Hi
You can pad some NA to the result before transforming it to required matrix but
as Frede pointed out, what do you want to do with such result?
tm.1=rbind(c(1,-3,2,-4), c(1,-3,2,-4),c(1,-3,2,-4))
xx-which(tm.1 0, arr.ind=TRUE)
res-apply(xx, 1, paste0, collapse=,)
dim(res)-c(3,2)
res
Hi,
Is there a way to extract a subset of non-contiguous elements of a matrix
elegantly and with 1 or very few scripts?
Suppose I have a matrix of positive and negative numbers (m) and I want to
retrieve only the positive number. This I can do
which(m0, arr.ind=T) which gives the indices of
On 19 Jun 2014, at 13:19, carol white wht_...@yahoo.com wrote:
Hi,
Is there a way to extract a subset of non-contiguous elements of a matrix
elegantly and with 1 or very few scripts?
Suppose I have a matrix of positive and negative numbers (m) and I want to
retrieve only the positive
The extracted values don't form a matrix and that's the question how to extract
because which returns the indexes? that is, from
1,1
2,1
1,2
how to retrieve values?
Or if at the position 2,1, there is a negative value, how to retrieve
1,1
1,2
Carol
On Thursday, June 19, 2014 1:29 PM, Bart
This _was_ in the answer you got, but to clarify things, perhaps try this:
(M - matrix(1:9,3,3))
(ix - rbind(c(3,2),c(1,3)))
M[3,2]
M[1,3]
M[ix]
-pd
On 19 Jun 2014, at 14:12 , carol white wht_...@yahoo.com wrote:
The extracted values don't form a matrix and that's the question how to
If you give an example of input and desired output I can think about this. But
at this
point I do not understand what you want. In the example I gave the positive
elements do
not form a submatrix in any way I can think of.
On 19 Jun 2014, at 15:04, carol white wht_...@yahoo.com wrote:
well
 tm.1=rbind(c(1,-3,2,-4), c(1,-3,2,-4),c(1,-3,2,-4))
which(tm.1 0, arr.ind=TRUE)
    row col
[1,]Â Â 1Â Â 1
[2,]Â Â 2Â Â 1
[3,]Â Â 3Â Â 1
[4,]Â Â 1Â Â 3
[5,]Â Â 2Â Â 3
[6,]Â Â 3Â Â 3
so the answer should have the elements of tm.1 with the following indexes
1,1 1,3
2,1 2,3
On 19 Jun 2014, at 15:16, carol white wht_...@yahoo.com wrote:
tm.1=rbind(c(1,-3,2,-4), c(1,-3,2,-4),c(1,-3,2,-4))
which(tm.1 0, arr.ind=TRUE)
row col
[1,] 1 1
[2,] 2 1
[3,] 3 1
[4,] 1 3
[5,] 2 3
[6,] 3 3
so the answer should have the elements of
As Peter and Bart I really have a problem understanding you.
Perhaps if you tell us what your desired result is going to used for we can be
more helpful. You can do that using your latest example.
In that example you want a matrix of sets of row and column indices. This will
probably have to
I realize that that the problem arises if there is a different number of
negative numbers in the rows and columns of the original matrix. In this case,
the resulting matrix won't have the same number of rows for all columns. The
problem for ex doesn't arise for my example but for Bart's example
You are really not helpful here. Are we talking the same scientific language?
Br. Frede
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Oprindelig meddelelse
Fra: carol white
Dato:19/06/2014 15.46 (GMT+01:00)
Til: Frede Aakmann Tøgersen ,Bart Kastermans
Cc: r-help@r-project.org
Emne: Re: [R]
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