[R] array
Hello, I have some files generated from microarray experiments. I used scan() to read the files, and assigned each file to a unique name with many rows and columns. Now I want to create a array (ArrayA) with unique names, and I can use ArrayA[1,2][[6]] to refer the data in each file. Is there any packages available for array of array? Thanks! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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
Hi, Tiandao Li wrote: Hello, I have some files generated from microarray experiments. I used scan() to read the files, and assigned each file to a unique name with many rows and columns. Now I want to create a array (ArrayA) with unique names, and I can use ArrayA[1,2][[6]] to refer the data in each file. Is there any packages available for array of array? if all of your initial arrays only consist of rows and columns (i.e. matrices) and have the same number of rows and columns, you can store it conveniently in a three dimensional array: mat1a - matrix(1:10, ncol=2) mat2a - matrix(11:20, ncol=2) mat1a mat2a arr3D - array(numeric(0), dim=c(nrow(mat1a), ncol(mat1a), 2)) arr3D[,,1] - mat1a arr3D[,,2] - mat2a arr3D arr3D[,,2] if your initial arrays (assuming again that you have matrices) have varying amount of rows and columns, I would suggest to use lists: mat1b - matrix(1:10, ncol=2) mat2b - matrix(101:133, ncol=3) mat1b mat2b list3D - list(numberone=mat1b, numbertwo=mat2b) list3D list3D[[1]] list3D[[1]] list3D[[numberone]] list3D[[numbertwo]] I hope this helps. Best, Roland __ R-help@stat.math.ethz.ch 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] array
Hello, I have 30 GenePix tab-delimited files generated from 10 microarray experiments, each has 3 replicates. I used scan() to read the files, and assigned each file to a unique name (such as A1, A2, A3 for experment A, and B1, B2, and B3 for experiment B) with 1000 rows and 56 columns. Now I want to create a array (ArrayA) with these file names, ArrayA [,1][,2]... [,10] [1,]A1 B1 C1 [2,]A2 B2 C2 [3,]A3 B3 C3 so I can use ArrayA[1,2][[6]] to refer the data in column6 of experimentB, replicate 1. Is there any packages available for array of array? Thanks! __ R-help@stat.math.ethz.ch 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 loop
Hi
Dong GUO 郭东 [EMAIL PROTECTED] napsal dne 31.07.2007 15:27:35:
Thanks, Petr.
I changed the equation mark from = to -, then, it works fine. Dont
know
what difference it has made between the = and -..
from help page
The operators - and = assign into the environment in which they are
evaluated. The operator- can be used anywhere, whereas the operator = is
only allowed at the top level (e.g., in the complete expression typed at
the command prompt) or as one of the subexpressions in a braced list of
expressions.
Although I do not fully understand where I can use - and where =, to be
on safe side I use - everywhere when I want to do assignment of some
value(s).
Regards
Petr
Regards,
Dong
On 7/31/07, Petr PIKAL [EMAIL PROTECTED] wrote:
Hi
as you say that the computing is part of a function than the best way to
see what is hapenning is to use
debug(your.function)
see ?debug for options.
Regards
Petr
[EMAIL PROTECTED]
[EMAIL PROTECTED] napsal dne 31.07.2007 00:11:00:
Dear all,
here are two arrays: region(26,31,8), nation(8)
I tried to get a new array, say, giGi(26,31,8)
giGi - array(0,dim = c(region_dim))
for (i in (1:region_dim[3]))
{
giGi[,,i] = region[,,i]-nation[,i]
}
As the above is part of function, but results shows only giGi[,,1] has
the
right answers, all the others (giGi[,,2],..giGi[..8]) are zeros. I
have
checked array of region and nation, they are not zeros at all
when I do manually, it is not the case, giGi has meanful numbers.
can some one tell me the trick in this process??
Many thanks in advance.
Dong
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] array loop
Thanks again, Petr. Following the reference, that would be true that =
only assign values to the top level...So apparently using '-' is the safe
all the time to assign values.
Dong
On 8/1/07, Petr PIKAL [EMAIL PROTECTED] wrote:
Hi
Dong GUO ¹ù¶« [EMAIL PROTECTED] napsal dne 31.07.2007 15:27:35:
Thanks, Petr.
I changed the equation mark from = to -, then, it works fine. Dont
know
what difference it has made between the = and -..
from help page
The operators - and = assign into the environment in which they are
evaluated. The operator- can be used anywhere, whereas the operator = is
only allowed at the top level (e.g., in the complete expression typed at
the command prompt) or as one of the subexpressions in a braced list of
expressions.
Although I do not fully understand where I can use - and where =, to be
on safe side I use - everywhere when I want to do assignment of some
value(s).
Regards
Petr
Regards,
Dong
On 7/31/07, Petr PIKAL [EMAIL PROTECTED] wrote:
Hi
as you say that the computing is part of a function than the best way to
see what is hapenning is to use
debug(your.function)
see ?debug for options.
Regards
Petr
[EMAIL PROTECTED]
[EMAIL PROTECTED] napsal dne 31.07.2007 00:11:00:
Dear all,
here are two arrays: region(26,31,8), nation(8)
I tried to get a new array, say, giGi(26,31,8)
giGi - array(0,dim = c(region_dim))
for (i in (1:region_dim[3]))
{
giGi[,,i] = region[,,i]-nation[,i]
}
As the above is part of function, but results shows only giGi[,,1] has
the
right answers, all the others (giGi[,,2],..giGi[..8]) are zeros. I
have
checked array of region and nation, they are not zeros at all
when I do manually, it is not the case, giGi has meanful numbers.
can some one tell me the trick in this process??
Many thanks in advance.
Dong
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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]]
__
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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 loop
Thanks, Petr.
I changed the equation mark from = to -, then, it works fine. Dont know
what difference it has made between the = and -..
Regards,
Dong
On 7/31/07, Petr PIKAL [EMAIL PROTECTED] wrote:
Hi
as you say that the computing is part of a function than the best way to
see what is hapenning is to use
debug(your.function)
see ?debug for options.
Regards
Petr
[EMAIL PROTECTED]
[EMAIL PROTECTED] napsal dne 31.07.2007 00:11:00:
Dear all,
here are two arrays: region(26,31,8), nation(8)
I tried to get a new array, say, giGi(26,31,8)
giGi - array(0,dim = c(region_dim))
for (i in (1:region_dim[3]))
{
giGi[,,i] = region[,,i]-nation[,i]
}
As the above is part of function, but results shows only giGi[,,1] has
the
right answers, all the others (giGi[,,2],..giGi[..8]) are zeros. I have
checked array of region and nation, they are not zeros at all
when I do manually, it is not the case, giGi has meanful numbers.
can some one tell me the trick in this process??
Many thanks in advance.
Dong
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] array writing and their filenames
Hi, I want to save a array (say, array[6,7,8]) write a cvs file. How can I do that??? can I write in one file? if I could not write in one file, i want to use a loop to save in different files (in the matrix[6,7,8], should be 8 csv files), such as the filename structure should be: file =filename +str(i) +. +csv Many thanks. Dong [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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 writing and their filenames
Many thanks to Edna and Roland.
I followed the suggestions, and it worked well.
Regards,
Dong
On 7/30/07, Edna Bell [EMAIL PROTECTED] wrote:
#Original 3x4x2 array
xb
, , 1
[,1] [,2] [,3] [,4]
[1,] 0.4 0.9 5.6 0.1
[2,] 0.3 2.3 3.3 0.7
[3,] 0.6 0.8 0.2 0.7
, , 2
[,1] [,2] [,3] [,4]
[1,] 0.4 0.9 5.6 0.1
[2,] 0.3 2.3 3.3 0.7
[3,] 0.6 0.8 0.2 0.7
for(i in 1:2) {
+ write(file=stuff4,t(xb[,,i]),ncol=4,append=T)
+ xx - paste(end of array ,i,sep=)
+ write(file=stuff4,xx,append=T)
+ }
file.show(stuff4)
#All is well
On 7/29/07, Dong GUO ¹ù¶« [EMAIL PROTECTED] wrote:
Hi,
I want to save a array (say, array[6,7,8]) write a cvs file. How can I
do
that??? can I write in one file?
if I could not write in one file, i want to use a loop to save in
different
files (in the matrix[6,7,8], should be 8 csv files), such as the
filename
structure should be: file =filename +str(i) +. +csv
Many thanks.
Dong
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] array loop
Dear all,
here are two arrays: region(26,31,8), nation(8)
I tried to get a new array, say, giGi(26,31,8)
giGi - array(0,dim = c(region_dim))
for (i in (1:region_dim[3]))
{
giGi[,,i] = region[,,i]-nation[,i]
}
As the above is part of function, but results shows only giGi[,,1] has the
right answers, all the others (giGi[,,2],..giGi[..8]) are zeros. I have
checked array of region and nation, they are not zeros at all
when I do manually, it is not the case, giGi has meanful numbers.
can some one tell me the trick in this process??
Many thanks in advance.
Dong
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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 writing and their filenames
Dong GUO 郭东 wrote: Hi, I want to save a array (say, array[6,7,8]) write a cvs file. How can I do that??? can I write in one file? For array[6,7,8], you don't need a csv(!) file since it is only a scalar. If this is what you want, check ?write.table But what you probably meant is how to write a three-dimensional array to disk. Have a look at this code: ## roland - array(1:(6*7*8), dim=c(6,7,8)) roland dump(list=roland, file = H:\\dumpdata.R) ls() rm(list=ls()) ls() source(H:\\dumpdata.R) ls() roland ## Does this help? Roland __ R-help@stat.math.ethz.ch 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 searches
Hi, This is truly amazing stuff. Inspired by Jim's and Olivier's suggestions, I'm trying to expand it to work with a m x n matrix, where the first column is dates and the next columns are all signals. I dare say a suitable application of 'apply' should work. Thanks a ton. Murali From: jim holtman [EMAIL PROTECTED] To: Murali Menon [EMAIL PROTECTED] CC: r-help@stat.math.ethz.ch Subject: Re: [R] array searches Date: Fri, 16 Feb 2007 10:21:40 -0500 try this: x - scan(textConnection(30/01/2007 0 + 31/01/2007 -1 + 01/02/2007 -1 + 02/02/2007 -1 + 03/02/2007 1 + 04/02/2007 1 + 05/02/2007 1 + 06/02/2007 1 + 07/02/2007 1 + 08/02/2007 1 + 09/02/2007 0 + 10/02/2007 0 + 11/02/2007 0 + 12/02/2007 1 + 13/02/2007 1 + 14/02/2007 1 + 15/02/2007 0 + 16/02/2007 0 + ), what=list(date=, value=0)) Read 18 records x$date - as.Date(x$date, %d/%m/%Y) # determine the breaks x.breaks - c(TRUE, diff(x$value) != 0) # determine the value at the break; assume that it is the minimum x.bdate - x$date[x.breaks] data.frame(date=x.bdate[cumsum(x.breaks)], value=x$value) date value 1 2007-01-30 0 2 2007-01-31-1 3 2007-01-31-1 4 2007-01-31-1 5 2007-02-03 1 6 2007-02-03 1 7 2007-02-03 1 8 2007-02-03 1 9 2007-02-03 1 10 2007-02-03 1 11 2007-02-09 0 12 2007-02-09 0 13 2007-02-09 0 14 2007-02-12 1 15 2007-02-12 1 16 2007-02-12 1 17 2007-02-15 0 18 2007-02-15 0 On 2/16/07, Murali Menon [EMAIL PROTECTED] wrote: Folks, I have a dataframe comprising a column of dates and a column of signals (-1, 0, 1) that looks something like this: 30/01/2007 0 31/01/2007 -1 01/02/2007 -1 02/02/2007 -1 03/02/2007 1 04/02/2007 1 05/02/2007 1 06/02/2007 1 07/02/2007 1 08/02/2007 1 09/02/2007 0 10/02/2007 0 11/02/2007 0 12/02/2007 1 13/02/2007 1 14/02/2007 1 15/02/2007 0 16/02/2007 0 What I need to do is for each signal *in reverse chronological order* to find the date that it first appeared. So, for the zero on 16/02/2007 and 15/02/2007, the 'inception' date would be 15/02/2007, because the day before, the signal was 1. Likewise, the 'inception' date for the signal 1 on 08/02/2007 and the five days prior, would be 03/02/2007. I need to create a structure of inception dates that would finally look as follows: -1 31/01/2007 -1 31/01/2007 -1 31/01/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 0 09/02/2007 0 09/02/2007 0 09/02/2007 1 12/02/2007 1 12/02/2007 1 12/02/2007 0 15/02/2007 0 15/02/2007 Is there a clever way of doing this? My sadly C-oriented upbringing can only think in terms of for-loops. Thanks! Murali _ The average US Credit Score is 675. The cost to see yours: $0 by Experian. __ R-help@stat.math.ethz.ch 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? _ fast as 1 year __ R-help@stat.math.ethz.ch 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] array searches
Folks, I have a dataframe comprising a column of dates and a column of signals (-1, 0, 1) that looks something like this: 30/01/2007 0 31/01/2007 -1 01/02/2007 -1 02/02/2007 -1 03/02/2007 1 04/02/2007 1 05/02/2007 1 06/02/2007 1 07/02/2007 1 08/02/2007 1 09/02/2007 0 10/02/2007 0 11/02/2007 0 12/02/2007 1 13/02/2007 1 14/02/2007 1 15/02/2007 0 16/02/2007 0 What I need to do is for each signal *in reverse chronological order* to find the date that it first appeared. So, for the zero on 16/02/2007 and 15/02/2007, the 'inception' date would be 15/02/2007, because the day before, the signal was 1. Likewise, the 'inception' date for the signal 1 on 08/02/2007 and the five days prior, would be 03/02/2007. I need to create a structure of inception dates that would finally look as follows: -1 31/01/2007 -1 31/01/2007 -1 31/01/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 0 09/02/2007 0 09/02/2007 0 09/02/2007 1 12/02/2007 1 12/02/2007 1 12/02/2007 0 15/02/2007 0 15/02/2007 Is there a clever way of doing this? My sadly C-oriented upbringing can only think in terms of for-loops. Thanks! Murali _ The average US Credit Score is 675. The cost to see yours: $0 by Experian. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] array searches
try this: x - scan(textConnection(30/01/2007 0 + 31/01/2007 -1 + 01/02/2007 -1 + 02/02/2007 -1 + 03/02/2007 1 + 04/02/2007 1 + 05/02/2007 1 + 06/02/2007 1 + 07/02/2007 1 + 08/02/2007 1 + 09/02/2007 0 + 10/02/2007 0 + 11/02/2007 0 + 12/02/2007 1 + 13/02/2007 1 + 14/02/2007 1 + 15/02/2007 0 + 16/02/2007 0 + ), what=list(date=, value=0)) Read 18 records x$date - as.Date(x$date, %d/%m/%Y) # determine the breaks x.breaks - c(TRUE, diff(x$value) != 0) # determine the value at the break; assume that it is the minimum x.bdate - x$date[x.breaks] data.frame(date=x.bdate[cumsum(x.breaks)], value=x$value) date value 1 2007-01-30 0 2 2007-01-31-1 3 2007-01-31-1 4 2007-01-31-1 5 2007-02-03 1 6 2007-02-03 1 7 2007-02-03 1 8 2007-02-03 1 9 2007-02-03 1 10 2007-02-03 1 11 2007-02-09 0 12 2007-02-09 0 13 2007-02-09 0 14 2007-02-12 1 15 2007-02-12 1 16 2007-02-12 1 17 2007-02-15 0 18 2007-02-15 0 On 2/16/07, Murali Menon [EMAIL PROTECTED] wrote: Folks, I have a dataframe comprising a column of dates and a column of signals (-1, 0, 1) that looks something like this: 30/01/2007 0 31/01/2007 -1 01/02/2007 -1 02/02/2007 -1 03/02/2007 1 04/02/2007 1 05/02/2007 1 06/02/2007 1 07/02/2007 1 08/02/2007 1 09/02/2007 0 10/02/2007 0 11/02/2007 0 12/02/2007 1 13/02/2007 1 14/02/2007 1 15/02/2007 0 16/02/2007 0 What I need to do is for each signal *in reverse chronological order* to find the date that it first appeared. So, for the zero on 16/02/2007 and 15/02/2007, the 'inception' date would be 15/02/2007, because the day before, the signal was 1. Likewise, the 'inception' date for the signal 1 on 08/02/2007 and the five days prior, would be 03/02/2007. I need to create a structure of inception dates that would finally look as follows: -1 31/01/2007 -1 31/01/2007 -1 31/01/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 1 03/02/2007 0 09/02/2007 0 09/02/2007 0 09/02/2007 1 12/02/2007 1 12/02/2007 1 12/02/2007 0 15/02/2007 0 15/02/2007 Is there a clever way of doing this? My sadly C-oriented upbringing can only think in terms of for-loops. Thanks! Murali _ The average US Credit Score is 675. The cost to see yours: $0 by Experian. __ R-help@stat.math.ethz.ch 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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 vs matrix vs dataframe?
On 10/1/2006 7:02 PM, r user wrote: What is the difference among an array, a dataframe and a matrix? Really, r, I think this is pretty well documented. Could you let us know what you've read that left this ambiguous? Why is the size of a dataframe so much larger? (see example below) I suspect it's the row names. Duncan Murdoch a-c(rep(1:100,1)) b-c(rep(1:100,1)) c1-cbind(a,b) cdf-as.data.frame(cbind(a,b)) cm-as.matrix(cbind(a,b)) object.size(a)/100 object.size(b)/100 object.size(c1)/100 object.size(cdf)/100 object.size(cm)/100 __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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 vs matrix vs dataframe?
On 10/1/2006 7:02 PM, r user wrote: What is the difference among an array, a dataframe and a matrix? Why is the size of a dataframe so much larger? (see example below) I forgot to mention: there will be very little difference in 2.4+; the row names were a problem in earlier releases, but Brian Ripley fixed that: a-c(rep(1:100,1)) b-c(rep(1:100,1)) c1-cbind(a,b) cdf-as.data.frame(cbind(a,b)) cm-as.matrix(cbind(a,b)) object.size(a)/100 [1] 4.24 object.size(b)/100 [1] 4.24 object.size(c1)/100 [1] 8.000296 object.size(cdf)/100 [1] 8.000448 object.size(cm)/100 [1] 8.000296 (These are in a recent release candidate of 2.4.0.) Duncan Murdoch __ R-help@stat.math.ethz.ch 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] Array
Dear all R users, I am wondering if there is any way to define a 3 dimentional or 4 dimentional array in R. Sincerely yours, thanks in advance Send instant messages to your online friends http://in.messenger.yahoo.com Stay connected with your friends even when away from PC. Link: http://in.mobile.yahoo.com/new/messenger/ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Array
Hi see ?array if you are interested array(0, c(5,5,5,5,5)) gives you 5 dimensional array. HTH Petr On 15 Jun 2006 at 15:34, stat stat wrote: Date sent: Thu, 15 Jun 2006 15:34:29 +0100 (BST) From: stat stat [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Subject:[R] Array Dear all R users, I am wondering if there is any way to define a 3 dimentional or 4 dimentional array in R. Sincerely yours, thanks in advance Send instant messages to your online friends http://in.messenger.yahoo.com Stay connected with your friends even when away from PC. Link: http://in.mobile.yahoo.com/new/messenger/ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] what are the limits on R array sizes?
I have some computers with a massive amount of memory, and I have some jobs that could use very large matrix sizes. Can R handle matrices of larger than 2GB? If I were to create a matrix of 1,000,000 x 1,000, it would use about 8GB. Can R work with an array of that size if I have compiled R on an IA64 Linux system with 15GB of RAM? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] what are the limits on R array sizes?
On Wed, 8 Feb 2006, Mike Miller wrote: I have some computers with a massive amount of memory, and I have some jobs that could use very large matrix sizes. Can R handle matrices of larger than 2GB? If I were to create a matrix of 1,000,000 x 1,000, it would use about 8GB. Can R work with an array of that size if I have compiled R on an IA64 Linux system with 15GB of RAM? Yes. See ?Memory-limits. help.search(limits) gets you there. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Converting a Perl Array of Arrayrefs to an R array or matrix using RS Perl
Dear R/RS-Perl users, I have a perl script in which I parse a large number of files and construct an array of arrayrefs from the data in the files. I then pass that construct to R using the RS Perl interface. I want to be able to use the construct as an R array or matrix so that I can use the R function colSums. So far, I've tried constructing an R matrix with dummy values, and then populating each row of the new matrix with the nested Arrays of the passed construct. But that just converts the R matrix to type list, so I get an error when I try the function colSums. I then tried converting each nested Array using as.numeric() before putting the values into the matrix. But for some reason as.numeric doesn't convert the nested arrays to numeric, so I still get an error when I try colSums. If I check the values (say matrix[1][1]) the correct values are in the correct locations, but it still won't perform colSums. I've been hammering away on this for longer than I would like to admit. Any help you could provide would be greatly appreciated. Best regards, Aaron __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array question
Hi all, I want to create an array of datetime. If I have a datetime object dt dt - strptime(10Jan2006 00:00:15, %d%b%Y %H:%M:%S) dt [1]2006-01-10 00:00:15 I want to make an array of dt, say 100 size. I got those error. [1] 2006-01-10 00:00:15 dtarray-array(dt, dim=c(100)); Error in array(dt, dim = c(100)) : dim- : dims [product 100] do not match the length of object [9] dtarray-array(, dim=c(100)); dtarray[1]-dt; Warning message: number of items to replace is not a multiple of replacement length Any help? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array question
Please use POSIXct and not POSIXlt objects, which are lists. On Tue, 17 Jan 2006, Chang Shen wrote: Hi all, I want to create an array of datetime. If I have a datetime object dt dt - strptime(10Jan2006 00:00:15, %d%b%Y %H:%M:%S) dt [1]2006-01-10 00:00:15 I want to make an array of dt, say 100 size. I got those error. [1] 2006-01-10 00:00:15 dtarray-array(dt, dim=c(100)); Error in array(dt, dim = c(100)) : dim- : dims [product 100] do not match the length of object [9] dtarray-array(, dim=c(100)); dtarray[1]-dt; Warning message: number of items to replace is not a multiple of replacement length Any help? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array of lists? is this the best way to do it?
[Q.] How to create an array of lists, or structures the most elegant way?
There have been questions in the past but none too recently...I want to know
if the following looks OK to you guys or if there is a better way to create an
array of lists:
# PREAMBLE ... JUST TO GET THINGS GOING
makeList- function(data, anythingElse) {
rval - list( data = data,
anythingElse = anythingElse
)
class(rval) - myListOfArbitraryThings
return(rval)
}
# make up some arbitrary data
payload- list( as.matrix(cbind(1,1:3)),
10:15,
data.frame(cbind(x=1, y=1:10), fac=sample(LETTERS[1:3], 10, repl=TRUE))
)
# HERE'S THE ARRAY-CONSTRUCTION PART THAT I WANT CRITIQUED:
n- 3 # number of lists in the array of lists
v- vector(list, n) # --- IS THIS THE BEST WAY TO CREATE AN ARRAY OF LISTS?
# fill the array with essentially arbitrary stuff:
for (i in 1:n) v[[i]]- makeList(payload[[i]], i)
Thanks,
Jack.
-
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Re: [R] array of lists? is this the best way to do it?
On 12/6/05, John McHenry [EMAIL PROTECTED] wrote:
[Q.] How to create an array of lists, or structures the most elegant way?
There have been questions in the past but none too recently...I want to know
if the following looks OK to you guys or if there is a better way to create
an array of lists:
# PREAMBLE ... JUST TO GET THINGS GOING
makeList- function(data, anythingElse) {
rval - list( data = data,
anythingElse = anythingElse
)
class(rval) - myListOfArbitraryThings
return(rval)
}
# make up some arbitrary data
payload- list( as.matrix(cbind(1,1:3)),
10:15,
data.frame(cbind(x=1, y=1:10), fac=sample(LETTERS[1:3], 10, repl=TRUE))
)
# HERE'S THE ARRAY-CONSTRUCTION PART THAT I WANT CRITIQUED:
n- 3 # number of lists in the array of lists
v- vector(list, n) # --- IS THIS THE BEST WAY TO CREATE AN ARRAY OF
LISTS?
# fill the array with essentially arbitrary stuff:
for (i in 1:n) v[[i]]- makeList(payload[[i]], i)
You could use lapply to avoid having to set up the empty list:
lapply(1:n, function(i) makeList(payload[[i]], i)) # untested
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[R] Array reversed
Dear R-helper, Is there a command to get an array indexed 1 to T from T to 1? For example: a - c(1, 2, 3) and by applying such a command I can get a[1] = 3 a[2] = 2 a[3] = 1 Thanks a lot and best regards Julio - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Array reversed
Please use R's existing help system before posting to the list.
help.search('reverse') is an obvious first thing to try, don't you think
(and gives you the almost obvious answer immediately)?
-- Bert Gunter
Genentech
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Julio Thomas
Sent: Thursday, December 01, 2005 8:44 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Array reversed
Dear R-helper,
Is there a command to get an array indexed 1 to T from T to 1?
For example:
a - c(1, 2, 3)
and by applying such a command I can get
a[1] = 3
a[2] = 2
a[3] = 1
Thanks a lot and best regards
Julio
-
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[R] array indices in synced vectors
Let us start with the following definitions xxx-rep(c(1,2),times=5) yyy-rep(c(1,2),each=5) a-c(11,12) b-matrix(1:4,2,2) a[xxx] produces [1] 11 12 11 12 11 12 11 12 11 12 b[xxx,yyy] produces [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]111113333 3 [2,]222224444 4 [3,]111113333 3 [4,]222224444 4 [5,]111113333 3 [6,]222224444 4 [7,]111113333 3 [8,]222224444 4 [9,]111113333 3 [10,]222224444 4 so it does an implicit outer for the indices in xxx and yyy. sapply(1:length(xxx),function(x)b[xxx[x],yyy[x]]) does what I need and produces [1] 1 2 1 2 1 4 3 4 3 4 Is there a function taking xxx,yyy, and b as arguments producing the same result? Essentially, I am asking for a version of lapply and/or sapply which works with functions of more than one argument and takes the iteration arguments as vectors or lists of equal length. -- Erich Neuwirth, Didactic Center for Computer Science University of Vienna Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39902 Fax: +43-1-4277-9399 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array indices in synced vectors
On Thu, 8 Sep 2005, Erich Neuwirth wrote: sapply(1:length(xxx),function(x)b[xxx[x],yyy[x]]) does what I need and produces [1] 1 2 1 2 1 4 3 4 3 4 Is there a function taking xxx,yyy, and b as arguments producing the same result? b[cbind(xxx,yyy)] Essentially, I am asking for a version of lapply and/or sapply which works with functions of more than one argument and takes the iteration arguments as vectors or lists of equal length. More generally there is mapply(), but the matrix subscript solution is better in this example mapply(function(i,j) b[i,j], xxx,yyy) [1] 1 2 1 2 1 4 3 4 3 4 -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array indexing and which
Hi R friends! I am stuck with a stupid question: I can circumvent it but I would like to understand why it is wrong. It would be nice if you could give me a hint... I have an 2D array d and do the following: ids - which(d[,1]0) then I have a vector gk with same column size as d and do: ids2 - which(gk[ids]==1) but I can't interprete the result I get in ids2. I get the expected result when I use: which(gk==1 d[,1]0) Why is the first version wrong? The reason why I try to use the ids vectors is that I want to avoid recomputation. Thanks for your help! Werner __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array indexing and which
On Sun, 2005-04-17 at 19:13 +0200, Werner Wernersen wrote: Hi R friends! I am stuck with a stupid question: I can circumvent it but I would like to understand why it is wrong. It would be nice if you could give me a hint... Having a reproducible example, as per the posting guide, would be helpful here. We'll use a contrived example that hopefully explains what I can only presume you are seeing. I have an 2D array d and do the following: ids - which(d[,1]0) Here ids contains the indices of the values in the vector d[, 1] that are 0. For example: d - matrix(sample(0:1, 12, replace = TRUE), ncol = 2) d [,1] [,2] [1,]11 [2,]11 [3,]01 [4,]00 [5,]01 [6,]10 ids - which(d[, 1] 0) ids [1] 1 2 6 Note that c(1, 2, 6) are the indices into the vector: d[, 1] [1] 1 1 0 0 0 1 of the values that are 0. then I have a vector gk with same column size as d and do: ids2 - which(gk[ids]==1) Here ids2 contains the indices of the values in gk[ids] that equal 1. gk - sample(0:1, 6, replace = TRUE) gk [1] 1 1 1 0 1 1 gk[ids] # same as gk[c(1, 2, 6)] [1] 1 1 1 ids2 - which(gk[ids] == 1) ids2 [1] 1 2 3 All three of the values in gk[ids] == 1. but I can't interprete the result I get in ids2. I get the expected result when I use: which(gk==1 d[,1]0) Here you are getting the result of logically comparing the two vectors: gk == 1 [1] TRUE TRUE TRUE FALSE TRUE TRUE AND d[, 1] 0 [1] TRUE TRUE FALSE FALSE FALSE TRUE where the result of the comparison is the index value of each pair in the two vectors where both values are TRUE. Thus: which(gk == 1 d[, 1] 0) [1] 1 2 6 versus: ids2 [1] 1 2 3 Why is the first version wrong? It's not wrong. It is giving you what you asked for. Your question was wrong. :-) The reason why I try to use the ids vectors is that I want to avoid recomputation. Thanks for your help! Werner HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] array indexing and which
You need to think about it just a bit harder. [Hint: what happens if you leave out the first 'which' and just make ids - (d[, 1] 0) does it work then...?] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Werner Wernersen Sent: Monday, 18 April 2005 3:13 AM To: r-help@stat.math.ethz.ch Subject: [R] array indexing and which Hi R friends! I am stuck with a stupid question: I can circumvent it but I would like to understand why it is wrong. It would be nice if you could give me a hint... I have an 2D array d and do the following: ids - which(d[,1]0) then I have a vector gk with same column size as d and do: ids2 - which(gk[ids]==1) but I can't interprete the result I get in ids2. I get the expected result when I use: which(gk==1 d[,1]0) Why is the first version wrong? The reason why I try to use the ids vectors is that I want to avoid recomputation. Thanks for your help! Werner __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Array Manipulation
I have a data set that looks like the following: ID Responce 1 57 1 63 1 49 2 31 2 45 2 67 2 91 3 56 3 43 4 23 4 51 4 61 4 76 4 68 5 34 5 35 5 45 I used sample(unique(ID)) to select a sample if ID's, say, (1,4,5). Now I want to pull out the rows with ID's 1, 4, and 5. I've tried forceing the matrix into a vector but it does not create and appropriate vector. I've also tried the if statment but it didn't work right either. Any suggestions? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] Array Manipulation
Something like: dat[dat$ID %in% sample(unique(dat$ID), 3), ] Andy From: [EMAIL PROTECTED] I have a data set that looks like the following: ID Responce 1 57 1 63 1 49 2 31 2 45 2 67 2 91 3 56 3 43 4 23 4 51 4 61 4 76 4 68 5 34 5 35 5 45 I used sample(unique(ID)) to select a sample if ID's, say, (1,4,5). Now I want to pull out the rows with ID's 1, 4, and 5. I've tried forceing the matrix into a vector but it does not create and appropriate vector. I've also tried the if statment but it didn't work right either. Any suggestions? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Array Manipulation
Liaw, Andy wrote: Something like: dat[dat$ID %in% sample(unique(dat$ID), 3), ] or subset(dat, ID %in% sample(unique(ID), 3)) which I find to be more readable. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Array Manipulation
and something like that: dat[dat$ID == sample(unique(dat$ID), 3), 2] ? I'm not sure about the ,2 maybe you need the full matrix ? ps: first time, i forgot the list __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array problem and for looping
Hi Beter not to give a same name your values (variables) as is function name. R is quite clever and sample - rnorm(10) sample - sample(sample,3) works as expected, but it is not a rule. Cheers Petr On 28 Oct 2004 at 17:54, Kunal Shetty wrote: Dear R- users and Helpers Is there some way to re initialise or clear the array elements? Pardon me for being vague but the problem itself quite vague. I have attached the code along with email. When I run the saved r- code using source(random1.txt) , command. The program runs fine..but at times there is an error see below # ; but again after the error if re-excuted it would work fine?I probably missed some array detail in my program any suggestions #Error in var(parrX[1, ]) : missing observations in cov/cor parrX[] is an array . 2) Also pardon me for the lengthy procedural code but as you could see in it..i have used the for loop for finding the positions (indexes) of the missing values and later carry out updating the new array element values at those particular positions/ So how can I escape the for loop in this case ? i.e get the missing position indexes and save another object ay vector or array ? And also later wanted to use matrix or vector multiplication (%*%) for the updating statement newy[i]- u2 + covXY/varX * (sample$x[i] - u1) is any of the apply function good out here ? I really feel that I am doing something very routine and donkey work and I am most certain that powerful R ? functions could just execute the same 10 liner for loop condition to mere 4 lines ? but how?I am getting lost in the sea of functions here? Thank u for reading Regards Kunal Petr Pikal [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array problem and for looping
Dear R- users and Helpers
Is there some way to re initialise or clear the array elements? Pardon me for being
vague but the problem itself quite vague. I have attached the code along with email.
When I run the saved r- code using source(random1.txt) , command.
The program runs fine..but at times there is an error see below # ; but again after
the error if re-excuted it would work fine
I probably missed some array detail in my
program any suggestions
#Error in var(parrX[1, ]) : missing observations in cov/cor
parrX[] is an array .
2) Also pardon me for the lengthy procedural code
but as you could see in it..i have used the for loop for finding the positions
(indexes) of the missing values and later carry out updating the new array element
values at those particular positions/
So how can I escape the for loop in this case ? i.e get the missing position
indexes and save another object ay vector or array ?
And also later wanted to use matrix or vector multiplication (%*%) for the
updating statement
newy[i]- u2 + covXY/varX * (sample$x[i] - u1)
is any of the apply function good out here ?
I really feel that I am doing something very routine and donkey work and I am most
certain that powerful R functions could just execute the same 10 liner for loop
condition to mere 4 lines ? but how
I am getting lost in the sea of functions here
Thank u for reading
Regards
Kunal
# creation for random data set
x - rnorm(100,17,24)
y - rnorm(100,7,11)
A - matrix(c(10,25,8,40),nrow=2,ncol=2)
z - rbind(x, y) # now 2 x 100
w - A %*% z # 2 x 100
x - w[1,]
y - w[2,]
print(meanX meanY varX varY covar)
# mean of the variates
u1 - mean(x)
u2 - mean(y)
# Variances of the variates
varX - var(x)
varY - var(y)
# coVariances of the variates
covXY - cov(x,y)
# printing mean , var , covar
print(c(u1,u2,varX,varY,covXY))
# now replace randomly with NA
x[sample(1:length(x), 10)] - NA
y[sample(1:length(y), 10)] - NA
# just a variable
sample - data.frame(x=x,y=y)
# Vector for missing values
missing - is.na(sample$x) | is.na(sample$y)
xNA - array(sample$x[missing], dim=c(1,length(sample$x[missing])))
yNA - array(sample$y[missing], dim=c(1,length(sample$y[missing])))
# find the mean of the non missing values in the array
mX - mean(sample$x,na.rm=TRUE)
mY - mean(sample$y,na.rm=TRUE)
# Imputing the missing values with current estimated means
x-array(ifelse(is.na(sample$x), mX, sample$x),dim=c(1,length(sample$x)))
y-array(ifelse(is.na(sample$y), mY, sample$y),dim=c(1,length(sample$y)))
print( meanX meanY varX varY covar)
# algorithm function to find out new estimate values
algoResult - function(parrX,parrY,parrXNA,parrYNA,ictr) {
# Variables for New X and Y
newx - array(parrX, dim=c(1,length(parrX)))
newy - array(parrY, dim=c(1,length(parrY)))
# mean of the variates
u1 - mean(parrX)
u2 - mean(parrY)
# Variances of the variates
varX - var(parrX[1,])
varY - var(parrY[1,])
# coVariances of the variates
covXY - cov(parrX[1,],parrY[1,])
# printing mean , var , covar
print(c(u1,u2,varX,varY,covXY))
# Expected or updated values for the missing values
# loop for finding positon of missing vectors
for (i in 1:length(missing)) {
if (missing[i]==TRUE)
{
if (is.na(sample$x[i]== TRUE))
{
newx[i] - u1 + covXY/varY * (sample$y[i] -u2)
}
if (is.na(sample$y[i]== TRUE))
{
newy[i]- u2 + covXY/varX * (sample$x[i] - u1)
}
}
}
l - list(c(newx),c(newy))
names(l)-c(X1,Y1); as.data.frame(l)
return(l)
}
# the iteration loop to do the algorithm
for (ctr in 1:50) {
Output - algoResult(x,y,xNA,yNA,ctr); Output
x - array(Output$X1, dim=c(1,length(Output$X1)))
y - array(Output$Y1, dim=c(1,length(Output$Y1)))
}
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RE: [R] array problem and for looping
First, the condition if (is.na(sample$x[i]== TRUE)) asks if sample$x[i] is equal to TRUE, and then checks whether the result of this comparison is NA. Because the comparison returns NA when one or the other argument is NA, this works, but note that it would work as well with FALSE in place of TRUE. I suppose you meant to say if (is.na(sample$x[i]) == TRUE) which is the same as if(is.na(sample$x[i])). Second, your code could generate data with *both* x and y missing simultaneously. That would produce missing results in your regression imputation. You probably want to check for these and drop them or otherwise handle them. Third, you might gain some insight by running your function algoResult on known data, and examining the results. The missing values have to come from somewhere, but because you run the entire script which generates random data each time you never get this check. Fourth, you function doesn't use its last 3 arguments, but does use the dataframe sample and the vector missing. As these are not passed as arguments, they are searched for in the enclosing environment. Do you want that? Fifth, yes, you don't need the loop. Just use subscripting for example, as you did to insert NAs randomly, or ifelse as you used to impute by means. Reid Huntsinger -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Kunal Shetty Sent: Thursday, October 28, 2004 1:55 PM To: [EMAIL PROTECTED] Subject: [R] array problem and for looping Dear R- users and Helpers Is there some way to re initialise or clear the array elements? Pardon me for being vague but the problem itself quite vague. I have attached the code along with email. When I run the saved r- code using source(random1.txt) , command. The program runs fine..but at times there is an error see below # ; but again after the error if re-excuted it would work fine...I probably missed some array detail in my program any suggestions #Error in var(parrX[1, ]) : missing observations in cov/cor parrX[] is an array . 2) Also pardon me for the lengthy procedural code but as you could see in it..i have used the for loop for finding the positions (indexes) of the missing values and later carry out updating the new array element values at those particular positions/ So how can I escape the for loop in this case ? i.e get the missing position indexes and save another object ay vector or array ? And also later wanted to use matrix or vector multiplication (%*%) for the updating statement newy[i]- u2 + covXY/varX * (sample$x[i] - u1) is any of the apply function good out here ? I really feel that I am doing something very routine and donkey work and I am most certain that powerful R - functions could just execute the same 10 liner for loop condition to mere 4 lines ? but how...I am getting lost in the sea of functions here... Thank u for reading Regards Kunal __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array construction
Hi all, I've got a file of the following format: X Y Z u v w 0 0 0 x x x 0 0 1 x x x 0 0 2 x x x .. .. .. .. .. .. 0 1 0 x x x 0 1 1 x x x 0 1 2 x x x .. .. .. .. .. .. 1 0 0 x x x 1 0 1 x x x 1 0 2 x x x etc x stand for decimal values X coordinate is 3h, Y and Z are h length. I read that file with: data - read.table(filename, fill=T) I extract coordinates with: coord - data[1:length,1:3]with length the number of lines of the file I get the x values with: values-data[1:length,4:6] I would like now to create an 3D array storing the values on the correct order: uvw - array(data[1:length],dim=c(3h,h,h)) with h predifened of course Is that command correct ? -- Cordialement ~~~ Emmanuel Poizot Cnam/Intechmer B.P. 324 50103 Cherbourg Cedex Tél: (33)(0)233887342 Fax: (33)(0)233887339 ~~~ __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] array addition doesn't recycle!
At 01:39 pm -0500 31/03/04, Raubertas, Richard wrote:
Another alternative is to use the underappreciated function
'sweep()':
sweep(A, 1:2, a, +)
I find the following helpful (it's not due to me but I cannot find
the original poster):
%.+% - function(a,x){sweep(a , 2:1 , x ,+ )}
%+.% - function(a,x){sweep(a , 1:2 , x ,+ )}
A - matrix(1:16,4,4)
x - 10^(0:3)
A %+.% x
[,1] [,2] [,3] [,4]
[1,]26 10 14
[2,] 12 16 20 24
[3,] 103 107 111 115
[4,] 1004 1008 1012 1016
A %.+% x
[,1] [,2] [,3] [,4]
[1,]2 15 109 1013
[2,]3 16 110 1014
[3,]4 17 111 1015
[4,]5 18 112 1016
For 3d arrays this generalizes to
%+..% - function(a,x){sweep(a , 1 , x ,+ )}
%.+.% - function(a,x){sweep(a , 2 , x ,+ )}
%..+% - function(a,x){sweep(a , 3 , x ,+ )}
Then if A - array(1:8,rep(2,3)) and x - c(10,100)
A %+..% x et seq give you a consistent method for addition.
--
Robin Hankin
Uncertainty Analyst
Southampton Oceanography Centre
SO14 3ZH
tel +44(0)23-8059-7743
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[R] array addition doesn't recycle!
Hi, I have noticed the following: a - array(1:4, c(2, 2)) A - array(1:4, c(2,2,2)) A + a Error in A + a : non-conformable arrays It works with a matrix + a vector, why doesn't it work with arrays? Am I missing something? How would you do the above operation efficiently (ie I need to add a matrix to each plane of 3-dim array)? At the moment I am using something like A + array(a, c(2,2,2)) but it doesn't seem that efficient. Thanks Tamas -- Tamás K. Papp E-mail: [EMAIL PROTECTED] (preferred, especially for large messages) [EMAIL PROTECTED] Please try to send only (latin-2) plain text, not HTML or other garbage. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] array addition doesn't recycle!
Another alternative is to use the underappreciated function 'sweep()': sweep(A, 1:2, a, +) Internally this is about the same as your 'A + array(a, c(2,2,2))'. But it has the advantage that it makes explicit what the relationship between the dimensions of 'A' and 'a' is. I find that relying on implicit recycling tends to produce errors that are hard to trace, and code that is hard to understand six months later. Rich Raubertas Merck Co. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian Ripley Sent: Wednesday, March 31, 2004 10:00 AM To: Tamas Papp Cc: R-help mailing list Subject: Re: [R] array addition doesn't recycle! The recycling rules are documented and this is not amongst them. Computer packages do have a tendency to follow their rules rather than read your mind. I suspect A + as.vector(a) is what you intended. On Wed, 31 Mar 2004, Tamas Papp wrote: Hi, I have noticed the following: a - array(1:4, c(2, 2)) A - array(1:4, c(2,2,2)) A + a Error in A + a : non-conformable arrays It works with a matrix + a vector, why doesn't it work with arrays? Am I missing something? How would you do the above operation efficiently (ie I need to add a matrix to each plane of 3-dim array)? At the moment I am using something like A + array(a, c(2,2,2)) but it doesn't seem that efficient. Why do you think is `not that efficient'? Does you have a need to save microseconds? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array of variable length vectors
Hi, I'd like to store N vectors of different lengths, and to be able to access them with an index, and eventually free the memory for one of them without modifying the indexes to the others. In C this would be a vector of N pointers that point to memory cells independently allocated. For example int *pv[3]; pv[0] = (int *) malloc(13 * sizeof(int)); pv[1] = (int *) malloc(7 * sizeof(int)); pv[2] = (int *) malloc(110 * sizeof(int)); free(pv[1]) ... What is the best data type (or class) in R to do such a thing? Thank you! Giampiero _ Giampiero Salvi, M.Sc. www.speech.kth.se/~giampi Speech, Music and Hearing Tel: +46-8-790 75 62 Royal Institute of Technology Fax: +46-8-790 78 54 Drottning Kristinasv. 31, SE-100 44, Stockholm, Sweden __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array of variable length vectors
Giampiero Salvi wrote: Hi, I'd like to store N vectors of different lengths, and to be able to access them with an index, and eventually free the memory for one of them without modifying the indexes to the others. int *pv[3]; pv[0] = (int *) malloc(13 * sizeof(int)); pv[1] = (int *) malloc(7 * sizeof(int)); pv[2] = (int *) malloc(110 * sizeof(int)); free(pv[1]) ... What is the best data type (or class) in R to do such a thing? A list, with vector elements (index starts at 1 in R): pv = list() pv[[1]] = real(13) pv[[2]] = real(7) pv[[3]] = real(110) then the equivalent of freeing the memory and keeping the indexing would be: pv[[2]] = real(0) and NOT pv[[2]] = NULL (which deletes element 2) *BUT* I dont know if R will really free() the memory at that point. You may need to force the garbage collection with gc() Baz __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array of variable length vectors
On Mon, 2 Feb 2004, Giampiero Salvi wrote: Hi, I'd like to store N vectors of different lengths, and to be able to access them with an index, and eventually free the memory for one of them without modifying the indexes to the others. In C this would be a vector of N pointers that point to memory cells independently allocated. For example int *pv[3]; pv[0] = (int *) malloc(13 * sizeof(int)); pv[1] = (int *) malloc(7 * sizeof(int)); pv[2] = (int *) malloc(110 * sizeof(int)); free(pv[1]) ... What is the best data type (or class) in R to do such a thing? Sounds like an R list. However, in R you cannot free memory, but what you can do (carefully) is to change the list element to NULL and then memory will be salvaged at a future garbage collection. z - vector(list, 3) z[[1]] - integer(13) z[[2]] - integer(7) z[[3]] - integer(110) then z[1] - list(NULL) # and not z[[1]] - NULL will potentially release the memory allocated for the first element. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array of variable length vectors
Giampiero Salvi wrote: Hi, I'd like to store N vectors of different lengths, and to be able to access them with an index, and eventually free the memory for one of them without modifying the indexes to the others. In C this would be a vector of N pointers that point to memory cells independently allocated. For example int *pv[3]; pv[0] = (int *) malloc(13 * sizeof(int)); pv[1] = (int *) malloc(7 * sizeof(int)); pv[2] = (int *) malloc(110 * sizeof(int)); free(pv[1]) ... What is the best data type (or class) in R to do such a thing? See ?list Uwe Ligges __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] array(list(),c(2,5)) gives error in R 1.8.1
Hi In R 1.7 the following worked fine: array(list(),c(2,5)) [,1] [,2] [,3] [,4] [,5] [1,] NULL NULL NULL NULL NULL [2,] NULL NULL NULL NULL NULL now in R 1.8.1 I get the error: Error in rep.int(data, t1) : invalid number of copies in rep In addition: Warning message: NAs introduced by coercion thanks for help, I need this possibility for storing objects (lm results) in an array cheers Christoph -- Christoph Lehmann [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] array(list(),c(2,5)) gives error in R 1.8.1
I confirmed this -- array(list(), c(2,2)) works in R 1.6.2 and R 1.7.1, but
not in R 1.8.0. This appears to be due to a change in array(): rep(data,
t1) was changed to rep.int(data, t1). When data=list(), t1==Inf, and
rep(data, t1) returns list(), while rep.int(data, t1) gives an
error. Here's a transcript from R 1.8.0:
array
function (data = NA, dim = length(data), dimnames = NULL)
{
data - as.vector(data)
vl - prod(dim)
if (length(data) != vl) {
t1 - ceiling(vl/length(data))
data - rep.int(data, t1)
if (length(data) != vl)
data - data[1:vl]
}
if (length(dim))
dim(data) - dim
if (is.list(dimnames) length(dimnames))
dimnames(data) - dimnames
data
}
environment: namespace:base
rep(list(), Inf)
list()
rep.int(list(), Inf)
Error in rep.int(list(), Inf) : invalid number of copies in rep
In addition: Warning message:
NAs introduced by coercion
array(numeric(3), 0,0)
numeric(0)
There's also the dangerous construct data[1:v1] in array()
(data[seq(len=v1)] would be much safer). However, it appears that the 1:0
trap doesn't occur under normal circumstances (because if v1=0, then t1
will be either 0 or Inf, and length(rep.int(data, t1)) will be 0 or an
error will have occurred (with most common data types at
least). However^2, given that functions in R don't always produce the
results one might expect, it might be safer to change this to
data[seq(len=v1)].
A workaround is to give array() a data value of the correct length:
array(list()[1:4], c(2,2))
[,1] [,2]
[1,] NULL NULL
[2,] NULL NULL
-- Tony Plate
At Wednesday 10:35 AM 1/14/2004 +0100, you wrote:
Hi
In R 1.7 the following worked fine:
array(list(),c(2,5))
[,1] [,2] [,3] [,4] [,5]
[1,] NULL NULL NULL NULL NULL
[2,] NULL NULL NULL NULL NULL
now in R 1.8.1 I get the error:
Error in rep.int(data, t1) : invalid number of copies in rep
In addition: Warning message:
NAs introduced by coercion
thanks for help, I need this possibility for storing objects (lm
results) in an array
cheers
Christoph
--
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[R] array problem
Dear all,
I define , for n=5 or any integer greater than 0.
A-array((1/2)^n , c(rep(2,n)))
then for any i not equal to j, and 1=i,j=n,
B-apply(a,c(i,j),sum)
now B is a 2 by 2 matrix, I also define another costant 2 by 2 matrix G,
How can I change the values of each elements of array A, according the rule
that,
for example, i=3,j=5,
A[i1,i2,m,i4,l]-A[i1,i2,m,i4,l]*G[m,l]/B[m,l] , where m,l=1,2 and
i1,i2,i4=1,2
I can control this given any i and j, however, I must do the iteration
for i in 1:(n-1) {
for j in (i,n)
{B-apply(a,c(i,j),sum)
here change the value of every elements of A according to my rule
}
}
Is there any easy way to change the value of A in the iteration? Thank you.
_
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RE: [R] array problem
Replace your line that updates A with this:
p - unique(c(i,j,1:5))
f - function(x) diag( matrix(x,4,4) )
AA - apply( outer(A,G/B), p[-(1:2)], f ) # form product
A - aperm( array( AA, dim(A) ), order(p) ) # reshape
Here are a couple of tests. You might want to do some
more tests yourself as well since these are the only
ones I did:
i - 3; j - 5
G - 100 * matrix(1:4,2)
B - 1+0*G # all ones
A - array(1:32,c(2,2,2,2,2))
A2 - A
A2[,,1,,1] - A2[,,1,,1] * G[1,1]
A2[,,1,,2] - A2[,,1,,2] * G[1,2]
A2[,,2,,1] - A2[,,2,,1] * G[2,1]
A2[,,2,,2] - A2[,,2,,2] * G[2,2]
test - function(A,i,j) {
+ p - unique(c(i,j,1:5))
+ f - function(x) diag( matrix(x,4,4) )
+ AA - apply( outer(A,G/B), p[-(1:2)], f ) # form product
+ A - aperm( array( AA, dim(A) ), order(p) ) # reshape
+ A
+ }
identical(test(A,i,j),A2)
[1] TRUE
i - 2; j - 4
A3 - A
A3[,1,,1,] - A3[,1,,1,] * G[1,1]
A3[,1,,2,] - A3[,1,,2,] * G[1,2]
A3[,2,,1,] - A3[,2,,1,] * G[2,1]
A3[,2,,2,] - A3[,2,,2,] * G[2,2]
identical(test(A,i,j),A3)
[1] TRUE
---
Date: Sun, 04 Jan 2004 21:42:55 +0800
From: Z P [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [R] array problem
Dear all,
I define , for n=5 or any integer greater than 0.
A-array((1/2)^n , c(rep(2,n)))
then for any i not equal to j, and 1=i,j=n,
B-apply(a,c(i,j),sum)
now B is a 2 by 2 matrix, I also define another costant 2 by 2 matrix G,
How can I change the values of each elements of array A, according the rule
that,
for example, i=3,j=5,
A[i1,i2,m,i4,l]-A[i1,i2,m,i4,l]*G[m,l]/B[m,l] , where m,l=1,2 and
i1,i2,i4=1,2
I can control this given any i and j, however, I must do the iteration
for i in 1:(n-1) {
for j in (i,n)
{B-apply(a,c(i,j),sum)
here change the value of every elements of A according to my rule
}
}
Is there any easy way to change the value of A in the iteration? Thank you.
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[R] Array Dimension Names
I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Array Dimension Names
Not that I know of. BUT dimnames can themselves have names attributes,
so a very small hack to apply() will do what you want.
I did
dump(apply,file=apply.R)
and added the following lines after dn - dimnames(X) (line 14) [this is
in R 1.7.1].
if (is.character(MARGIN)) {
if (is.null(dn) stop(dimnames(X) must have names)
MARGIN - match(MARGIN,names(dn))
}
and then did
source(apply.R)
x = array(1,dim=c(2,2,2))
dimnames(x) = list(a=1:2,b=1:2,c=1:2)
apply(x,a,sum)
apply(x,c(a,b),sum)
On Fri, 31 Oct 2003 [EMAIL PROTECTED] wrote:
I would like to reference array dimensions by name in an apply and a summary
function. For example:
apply(x, workers, sum)
Is there a better way to do this than creating a new attribute for the array
and then creating new methods for apply and summary? I don't want to name
the individual elements of each dimension (such as with dimnames) but rather
name the dimensions. Thanks for your help.
Benjamin Stabler
Transportation Planning Analysis Unit
Oregon Department of Transportation
555 13th Street NE, Suite 2
Salem, OR 97301 Ph: 503-986-4104
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Box 118525 (ph) 352-392-5697
Gainesville, FL 32611-8525 (fax) 352-392-3704
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RE: [R] Array Dimension Names
Oh yeah, thanks. I thought I might write a function such as getNames() that returns the dimension number of the names of the dimnames of an object. That way I don't have to rewrite apply, sweep, and aperm. But even better would be for R to allow character names in addition to index numbers for the MARGIN argument to apply and sweep, and the perm argument to aperm. Thanks again, Ben Stabler -Original Message- From: Tony Plate [mailto:[EMAIL PROTECTED] Sent: Friday, October 31, 2003 1:07 PM To: STABLER Benjamin; [EMAIL PROTECTED] Subject: Re: [R] Array Dimension Names You can already name dimensions using standard arrays (but you can't use these names for the MARGIN argument of apply) e.g.: x - array(1:6, 3:2, dimnames=list(rows=letters[1:3],cols=LETTERS[24:25])) x cols rows X Y a 1 4 b 2 5 c 3 6 apply(x, 2, sum) X Y 6 15 apply(x, cols, sum) Error in -MARGIN : Invalid argument to unary operator You could pretty easily create your own version of apply() that checked if MARGIN was character, and if it were, matched it against names(dimnames(X)) hope this helps, Tony Plate At Friday 12:50 PM 10/31/2003 -0800, [EMAIL PROTECTED] wrote: I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Array Dimension Names
You could add attributes to your array when creating it and then retrieve them: x - matrix(1:8,2,4) attr(x,workers) - 1 attr(x,variables) - 2 apply(x,attr(x,variables),sum) or perhaps: y - matrix(1:8,2,4) attr(y,margins) - list(workers = 1, variables = 2) apply(y,attr(y,margins)$variables,sum) These are simple enough that you might not need to develop your own apply but you could pretty them up even more if you did: my.apply - function(x,dim,fn) apply(x,attr(x,dim),fn) my.apply(x,variables,sum) --- From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: [R] Array Dimension Names I would like to reference array dimensions by name in an apply and a summary function. For example: apply(x, workers, sum) Is there a better way to do this than creating a new attribute for the array and then creating new methods for apply and summary? I don't want to name the individual elements of each dimension (such as with dimnames) but rather name the dimensions. Thanks for your help. Benjamin Stabler Transportation Planning Analysis Unit Oregon Department of Transportation 555 13th Street NE, Suite 2 Salem, OR 97301 Ph: 503-986-4104 ___ No banners. No pop-ups. No kidding. Introducing My Way - http://www.myway.com __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Array of 3D
Hi, Can be created an Array of 3 dimensions in R? How? Tks, Francisco. ^^ Francisco Júnior, Computer Science - UFPE-Brazil One life has more value that the world whole ^^ __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Array of 3D
Hi, try ?array. So long, Winfried On 29-Jan-03 Francisco do Nascimento Junior wrote: Hi, Can be created an Array of 3 dimensions in R? How? Tks, Francisco. ^^ Francisco Júnior, Computer Science - UFPE-Brazil One life has more value that the world whole ^^ __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help - E-Mail: Winfried Theis [EMAIL PROTECTED] Date: 29-Jan-03 Dipl.-Math. Winfried Theis SFB 475, Fachbereich Statistik, Universitat Dortmund, 44221 Dortmund Tel.: +49-231-755-5903 FAX: +49-231-755-4387 -- __ [EMAIL PROTECTED] mailing list http://www.stat.math.ethz.ch/mailman/listinfo/r-help
