On May 6, 2011, at 4:24 AM, Petr Savicky wrote:
On Fri, May 06, 2011 at 02:28:57PM +1000, andre bedon wrote:
Hi,
I'm hoping someone can offer some advice:I have a matrix "x" of
dimensions 160 by 10000. I need to create a matrix "y", where the
first 7 elements are equal to x[1]^1/7, then the next 6 equal to
x[2]^1/6, next seven x[3]^1/7 and so on all the way to the
10400000th element. I have implemented this with a for loop an hour
ago and it is still loading, can anyone offer any suggestions as to
how I can create this matrix without using loops? I would really
appreciate any suggestions.
Hi.
Since indexing x[1], x[2], ... is used and also the description
of y corresponds more to a vector, let me first suggest a solution
for vectors.
x <- rep(42, times=4) # any vector of even length
x <- x/c(7, 6)
rep(x, times=rep(c(7, 6), length=length(x)))
[1] 6 6 6 6 6 6 6 7 7 7 7 7 7 6 6 6 6 6 6 6 7 7 7 7 7 7
The input vector may be obtained using c() from a matrix. The
output vector may be reformatted using matrix(). However, for
a matrix solution, a more precise description of the question
is needed.
I certainly agree there, but you have pointed the way to a possible
solution if the OP wants a solution that works along column-wise
application of the 7 alternate with 6 rule applied to a smaller
version that might be called "minimal":
mtx <- matrix(1:160, 16, 10)
mtx^(1/c(7,7,7,7,7,7,7,6,6,6,6,6,6) ) # argument recycling results in
alternating lengths of 7 and 6
(takes less than 5 seconds, and you should expect a warning message
since 160*10000 is not evenly divisible by 13.)
I am not smart enough to know whether mtx2[160,10000] should be
1600000^(1/6) or (.)^(1/7) but this suggests it is (1/6):
> all.equal( mtx2[160,10000] , 1600000^(1/7) )
[1] "Mean relative difference: 0.2883231"
> all.equal( mtx2[160,10000] , 1600000^(1/6) )
[1] TRUE
If he wanted it to be row-wise then he could transpose, operate, and
back-transpose. This, of course, assumes that the OP did not really
mean x[1]^1/7 but rather meant x[1]^(1/7), since I know of no computer
language where exponentiation is lower in precedence than division.
Hope this helps.
Petr Savicky.
--
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
