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] extract a subset of non-contiguous elements of a matrix

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

> [1,]  1  1
> [2,]  2  1
> [3,]  1  2

how to combine these elemnts? If the 2nd col contains the number of the resulting matrix, then, the number of rows are different and the matrix can't be completed.


On Thursday, June 19, 2014 3:39 PM, Frede Aakmann Tøgersen <frtog@vestas.com> wrote:


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 be a matrix of characters. There you go from a 3×4 matrix to a 3×2 matrix.  What do you want in case of Barts 2×2 matrix? A 3×1 or 1×3 matrix? And in a more general case?

Best regards

Frede


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-------- Oprindelig meddelelse --------
Fra: carol white
Dato:19/06/2014 15.18 (GMT+01:00)
Til: Bart Kastermans
Cc: r-help@r-project.org
Emne: Re: [R] extract a subset of non-contiguous elements of a matrix

 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 tm.1 with the following indexes

1,1 1,3
2,1 2,3
3,1 3,3



On Thursday, June 19, 2014 3:08 PM, Bart Kastermans <kasterma@kasterma.net> wrote:



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_crl@yahoo.com> wrote:

> well it gives a vector which is useless as I want a matrix.
>
>
> On Thursday, June 19, 2014 2:40 PM, Bart Kastermans <kasterma@kasterma.net> wrote:
>
>
> > tm.1 <- matrix(c(11,22,33,-4), ncol=2)
> > which(tm.1 > 0, arr.ind=TRUE)
>   row col
> [1,]  1  1
> [2,]  2  1
> [3,]  1  2
> > tm.1[which(tm.1 > 0, arr.ind=TRUE)]
> [1] 11 22 33
>
> This last command does what you ask I think.
>
> On 19 Jun 2014, at 14:12, carol white <wht_crl@yahoo.com> wrote:
>
> > 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 Kastermans <kasterma@kasterma.net> wrote:
> >
> >
> > On 19 Jun 2014, at 13:19, carol white <wht_crl@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 number. This I can do
> > >
> > > which(m>0, arr.ind=T) which gives the indices of positive elements like (37,1), (80,1), ..., (54,2) etc. How can I extract positive numbers without looping on the indexes provided by which to make a new matrix?
> >
> > What matrix do you want?  For e.g.
> >
> > > tm.1 <- matrix(c(11,22,33,-4), ncol=2)
> > > which(tm.1 > 0, arr.ind=TRUE)
> >    row col
> > [1,]  1  1
> > [2,]  2  1
> > [3,]  1  2
> > > tm.1[which(tm.1 > 0, arr.ind=TRUE)]
> > [1] 11 22 33
> >
> > The extracted values do not form a matrix.
> >
> > Either the above contains the answer, or I don’t understand the question.
> >
> > Best,
> > Bart
> >
> >
> > >
> > > thanks,
> > >
> > > Carol
> >
> >
>
>
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