On 21/04/2008, at 9:54 PM, William Simpson wrote:
> Hi Everyone,
>
> I am running into a problem with matrices. I use R version 2.4.1 and
> an older version.
>
> The problem is this:
> m<-matrix(ncol=3,nrow=4)
> m[,1:3]<-runif(n=4)
>
> That does what I expect; it fills up the rows of the matrix with the
> data vector
>> m
> [,1] [,2] [,3]
> [1,] 0.2083071 0.2083071 0.2083071
> [2,] 0.5865763 0.5865763 0.5865763
> [3,] 0.7901782 0.7901782 0.7901782
> [4,] 0.8298317 0.8298317 0.8298317
>
> But this doesn't work:
> m[1:4,]<-runif(n=3)
>> m
> [,1] [,2] [,3]
> [1,] 0.96864939 0.11656740 0.06182311
> [2,] 0.11656740 0.06182311 0.96864939
> [3,] 0.06182311 0.96864939 0.11656740
> [4,] 0.96864939 0.11656740 0.06182311
>
> I want it to fill up the columns of the matrix with the data vector.
>
> Maybe there is a better way to do what I want. I need to do both of
> the above. The matrices are large, so I need a fast method.
R fills arrays in ``reverse odometer order'' --- the first index (row
index in matrices) ticks over fastest; the second index second fastest.
Unless you tell it to do otherwise.
In other words matrices get filled up column by column. Unless other
instructions are given.
Thus m[] <- runif(3) (note that the ``1:4'' is redundant) puts
the generated 3-vector into the first 3 entries of column 1, then
starts over again, putting the first entry of that vector into the
last entry of column 1, then the next two entries of the vector
into the first two entries of column 2, and so on.
What you want to do is
m <- matrix(runif(3),nrow=4,ncol=3,byrow=TRUE)
i.e. it is unnecessary and counter-productive to create m beforehand
and then try to fill it up.
cheers,
Rolf Turner
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