[R] simple matrix calculation
David Winsemius
dwinsemius at comcast.net
Thu Mar 29 00:01:46 CEST 2012
On Mar 28, 2012, at 4:46 PM, Kehl Dániel wrote:
> Dear list-members,
>
> I have a 9-by-9 matrix lets call it A with first row a11, a12,
> a13,..., a19 etc.
> I also have a vector of length 3 (B).
> I want to construct a matrix of size 3x3 in the following way:
> - divide matrix A to 9 3x3 blocks
> - first is
> a11, a12, a13
> a21, a22, a23
> a31, a32, a33
> - I want to get rowSums of this A1 matrix
> - Multiply A1*B and get a scalar, the first element of my new 3x3
> matrix.
>
You're going to need some sort of loop, but maybe you can make it
more elegant my noticing that you can access the 3x3 elements more
elegantly using single subscript values rather than ranges:
amat <- matrix(paste("a", rep(1:9, time=9), rep(1:9, each=9),
sep=""), 9,9)
amat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] "a11" "a12" "a13" "a14" "a15" "a16" "a17" "a18" "a19"
[2,] "a21" "a22" "a23" "a24" "a25" "a26" "a27" "a28" "a29"
[3,] "a31" "a32" "a33" "a34" "a35" "a36" "a37" "a38" "a39"
[4,] "a41" "a42" "a43" "a44" "a45" "a46" "a47" "a48" "a49"
[5,] "a51" "a52" "a53" "a54" "a55" "a56" "a57" "a58" "a59"
[6,] "a61" "a62" "a63" "a64" "a65" "a66" "a67" "a68" "a69"
[7,] "a71" "a72" "a73" "a74" "a75" "a76" "a77" "a78" "a79"
[8,] "a81" "a82" "a83" "a84" "a85" "a86" "a87" "a88" "a89"
[9,] "a91" "a92" "a93" "a94" "a95" "a96" "a97" "a98" "a99"
> dim(amat) <- c(3,3,3,3)
> amat[,1,,1] # upper/left 3x3
[,1] [,2] [,3]
[1,] "a11" "a12" "a13"
[2,] "a21" "a22" "a23"
[3,] "a31" "a32" "a33"
> amat[,2,,2] # middle/middle 3x3
[,1] [,2] [,3]
[1,] "a44" "a45" "a46"
[2,] "a54" "a55" "a56"
[3,] "a64" "a65" "a66"
If speed is an issue than perhaps you could use `aperm` to rearrange
this 3x3x3x3 array so that the proposed rowSums operation would be
correct (within three row segments.)
> newmat <- aperm(amat, c(2,4,3,1))
> dim(newmat) <- c( 3*9,3)
> newmat[1:6,1:3]
[,1] [,2] [,3]
[1,] "a11" "a21" "a31"
[2,] "a41" "a51" "a61"
[3,] "a71" "a81" "a91"
[4,] "a14" "a24" "a34"
[5,] "a44" "a54" "a64"
[6,] "a74" "a84" "a94"
> I could do that with loop. Can you suggest something that is more
> elegant and faster?
>
> Thank you
> Daniel
>
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David Winsemius, MD
West Hartford, CT
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