4]...q 1 SOME DATA WILL FILL THIS . 2 . . .. 3 . . . . 4 .. . . . . . . . . . . .. . . . . . P(3) ... Again, how to get those [,1], [,2]... to read (and operate) in the same fashion as the column names in the data matrix? Also, am I interpreting the dimensions of the array incorrectly? Please feel free to post any helpful links on the subject, as I have found dimnames and array in the R-help documentation unhelpful. Any help is greatly appreciated. Dave Mitchell Undergraduate: Statistics and Mathematics, University of Illinois, Urbana-Champaign [[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] matrix graph
Try these: x - matrix(rnorm(100), ncol=10) persp(x) contour(x) Also, look at the R graph gallery: http://addictedtor.free.fr/graphiques/ -- Tony Plate threshold wrote: Hi All, simple question: do you know how to graph the following object/matrix in a 'surface manner': [,1] [,2] [,3][,4] [,5][,6] [1,] -0.154 -0.065 0.129 0.637 0.780 0.221 [2,] 0.236 0.580 0.448 0.729 0.859 0.475 [3,] 0.401 0.506 0.310 0.650 0.822 0.448 [4,] 0.548 0.625 0.883 0.825 0.945 0.637 [5,] 0.544 0.746 0.823 0.877 0.861 0.642 [6,] 0.262 0.399 0.432 0.620 0.711 0.404 will be very grateful for hints. rob __ 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] How to know created time of object in R?
The trackObjs package will do this, at the level of objects in an environment. E.g., from the docs: library(trackObjs) track.start(tmp1) x - 123 # Not yet tracked track(x) # Variable 'x' is now tracked track(y - matrix(1:6, ncol=2)) # 'y' is assigned tracked z1 - list(a, b, c) z2 - Sys.time() track(list=c(z1, z2)) # Track a bunch of variables track.summary(size=F) # See a summary of tracked vars classmode extent lengthmodified TA TW x numeric numeric[1] 1 2007-09-07 08:50:58 0 1 y matrix numeric [3x2] 6 2007-09-07 08:50:58 0 1 z1 listlist [[3]] 3 2007-09-07 08:50:58 0 1 z2 POSIXt,POSIXct numeric[1] 1 2007-09-07 08:50:58 0 1 # (TA=total accesses, TW=total writes) (creation time is tracked too, but not displayed with default settings) For more info, look in the trackObjs package. -- Tony Plate Dong-hyun Oh wrote: Dear UseRs, I would like to know the created time and date of specific object. Is there any function for it? Thank you in advance. Sincerely, __ 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] timezone conversion difficulties with the new US daylight saving time switch over
Mike Waters wrote: Tony Plate wrote: [...] You don't say if this an R-specific problem, or if it afflicts your computer system clock as well. Thanks, I should have noted that my computer system clock is fine, and as far as I can tell it (correctly) believes we are still in Daylight Saving mode. It did not incorrectly fall back last Sunday at the date that would have been the end of Daylight Saving under the old rules. In case it matters, I do have a program running on my computer that synchronizes time: Domain Time II version 3.1.b.20040724R -- Tony Plate __ 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] timezone conversion difficulties with the new US daylight saving time switch over
I'm having difficulties with daylight saving times in US time zones. (Apologies for the long post, but the problem seems subtle and complex, unless I'm doing something completely wrong, in which case it should be evident from the first 10 lines below.) This is what I see, using a (slightly modified) example from ?as.POSIXlt : as.POSIXlt((d - Sys.time()), EST5EDT) # the current time in New York [1] 2007-10-30 12:38:47 EST d [1] 2007-10-30 11:38:47 Mountain Daylight Time The problem is that Mountain Time is 2 hours behind Eastern Time and the US is still on Daylight Saving Time - the current time in New York should be reported as 13:38:47 EDT, not 12:38:47 EST. This is running on Windows XP 64 bit (SP2), and I see the same behavior on Windows XP 32 bit and on Windows 2000 Server (SP4). AFAICS, this problem only occurs this week, and this week is unusual in that it is the first year that Daylight Saving Time in the US ends on the first Sunday in November rather than the last Sunday in October (I don't know whether this is the cause of the problem, but it seems likely). I see the same problem around the same week last year, but before and after this week in both years, the conversions are fine: # *** problem in 2007 as.POSIXlt(as.POSIXct(2007-10-30 11:38:47), EST5EDT) [1] 2007-10-30 12:38:47 EST # before the problem week 2007 as.POSIXlt(as.POSIXct(2007-10-20 11:38:47), EST5EDT) [1] 2007-10-20 13:38:47 EDT # after the problem week 2007 as.POSIXlt(as.POSIXct(2007-11-05 11:38:47), EST5EDT) [1] 2007-11-05 13:38:47 EST # *** problem in 2006 as.POSIXlt(as.POSIXct(2006-10-30 11:38:47), EST5EDT) [1] 2006-10-30 12:38:47 EST # before the problem week 2006 as.POSIXlt(as.POSIXct(2006-10-27 11:38:47), EST5EDT) [1] 2006-10-27 13:38:47 EDT # after the problem week 2006 as.POSIXlt(as.POSIXct(2006-11-07 11:38:47), EST5EDT) [1] 2006-11-07 13:38:47 EST My computer is set to the Mountain Daylight timezone, and is set to automatically adjust for Daylight Saving Time changes. sessionInfo() R version 2.6.0 Patched (2007-10-11 r43143) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base Sys.timezone() [1] Mountain Daylight Time If I explicitly set env var TZ, the conversion problems go away, but the time reported by Sys.time() is inappropriately not in daylight saving time: Sys.time() [1] 2007-10-30 13:14:38 Mountain Daylight Time Sys.setenv(TZ=MST7MDT) Sys.time() [1] 2007-10-30 12:14:51 MST If I set my system timezone to Eastern Daylight Time, and restart R, I also get problematic behavior (as.POSIXlt inappropriately adjusting a time by an hour on the day after Daylight saving time ends): Sys.timezone() [1] Eastern Daylight Time as.POSIXlt((d - Sys.time()), EST5EDT) [1] 2007-10-30 12:57:40 EST d [1] 2007-10-30 13:57:40 Eastern Daylight Time # *** problem week 2007 as.POSIXlt(as.POSIXct(2007-10-30 11:38:47), EST5EDT) [1] 2007-10-30 10:38:47 EST # before the problem week 2007 as.POSIXlt(as.POSIXct(2007-10-20 11:38:47), EST5EDT) [1] 2007-10-20 11:38:47 EDT # after the problem week 2007 as.POSIXlt(as.POSIXct(2007-11-05 11:38:47), EST5EDT) [1] 2007-11-05 11:38:47 EST # *** problem week 2006 - the day is after the switch, but # the time gets adjusted by one hour as.POSIXlt(as.POSIXct(2006-10-30 11:38:47), EST5EDT) [1] 2006-10-30 10:38:47 EST # before the problem week 2006 as.POSIXlt(as.POSIXct(2006-10-27 11:38:47), EST5EDT) [1] 2006-10-27 11:38:47 EDT # after the problem week 2006 as.POSIXlt(as.POSIXct(2006-11-10 11:38:47), EST5EDT) [1] 2006-11-10 11:38:47 EST The problem in 2006 goes away if I set TZ=EST5EDT: Sys.setenv(TZ = EST5EDT) Sys.timezone() [1] EST as.POSIXlt(as.POSIXct(2006-10-30 11:38:47), EST5EDT) [1] 2006-10-30 11:38:47 EST Questions: (1) should I be using a different way to convert times between time zones? (2) is there a problem in how R is interacting with the system time conversion facilities? (3) if the answer to (2) is yes, does anyone have any good hints for where to start looking for a solution? It looks to me like unclass(as.POSIXlt(as.POSIXct(2007-10-30 11:38:47), EST5EDT))$isdst is inappropriately zero. I tried following the code in src/main/datetime.c:do_asPOSIXlt, and it looks like this isdst value comes from a call to localtime() from localtime0() ... thanks in advance for any help, Tony Plate PS. I do not see the same problem in an old version of R (2.2.1) running under Ubuntu Linux: Sys.timezone() [1] as.POSIXlt((d - Sys.time()), EST5EDT) [1] 2007-10-29 14:56:49 EDT d [1] 2007-10-29 12:56:49 MDT # ok in 2007 as.POSIXlt(as.POSIXct(2007-10-30 11:38:47), EST5EDT) [1] 2007-10-30 13:38:47 EDT # before the problem week 2007
Re: [R] Input appreciated: R teaching idea + a way to improve R-wiki
hadley wickham wrote: On 10/23/07, Philippe Grosjean [EMAIL PROTECTED] wrote: Hi Matt, The R-Wiki is actively maintained... the addition of material to it is up to R users with any kind of initiative like this being warmly welcome. As for Bill Venable's comment, I totally agree: you should better test your concept first, and be ready to have very poor, as well as probably some excellent documents. I think it should be wise to announce to your students that the best documents will be posted to the R wiki, so that you may place a filter somewhere. As for the format, PDF is interesting as the student could learn Sweave too. However, the R Wiki allows for further corrections and additions to the documents. For the possible section in the Wiki, may be, a dedicated section like Users' guide (written by users) could be created, and then, you will organize material inside as you like. Otherwise, the existing Guides section should be fine (feel free to create subdirectories). I tend to give a lot of attention to documents written by beginners, because they are the best people to tell what is difficult and what is not in R! It is the starting motivation for the R Wiki, indeed. But they are simultaneously the worst people to provide good advice. The wiki seems to be riddled with poor practice and hacks to get around misunderstandings of the way R works. Is there any way on the R-Wiki for people to quickly and easily add an annotation indicating that they believe some particular advice is poor practice? Ideally, these annotations would be easily searchable so that other users could find and fix or respond to them. -- Tony Plate Hadley __ 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] functions applied to two vectors
a - c(2, 3, 7, 5) b - c(4, 7, 8, 9) mapply(seq, a, b) [[1]] [1] 2 3 4 [[2]] [1] 3 4 5 6 7 [[3]] [1] 7 8 [[4]] [1] 5 6 7 8 9 mapply(sum, a, b) [1] 6 10 15 14 hope this helps, Tony Plate Anya Okhmatovskaia wrote: Hi, I am very new to R, so I apologize if I have missed some trivial thing in the manuals/archives. I am trying to get rid of for loops in my code and replace them with R vector magic (with no success). Let's say I have 2 vectors of the same length: a - c(2, 3, 7, 5) b - c(4, 7, 8, 9) What I'd like to do is to generate a list(?) of 4 sequences using a[i] as a start indices, and b[i] as end indices: 2,3,4 3,4,5,6,7 7,8 5,6,7,8,9 My first guess: a:b, of course, does not work - only one sequence gets generated using the first values from both vectors, plus a warning. Is there a special syntax I can use to make : treat its operands as vectors? More generally, is there a standard way to apply some arbitrary functions to two or more vectors in an element-by-element fashion? For instance, sum(a,b) will sum all values in both vectors; how can I make it produce a vector of pairwise sums instead? Thank you, Anuta [[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] Save and load workspace in R: strange error.
Did you check whether 'junk4.RData' was created and what its length was
- maybe an empty file is being created. Is there some sort of quota or
permissions problem? My suggestion would be to look at the size and
permissions on the directory and the file. If you need more help, I
would suggest posting more details back to the list, e.g., what OS you
are using, and a directory listing that shows file sizes and permissions
(i.e., as you get with 'ls -l' on Unix systems.)
-- Tony Plate
Hongxiao Zhu wrote:
Hi,
I tried to load a .RData object on unix system using R, it gives error:
Error: restore file may be empty -- no data loaded
In addition: Warning message:
file 'junk3.RData' has magic number ''
Use of save versions prior to 2 is deprecated
This happens only for using MY user account for the Unix system. I
tried to use a friends's user account to load the same data object, it is
fine. And it never happened to me before until sometime last week.
And This error happens even when I generate a simple random number
from my user account and save it, and load it again.(So obviously it is
not a R version mismatch problem). Does anybody know what happened?
Here is an example what happened:
x=rnorm(100)
save.image('junk4.RData')
load('junk4.RData')
Error: restore file may be empty -- no data loaded
In addition: Warning message:
file 'junk4.RData' has magic number ''
Use of save versions prior to 2 is deprecated
Thanks for any suggestion.
Hongxiao
**
* Hongxiao Zhu *
* Department of Statistics, Rice Univeristy *
* Office: DH 3136, Phone: 713-348-2839 *
* http://www.stat.rice.edu/~hxzhu/ *
__
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Re: [R] Save and load workspace in R: strange error.
Sounds like you are having permissions problems. And as you're using a
mix of Unix and WinXP, you might be suffering from some strange
permissions settings. WinXP allows a very rich set of permissions which
for many exotic combinations have no corresponding mapping to the much
simpler 9-octet Unix permissions space -- so the limited view provided
by the Unix permission string -rw-r--r-- may not really reflect what you
can do with the file. On my own system, where I use WinXP and cygwin,
I've sometimes seem very strange Windows permission sets that look OK in
cygwin, but essentially disable use of a particular file. I generally
fix this by resetting ownership and all permissions using Windows
dialogs. It sounds like you need to get your sysadmins to help you sort
this problem out.
-- Tony Plate
Hongxiao Zhu wrote:
Tony,
Thanks for return. Actually, the data object 'junk4.RData' was created
but have size 0. It seems no data was saved. But the real data that I
want to load have data in it, which I can't load use my own user
account. But if using other people's user account under the same
system, it can be loaded.
All the files has the following property if I use ls -l:
-rw-r--r--
The OS I used is windows xp. But I use SSH to connect to the unix server.
I have been using this server for a long time, this error happened
since some day and from then on, I can never load/save workspace.
Hong
**
* Hongxiao Zhu *
* Department of Statistics, Rice Univeristy *
* Office: DH 3136, Phone: 713-348-2839 *
* http://www.stat.rice.edu/~hxzhu/ *
**
On Wed, 3 Oct 2007, Tony Plate wrote:
Did you check whether 'junk4.RData' was created and what its length
was - maybe an empty file is being created. Is there some sort of
quota or permissions problem? My suggestion would be to look at the
size and permissions on the directory and the file. If you need more
help, I would suggest posting more details back to the list, e.g.,
what OS you are using, and a directory listing that shows file sizes
and permissions (i.e., as you get with 'ls -l' on Unix systems.)
-- Tony Plate
Hongxiao Zhu wrote:
Hi,
I tried to load a .RData object on unix system using R, it gives error:
Error: restore file may be empty -- no data loaded
In addition: Warning message:
file 'junk3.RData' has magic number ''
Use of save versions prior to 2 is deprecated
This happens only for using MY user account for the Unix system. I
tried to use a friends's user account to load the same data object,
it is
fine. And it never happened to me before until sometime last week.
And This error happens even when I generate a simple random number
from my user account and save it, and load it again.(So obviously it
is not a R version mismatch problem). Does anybody know what happened?
Here is an example what happened:
x=rnorm(100)
save.image('junk4.RData')
load('junk4.RData')
Error: restore file may be empty -- no data loaded
In addition: Warning message:
file 'junk4.RData' has magic number ''
Use of save versions prior to 2 is deprecated
Thanks for any suggestion.
Hongxiao
**
* Hongxiao Zhu *
* Department of Statistics, Rice Univeristy *
* Office: DH 3136, Phone: 713-348-2839 *
* http://www.stat.rice.edu/~hxzhu/ *
__
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.
!DSPAM:4703b16f15261021468!
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Re: [R] how can I attach a variable stored in
Here's a function that does what I think you want to do:
attach.firstvar - function(file) {
+ tmpenv - new.env()
+ vars - load(file, envir=tmpenv)
+ x - get(vars[1], envir=tmpenv, inherits=FALSE)
+ if (is.list(x))
+ attach(x, name=vars[1])
+ return(vars)
+ }
x - list(xa=1, xb=2, xc=3)
save(list=x, file=tmp1.rda)
remove(list=x)
attach.firstvar(tmp1.rda)
[1] x
ls(pos=2)
[1] xa xb xc
find(xa)
[1] x
search()
[1] .GlobalEnvx package:stats
[4] package:graphics package:grDevices package:utils
[7] package:datasets package:methods Autoloads
[10] package:base
xa
[1] 1
xb
[1] 2
Peter Waltman wrote:
Hi Mark -
Thanks for the reply. Sorry I didn't really clarify too well what I'm
trying to do. The issue is not that I can't see the variable that gets
loaded.
The issue is that the variable is a list variable, and I'd like to write
a function that will take the .RData filename and attach the variable it
contains so that I can more easily access its contents, i.e.
foo.bar - list( a= a, b=1 )
save( file=foo.bar.RData, foo.bar )
rm( foo.bar )
my.fn - function( fname ) {
load( fname )
attach( ls( pat=foo ) ) # I want to attach( foo.bar ), but
this doesn't work
}
ls() # prints out foo.bar
attach( ls( ) ) # still doesn't work
attach( foo.bar ) # works
So, basically, the question is how can I attach the variable that's
stored in a file if I don't already know it's name?
Thanks again!
Peter
Leeds, Mark (IED) wrote:
I don't think I understand your question but John Fox has written a very
nice documentat about scoping and environments on his website.
It's probably easy to find the site by googling John Fox but, if you
can't find it, let me know.
As I said, I don't think that I understand your question but, if you
loaded a list variable using load(whatever.Rdata), the variable will
just be suitting in your workspace. You don't need to attach anything
because load just loads the data right into the workspace.
So typing the variable name should show the data.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Peter Waltman
Sent: Thursday, September 20, 2007 1:32 PM
To: r-help@r-project.org
Subject: [R] how can I attach a variable stored in
Hi -
Any help would be greatly appreciated.
I'm loading a list variable that's stored in an .RData file and would
like attach it.
I've used attach( file_name ), but that only lets me see the variable
that's stored in the file.
As the variable name is of the form comp.x.x, I've tried using attach(
ls( pat=comp ) ), but get an error as ls() just gives back a string.
I've also played around with eval(), but don't really quite get what
that function does since it seems to get into the R internals which I
don't entirely understand and I haven't found any great unified
documentation on R's handling environment and scoping.
Thanks,
Peter Waltman
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