A great example of why you need to read and follow the posting guide
-- this is a plain text list: NO HTML.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic st
On 04/03/17 06:59, exponential wrote:
Hi! I've tried on SO, but without success. Maybe you will be
able
to help. I have the following array (I use 3-dimensional in this example
for simplicity, but it can be 3+ dims - 10,11 or more): a <-
c('a1', 'a2', 'a3') b <- c('bb',
'bbb') c <- c('C', 'CC',
Hi! I've tried on SO, but without success. Maybe you will be able to
help. I have the following array (I use 3-dimensional in this example for
simplicity, but it can be 3+ dims - 10,11 or more): a <- c('a1',
'a2', 'a3') b <- c('bb', 'bbb') c <-
c('C', 'CC', 'CCC') dimNa <- list('a' =
slice.index() in base
On 4/2/2013 9:53 AM, Enrico Bibbona wrote:
Great!
Thanks a lot, Enrico
2013/4/2 Duncan Murdoch
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
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]]
___
Great!
Thanks a lot, Enrico
2013/4/2 Duncan Murdoch
> 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 k
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
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]]
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I will try again.
1. If you have 5 groups, and split each in half, then you have 10 groups.
If MH is applicable to 5 groups, then it should be applicable to the 10
groups as
long as they are disjoint groups.
2. I don't think MH applies to your example because the groups do not have
similar behavi
On Jun 24, 2012, at 5:30 AM, francogrex wrote:
Thanks for your answer. The answer advertises your VCD package,
which by the
way is a very nice package that I use and recommend for everyone
doing such
kind of data analysis. However if you really examine the answer you
gave me,
it does not
Thanks for your answer. The answer advertises your VCD package, which by the
way is a very nice package that I use and recommend for everyone doing such
kind of data analysis. However if you really examine the answer you gave me,
it does not really or specifically answer my question.
--
View this
You have a very nice graph of a dose-response function here.
library(vcd)
library(RColorBrewer)
Pen <- array(c(0, 0, 6, 5,
3, 0, 3, 6,
6, 2, 0, 4,
5, 6, 1, 0,
2, 5, 0, 0),
dim = c(2, 2, 5),
dimnames = list(
Delay = c("None", "1.5h"),
If we take the matel-haenszel test on these data of five 2x2 tables
stratified along Penicillin.Levels
array(c(0, 0, 6, 5,
3, 0, 3, 6,
6, 2, 0, 4,
5, 6, 1, 0,
2, 5, 0, 0),
dim = c(2, 2, 5),
dimnames = list(
Delay = c("None", "1.5h"),
0.1931739 -0.5358076 -0.1318054
[1] 0.4009829 0.8228106 0.9057512
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for (j in 1:180) {
ssta_winter[i,j,k] <- mean(temp[i,j,])
}
}
}
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Sent from the R help mailing l
800321 1.996305
[1] 1.046745 1.335898 1.245474
[1] 0.6113644 0.9268212 1.2416964
[1] 1.220745 1.864054 1.566535
[1] -0.5436361 0.4582486 1.8623419
[1] 0.1931739 -0.5358076 -0.1318054
[1] 0.4009829 0.8228106 0.9057512
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.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
[1] 0.583
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PLEASE do read the posting gu
Asher Meir wrote on 12/30/2011 03:45:48 AM:
> Yes this is the solution to my problem.
>
> I think there is a small typo in old.index, instead of:
>
> k=rep(1:nks, rep(njs*nks, nks)))
>
> it should be:
> k=rep(1:nks, rep(nis*njs, nks)))
> That is, the "k" index should be repli
Yes this is the solution to my problem.
I think there is a small typo in old.index, instead of:
k=rep(1:nks, rep(njs*nks, nks)))
it should be:
k=rep(1:nks, rep(nis*njs, nks)))
That is, the "k" index should be replicated by nis*njs,, not nks*njs.
Many thanks
Asher
On Fri, Dec 3
Asher Meir wrote on 12/29/2011 04:47:21 PM:
> Hello Jean.
>
> Thank you for your prompt response.
>
> I have an economic model. I start out with a list of asset levels
> 1:nks. For every asset level, a person has nis possible income
> levels and njs possible expense levels, with probabilities
Asher Meir wrote on 12/29/2011 02:58:00 PM:
> I want to create a new array which selects values from an original array
> based on a function of the indices. That is:
>
> I want to create a new matrix Vnew[i,j,k]=Vold[i,j,ks] where ks is a
> function of the index elements i,j,k. I want to do this
rg] On Behalf Of Asher Meir
> Sent: Thursday, December 29, 2011 1:58 PM
> To: r-help@r-project.org
> Subject: [R] Array element is function of its position in the array
>
> I want to create a new array which selects values from an original
> array
> based on a function of the in
I want to create a new array which selects values from an original array
based on a function of the indices. That is:
I want to create a new matrix Vnew[i,j,k]=Vold[i,j,ks] where ks is a
function of the index elements i,j,k. I want to do this WITHOUT a loop.
Call the function "ksfunction", and th
This worked example, hoping to be helpful, has been requested after a (my)
further enquiry about array manipulation.
I was looking for a command that is equivalent to repmat() in matlab and
that could also be applied to array.
(for Matlab users)
The Matlal code was the following:
1)temp_u=zeros
This worked example, hoping to be helpful, has been requested after a (my)
further enquiry about array manipulation.
I was looking for a command that is equivalent to repmat() in matlab and
that could also be applied to array.
(for Matlab users)
The Matlal code was the following:
temp_u=zeros(d
You seem to be looking for 'aperm'.
There is a chapter in 'S Poetry' (available
on http://www.burns-stat.com) that talks
about working with higher dimensional
arrays. I don't think any changes need
to be made for R.
On 02/11/2011 16:16, Simone Salvadei wrote:
Hello,
I'm at the very beginning o
On Nov 2, 2011, at 12:16 PM, Simone Salvadei wrote:
Hello,
I'm at the very beginning of the learning process of this language.
Sorry in advance for the (possible but plausible) stupidity of my
question.
I would like to find a way to permute the DIMENSIONS of an array.
Something that sounds
Hello,
I'm at the very beginning of the learning process of this language.
Sorry in advance for the (possible but plausible) stupidity of my question.
I would like to find a way to permute the DIMENSIONS of an array.
Something that sounds like the function "permute()" in matlab.
Given an array C
Understood now. Thanks Duncan.
Muhammad
On Sat, 18 Jun 2011, Duncan Murdoch wrote:
On 11-06-18 10:45 AM, Muhammad Rahiz wrote:
Hi,
Can someone advise why the followind did not produce the array, given
the condition specified?
s<- 1
a1<- array(dim=c(1,4))
a2<- array(dim=c(2,4))
arr<- ifelse(s
Try using 'if':
> s <- 1
> a1 <- array(dim=c(1,4))
> a2 <- array(dim=c(2,4))
> arr <- if (s == 1) a1 else a2
> str(arr)
logi [1, 1:4] NA NA NA NA
>
On Sat, Jun 18, 2011 at 10:45 AM, Muhammad Rahiz
wrote:
> Hi,
> Can someone advise why the followind did not produce the array, given the
> condi
On 11-06-18 10:45 AM, Muhammad Rahiz wrote:
Hi,
Can someone advise why the followind did not produce the array, given
the condition specified?
s<- 1
a1<- array(dim=c(1,4))
a2<- array(dim=c(2,4))
arr<- ifelse(s==0,a1,a2)
See the Value section of ?ifelse.
Duncan Murdoch
__
Hi,
Can someone advise why the followind did not produce the array, given
the condition specified?
s <- 1
a1 <- array(dim=c(1,4))
a2 <- array(dim=c(2,4))
arr <- ifelse(s==0,a1,a2)
Thanks.
Muhammad
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abind works well for this example.
a1 <- array(1:18, dim=c(3,3,2))
a2 <- array(101:150, dim=c(5,5,2))
a1b <- abind(a1, array(400, dim=c(2,3,2)), along=1)
a1c <- abind(a1b, array(500, dim=c(5,2,2)), along=2)
dim(a1c)
a1
a2
a1c
On Mon, Mar 7, 2011 at 6:54 AM, Usman Munir wrote:
> Hi,
>
> I ha
Hi,
I have two 3 D arrays. Both are of this form
array_1<- array[n,n,k]
array_2<-array[m,m,k]
Lets say n=83 and m=80
Since n>m. I would like to add rows and columns to array_2 to make them
equal. I want to keep the size of the third dimension fixed i.e.. k.
i.e.
if (nrow(array_1)>nrow(array_2))
Thank you Joshua!! StatTemps[c(1:7, 22:34), ] that's what i needed! I
needed different ranges of rows but it is exactly what i was looking for.
Thanks so much!! R can be kind of fun! hehe
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Hi Brian,
I believe there was some miscommunication earlier due to R's array
class for objects and the colloquial usage of array (the idea that
'array' is used colloquially is a bit odd, but I digress). In any
case, here are some steps I take (certainly not the only ones) when
exploring a new dat
if you can load the PASWR package and pull up StatTemps you will see what I
am talking about. Otherwise I fear that my question will just be confusing.
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Instead of:
7:11&23:34
I think you mean:
c(7:11, 23:34)
Using '&' for concatenation
is not an unreasonable idea,
but it is decidedly not what
R does.
It would be instructive to do:
7:11 & 23:34
at the R prompt to see what you
get.
On 29/11/2010 03:56, bfhancock wrote:
Josh, the d
On Nov 28, 2010, at 10:56 PM, bfhancock wrote:
Josh, the data set is called StatTemps and is in the PASWR package.
I want
to make an array that involves only the 8 a.m. and a separate array
that
involves only the 9 a.m. so i can get info on the temperatures in
those
groups. So I still w
Josh, the data set is called StatTemps and is in the PASWR package. I want
to make an array that involves only the 8 a.m. and a separate array that
involves only the 9 a.m. so i can get info on the temperatures in those
groups. So I still want it in the format of StatTemps but in two arrays that
Hi B,
What you need to do is pass a vector with the indices you want to
extract. So, you have 1:7 (which expands to 1, 2, 3, ... 7) and 12:34
(which again expands). How would you combine two sets of numbers?
c(), the combine or concatenate function. Putting this in action:
mya <- array(1:510,
Hi! I am learning R and have a question that is probably fairly simple for
those of you much more learned than I.
I am messing with Arrays and am doing some simple stuff to get the hang of
them. I will have a seperate array already pulled up and it will have
columns and rows. I figured out th
Hello everyone.
I would like to create some agents that span over a specific area map.Every
agent needs to have its own data structures like one or two matrices and one
list.
I think that the best way to do this is to create objects and every instance of
an object will be used for a single agen
On 08/18/2010 06:16 AM, navishkumarb wrote:
Hello
Can any one let me know how to delete a value from an array and then push
back rest of the remaining elements up into an array.
Thanks
Reproducible examples help us understand what you mean.
x <- 1:10
x[-1]
Is that what you mean?
Hello
Can any one let me know how to delete a value from an array and then push
back rest of the remaining elements up into an array.
Thanks
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Hello.
Can I just know how to compare two different arrays, but take only the
elements from first array (both arrays are multidimensional).
Its just opposite of "intersect" option in R.
Thanks in advance
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-project.org
Subject: [R] array manipulation
Hello listeRs,
I'm trying to make a square radius around a given reference point. So
given the following array, how can I manipulate it so that
x0 <- array(1,dim=c(5,5))
x0
1 1 1 1 1
1 1 1 1 1
1 1 *1* 1 1
1 1 1 1 1
1 1 1 1 1
becomes
into
3
Hello listeRs,
I'm trying to make a square radius around a given reference point. So
given the following array, how can I manipulate it so that
x0 <- array(1,dim=c(5,5))
x0
1 1 1 1 1
1 1 1 1 1
1 1 *1* 1 1
1 1 1 1 1
1 1 1 1 1
becomes
into
3 3 3 3 3
3 2 2 2 3
3 2 *1* 2 3
3 2 2 2 3
3 3 3 3 3
Hi:
Alternatively, you could use rollmean in the zoo package:
library(zoo)
x <- c(2, 4, 5, 5, 6, 4, 5, 2, 1)
rollmean(x, 4)
[1] 4.00 5.00 5.00 5.00 4.25 3.00
which avoids the NAs that get produced from filter:
filter(x, rep(1/4, 4))
Time Series:
Start = 1
End = 9
Frequency = 1
[1] NA 4.00 5.00
you mean something along the lines of
filter(x, rep(1/4, 4))
(which you can combine with na.omit)
?
On Wed, Feb 17, 2010 at 5:18 PM, Mohsen Jafarikia wrote:
> Hello All:
>
> If I do have:
>
> x = (2, 4, 5, 5, 6, 4, 5, 2, 1)
> y = (9, 11.5, 12.5, 13, 14, 19, 20, 21, 22)
>
> I wanted to find a s
Hello All:
If I do have:
x = (2, 4, 5, 5, 6, 4, 5, 2, 1)
y = (9, 11.5, 12.5, 13, 14, 19, 20, 21, 22)
I wanted to find a simple function in R which calculates the averages of X
and Y for every 4 unit increase of Y. The results should look like:
x = (4, 6, 3) #where (2+ 4+ 5+ 5)/ 4
See
Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation in
Plots. R News 2 (3), 32-34.
http://www.r-project.org/doc/Rnews/Rnews_2002-3.pdf
Uwe Ligges
Koen wrote:
Hello,
I am trying to include legend text with math symbols from a predefined
character variable that is read i
Hello,
I am trying to include legend text with math symbols from a predefined
character variable that is read in from a file.
If there is only one line of text in the legend, the following, although
cumbersome, works for me:
> LegendText = " 'U' [infinity], '=10 m/s' " # (read in from a fil
Works a charme; thanks much everybody. Daniel.
David Winsemius wrote:
>
>
> On Sep 27, 2009, at 6:33 PM, Daniel Malter wrote:
>
>>
>> Hi, I have an array of dimension 6:6:16. I want to stack the
>> 16
>> slices of the array into a matrix or a data frame of dimension
>> 6:(6*16=9
On 28/09/2009, at 11:51 AM, David Winsemius wrote:
apply(x, 2 , I)
***Much*** sexier than my solution!
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and co
Try this:
matrix(aperm(x, c(1,3,2)), nc = 6)
On Sun, Sep 27, 2009 at 6:33 PM, Daniel Malter wrote:
>
> Hi, I have an array of dimension 6:6:16. I want to stack the 16
> slices of the array into a matrix or a data frame of dimension
> 6:(6*16=96) in order to write it to a csv fil
On Sep 27, 2009, at 6:33 PM, Daniel Malter wrote:
Hi, I have an array of dimension 6:6:16. I want to stack the
16
slices of the array into a matrix or a data frame of dimension
6:(6*16=96) in order to write it to a csv file. It seems I
cannot
get the matrix or data.frame f
Hi, I have an array of dimension 6:6:16. I want to stack the 16
slices of the array into a matrix or a data frame of dimension
6:(6*16=96) in order to write it to a csv file. It seems I cannot
get the matrix or data.frame functions to put the values from the array in
the same order
Jaffe
Subject: Re: [R] array slice notation?
To: r-help@r-project.org
Message-ID: <24819883.p...@talk.nabble.com>
Content-Type: text/plain; charset=UTF-8
Very nice. Two questions:
1> Do you have any idea of the timing difference, if any, between this and
the vector-subscripting metho
-boun...@r-project.org] På vegne
af Steve Jaffe [sja...@riskspan.com]
Sendt: 5. august 2009 04:45
Til: r-help@r-project.org
Emne: Re: [R] array slice notation?
Very nice. Two questions:
1> Do you have any idea of the timing difference, if any, between this and
the vector-subscripting method?
2
Very nice. Two questions:
1> Do you have any idea of the timing difference, if any, between this and
the vector-subscripting method?
2> How do you generalize this to select multiple rows eg with indexes given
by a vector 'v'?
Søren Højsgaard wrote:
>
> You can do
>> A <- HairEyeColor
>> do.ca
Hi Steve,
I had the same problem when writing the hierobarp function in plotrix.
Have a look at the source code (v2.6-4 in my copy lines 128-133) as the
solution seems to work well.
Jim
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correction -- that would work for a single row, if you want the result to be
an array with one fewer dimensions. But in general you get an array of the
same dimension you started with (where the first dimension may be length 1).
So:
dim(slice) <- c(length(v), dim(arr)[-1])
Although you *do* hav
g [r-help-boun...@r-project.org] På vegne
af Steve Jaffe [sja...@riskspan.com]
Sendt: 4. august 2009 21:23
Til: r-help@r-project.org
Emne: [R] array slice notation?
Suppose I have an n-diml array A and I want to extract the first "row" -- ie
all elements A[1, ...]
Interactively if I know
Although you *do* have to re-assign the dimensions, otherwise the result is
just a flat vector, ie
slice <- A[ outer(v, dim(A)[1]*( 1:prod(dim(A)[-1])-1 ), '+') ]
dim(slice) <- dim(A)[-1]
Steve Jaffe wrote:
>
> A[ outer(v, dim(A)[1]*( 1:prod(dim(A)[-1])-1 ), '+') ]
>
--
View this message i
Yes, I was thinking more in terms of mental operations than physical.
I think the following works, but it doesn't seem entirely transparent :-)
Given array A, and a vector of row indices v (ie 1 <= v <= dim(A)[1]), the
slice of rows v is
A[ outer(v, dim(A)[1]*( 1:prod(dim(A)[-1])-1 ), '+') ]
Hi,
On Aug 4, 2009, at 3:23 PM, Steve Jaffe wrote:
Suppose I have an n-diml array A and I want to extract the first
"row" -- ie
all elements A[1, ...]
Interactively if I know 'n' I can write A[1,] with (n-1) commas.
How do I do the same more generally, eg in a script?
(I can think of
Suppose I have an n-diml array A and I want to extract the first "row" -- ie
all elements A[1, ...]
Interactively if I know 'n' I can write A[1,] with (n-1) commas.
How do I do the same more generally, eg in a script?
(I can think of doing this by converting A to a vector then extracting t
Bronagh Grimes wrote:
Hi there,
Just wondering if anyone has any tips for using arrays?
I am trying to convert the following SAS code to R:
You're probably going to get more useful responses in general if, instead of
posting code from SAS to be ported to R, you describe what the cod
Thanks for this, will have a look now.
Much appreciated,
-Original Message-
From: Peter Dalgaard [mailto:p.dalga...@biostat.ku.dk]
Sent: 24 April 2009 11:40
To: Bronagh Grimes
Cc: r-help@r-project.org
Subject: Re: [R] Array
Bronagh Grimes wrote:
> Just wondering if anyone has any t
Bronagh Grimes wrote:
> Just wondering if anyone has any tips for using arrays?
You're presupposing that R works like SAS. It doesn't.
This looks like a job for reshape (either Hadley Wickham's package or
the built-in function).
> I am trying to convert the following SAS code to R:
> data A2;
Hi there,
Just wondering if anyone has any tips for using arrays?
I am trying to convert the following SAS code to R:
data A2;
set A1;
by subject_id;
retain BX1-BX10i;
array b(1:10) BX1-BX10 ;
if first.subject_id then do ;
do j=1 to 10;
b(j) =
On Feb 24, 3:57 pm, jdeisenberg wrote:
> Λεωνίδας Μπαντής wrote:
>
> > 1. Suppose I have a<-c(1:10) (or a<-array(c(1:10),dim=c(1,10)))
>
> > and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
> > substitute alla elements that are <5 with 0 and >5 with 1.
>
> I think you
2009/2/23 Λεωνίδας Μπαντής :
>
> Hi there,
>
> I am pretty new to R. Actually I started using it yesterday. I have two
> questions:
>
> 1. Suppose I have a<-c(1:10) (or a<-array(c(1:10),dim=c(1,10)))
>
> and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
> substitute alla
Λεωνίδας Μπαντής wrote:
>
> 1. Suppose I have a<-c(1:10) (or a<-array(c(1:10),dim=c(1,10)))
>
> and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
> substitute alla elements that are <5 with 0 and >5 with 1.
>
I think you mean >=5, not > 5. In that case, try this:
Hi there,
I am pretty new to R. Actually I started using it yesterday. I have two
questions:
1. Suppose I have a<-c(1:10) (or a<-array(c(1:10),dim=c(1,10)))
and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
substitute alla elements that are <5 with 0 and >5 with
Hi all,
I have a data frame of coordinates, "coord". I need an "output"
(array) where the number of rows = number of rows in "coord".
I am trying to use the "which" function to extract indices from the
coordinates with the following code:
d=2
nd=length(coord[,1])
output=list(array)
...
...
which(
How about
c(rbind(A,B,C))
a<-1:5
> b<-11:15
> c<-21:25
> rbind(a,b,c)
[,1] [,2] [,3] [,4] [,5]
a12345
b 11 12 13 14 15
c 21 22 23 24 25
> c(rbind(a,b,c))
[1] 1 11 21 2 12 22 3 13 23 4 14 24 5 15 25
On Mon, Dec 29, 2008 at 4:01 AM, ykank wrote:
t; -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of ykank
> Sent: Monday, December 29, 2008 2:01 AM
> To: r-help@r-project.org
> Subject: [R] Array Making
>
>
> Dear R Users
>
> Suppose I've certain valu
Dear R Users
Suppose I've certain values of A[1], A[2], A[3]
,A[4].B[1],B[2],B[3]..C[1],C[2],C[3].so on..All A,B,Cs are some
numeric constant.
I want to make an array which will look like [A[1], B[1], C[1], A[2], B[2],
C[2], A[3], B[3] ,C[3]
Please suggest me any way to do i
Ok- it works now after flushing my objects. The package Oarray doesn't seem
to work under version 2.8.0. I emailed Jonathan about it and he said he
won't get to updating it until next year. In the meantime, I was trying to
produce a seat-of-the-pants workaround and I probably redefined "as.arra
On 3/11/2008, at 2:11 PM, David Stoffer wrote:
What happened? TIA.
In version 2.7.x:
(x <- array(1:4, c(2,2)))
[,1] [,2]
[1,]13
[2,]24
as.array(x)
[,1] [,2]
[1,]13
[2,]24
In version 2.8.0:
(x <- array(1:4, c(2,2)))
[,1] [,2]
[1,]1
What happened? TIA.
In version 2.7.x:
> (x <- array(1:4, c(2,2)))
[,1] [,2]
[1,]13
[2,]24
> as.array(x)
[,1] [,2]
[1,]13
[2,]24
In version 2.8.0:
> (x <- array(1:4, c(2,2)))
[,1] [,2]
[1,]13
[2,]24
> as.array(x)
Error: evaluati
If ind is the vector of rows of dat you are interested in then dat[ind,] is
what you need.
--- On Wed, 18/6/08, Paul Adams <[EMAIL PROTECTED]> wrote:
> From: Paul Adams <[EMAIL PROTECTED]>
> Subject: [R] array elements incorrect
> To: r-help@r-project.org
> Received:
Hello everyone,
I have a question as to what code should be used if
one wanted to subset from a dataframe of 7000rows by 38
columns.For example if you wanted to generate a array of
100 by 38 would you not use the following:
Z<-(dat,dim=c(100,38))
where dat is the dataframe of 7000by 38
Whenever I
Thanks to all three of you.
I used a 4 dimensional array as you suggested.
I will have to do a lot of loops with loop variables i and j. I should
perhaps save my results from time to time, no ? If I need to save "object1"
and "object2",
the command should be save(object1,object2,file=paste("save"
Is it possible to have one array for bias and one for variance? such as..
biasAr=
b1 b2
a1 biasbias
a2 biasbias
and
varAr=
b1 b2
a1 var var
a2 var var
then you can combine the two in a data frame?
thanks
y
Michael Prince wrote:
>
> Dear R users,
>
> I
Dear R users,
I want to calculate the bias and variance of an estimator for several values
of two parameters a and b.
For example :
b1 b2
a1 bias bias
variance variance
a2 bias bias
variance variance
Can one do array of arrays ? I have tried and it did not wor
On 5/13/2008 12:47 AM, Knut M. Wittkowski wrote:
Why does the assignment of a 3178x93 object to
another 3178x93 object remove the dimension attribute?
Your example is not reproducible. When I make one that is reproducible,
I don't see the error:
> GT <- array(dim = c(6,3178,93))
> dim(GT)
[
>Why does the assignment of a 3178x93 object to >another 3178x93 object remove
>the dimension attribute?>> GT <- array(dim = c(6,nrow(InData),ncol(InSNPs)))>>
>dim(GT)>[1] 6 3178 93>> SNP1 <- InSNPs[InData[,"C1"],]>> dim(SNP1)>[1] 3178
> 93>> SNP2 <- InSNPs[InData[,"C2"],]>> dim(SNP2)>[1] 31
Why does the assignment of a 3178x93 object to
another 3178x93 object remove the dimension attribute?
> GT <- array(dim = c(6,nrow(InData),ncol(InSNPs)))
> dim(GT)
[1]6 3178 93
> SNP1 <- InSNPs[InData[,"C1"],]
> dim(SNP1)
[1] 3178 93
> SNP2 <- InSNPs[InData[,"C2"],]
> dim(SNP2)
[1
Hello,
I need help to build an array within an array, i. e., I have this:
tt[,,c(1,2)]
, , metadados.class_7.R
CS WRC LRA
Inicial 1.000 1.000 1.000
Final 0.5974482 0.6095162 0.5866560
Indep 0.4335460 0.4799575 0.4169591
Inicial 0.9925572 0.9925572 0.99
Thank both of you for the suggestions!
Gang
On Mar 6, 2008, at 2:12 PM, Henrique Dallazuanna wrote:
> Hum you are rigth, I forgot of 'else'.
>
> On 06/03/2008, Benilton Carvalho <[EMAIL PROTECTED]> wrote:
>> no, it won't.
>>
>> you're doing the right math on the "valid" subset... but you're no
no, it won't.
you're doing the right math on the "valid" subset... but you're not
returning the zeros where needed therefore, the whole thing will
get recycled to match the dimensions.
b
On Mar 6, 2008, at 2:03 PM, Henrique Dallazuanna wrote:
I think this should work:
array(A[abs(B)
Hum you are rigth, I forgot of 'else'.
On 06/03/2008, Benilton Carvalho <[EMAIL PROTECTED]> wrote:
> no, it won't.
>
> you're doing the right math on the "valid" subset... but you're not
> returning the zeros where needed therefore, the whole thing will
> get recycled to match the dimension
I think this should work:
array(A[abs(B) > 10e-5]/B[abs(B) > 10e-5], dim=c(L, M, N, P))
On 06/03/2008, Gang Chen <[EMAIL PROTECTED]> wrote:
> I have two arrays A and B with dimensions of (L, M, N, P) and (L, M,
> N), and I want to do
>
> for (i in 1:L) {
> for (j in 1:M) {
> for (k in 1:N) {
apparently you forgot the "commented, minimal, self-contained,
reproducible code" part...
L = 10
M = 20
N = 30
P = 40
set.seed(1)
A = array(rnorm(L*M*N*P), dim=c(L, M, N, P))
B = array(rnorm(L*M*N), dim=c(L, M, N))
B[sample(100, 10)] = 0
C = array(0, dim=c(L, M, N, P))
for (i in 1:L) {
for (j i
I have two arrays A and B with dimensions of (L, M, N, P) and (L, M,
N), and I want to do
for (i in 1:L) {
for (j in 1:M) {
for (k in 1:N) {
if (abs(B[i, j, k]) > 10e-5) C[i, j, k,] <- A[i, j, k,]/B[i, j, k]
else C[i, j, k,] <- 0
}
}
}
How can I get C more efficiently than looping?
Thanks
Hello,
I am trying to run a piece of Fortran code from R, and I am having trouble
passing an array to the fortran subroutine. My attempts to pass an array into
the fortran are producing memory errors, and I cannot find an example that
performs the task. I have looked at "Writing R Extensions
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