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
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
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
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
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
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
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
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
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
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
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
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
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
+
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
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
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.
@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
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 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
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
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
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,
[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
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
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
] 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
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,]11111333
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
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 -
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
: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
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,
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
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
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
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
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
, 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
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
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 ,+ )}
%+.% -
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
: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
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
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 *
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
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
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
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.
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,
[,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
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
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
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
,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
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
^^
__
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
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