On Aug 28, 2010, at 2:54 AM, Joshua Wiley wrote:

Is this sufficiently single steppish for you?

D <- A <- matrix(1:16, 4)
D[3, ] <- 2 * D[1, ] + D[3, ]

# Alternately, you could do this
# but it is much messier, and I do not see how
# two steps is really an issue
# you want to end up with two matrices anyways
# so it's not like you save memory by only making one assignment
A2 <- matrix(1:16, 4)
D2 <- rbind(A2[1:2, ], 2 * A2[1, ] + A2[3, ], A2[4, ])

You can do it the "matrix-ey" way as well by creating a row extractor/ multiplier operator. This matrix multiplication copies 2 times the 1st row to the third row:
The operation:
> matrix( c(0,0,2,rep(0,13)), 4) %*% A2
     [,1] [,2] [,3] [,4]
[1,]    0    0    0    0
[2,]    0    0    0    0
[3,]    2   10   18   26
[4,]    0    0    0    0

So it can be used thusly:

 D2 <- A2 + matrix (c(0,0,2,rep(0,13)), 4) %*% A2
 D2
     [,1] [,2] [,3] [,4]
[1,]    1    5    9   13
[2,]    2    6   10   14
[3,]    5   17   29   41
[4,]    4    8   12   16

This method does not pull the "donor" matrix apart.

--
David

all.equal(A, A2)
all.equal(D, D2)

Cheers,

Josh

On Fri, Aug 27, 2010 at 10:32 PM, Cheng Peng <cp...@usm.maine.edu> wrote:

Sorry for possible misunderstanding:

I want to define a matrix (B) based on an existing matrix (A) in a single
step and keep A unchanged:

#Existing matrix
A=matrix(1:16,ncol=4)
A
    [,1] [,2] [,3] [,4]
[1,]    1    5    9   13
[2,]    2    6   10   14
[3,]    3    7   11   15
[4,]    4    8   12   16
# New matrix B is defined to be the submatrix after row1 and column1 are
deleted.
B=A[-1,-1] # this single step deletes row1 nad column 1 and assigns the
name to the resulting submatrix.
B                 # check the new matrix B
    [,1] [,2] [,3]
[1,]    6   10   14
[2,]    7   11   15
[3,]    8   12   16
A                 # check the original matrix A
    [,1] [,2] [,3] [,4]
[1,]    1    5    9   13
[2,]    2    6   10   14
[3,]    3    7   11   15
[4,]    4    8   12   16


Question: How can I do define a new matrix (D) by adding 2*row1 to row3 in A
in a single step as what was done in the above example?

If you do: A[3,]=2*A[1,]+A[3,], the new A is not the original A; if you
D=A first, then D[3,]=2*D[1,]+D[3,], you used two step!

Hope this clarifies my original question. Thanks again.





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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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