[R] Construct a lower-triangular matrix
peter dalgaard
pdalgd at gmail.com
Sat Oct 10 13:39:14 CEST 2015
> On 10 Oct 2015, at 10:49 , Berend Hasselman <bhh at xs4all.nl> wrote:
>
>>
>> On 10 Oct 2015, at 10:00, David Winsemius <dwinsemius at comcast.net> wrote:
>>
>>
>> On Oct 9, 2015, at 10:57 PM, Steven Yen wrote:
>>
>>> Dear
>>> How do you construct a lower triangular matrix from a vector.
>>>
>>> I want to make vector
>>>
>>> a <- 1:10
>>>
>>> into a triangular matrix
>>>
>>> 1 0 0 0
>>> 2 3 0 0
>>> 4 5 6 0
>>> 7 8 9 10
>>>
>>
>> I'm not sure this method with logical indexing will be the most elegant:
>>
>> ?lower.tri
>> ?col
>>
>>> b=matrix(0, sqrt(10)+1,sqrt(10)+1)
>>
>>> b[lower.tri(b)| row(b)==col(b)] <- 1:10
>>> b
>> [,1] [,2] [,3] [,4]
>> [1,] 1 0 0 0
>> [2,] 2 5 0 0
>> [3,] 3 6 8 0
>> [4,] 4 7 9 10
>>
>
> That doesn’t seem to be what the OP wanted.
>
> This should do it.
>
> a <- 1:10
> C <- matrix(0, sqrt(length(a))+1,sqrt(length(a))+1)
> i.upr <- which(upper.tri(C, diag = TRUE), arr.ind=TRUE)
> C[i.upr] <- a
> t(C)
>
> resulting in
>
> [,1] [,2] [,3] [,4]
> [1,] 1 0 0 0
> [2,] 2 3 0 0
> [3,] 4 5 6 0
> [4,] 7 8 9 10
>
> I found this here: http://stackoverflow.com/questions/24472060/indexing-upper-or-lower-triangle-in-matrix-with-diagonal
It's crossing the creek a couple of times too many, though. This'll do:
> M <- matrix(0,4,4)
> M[upper.tri(M,TRUE)] <- 1:10
> t(M)
[,1] [,2] [,3] [,4]
[1,] 1 0 0 0
[2,] 2 3 0 0
[3,] 4 5 6 0
[4,] 7 8 9 10
I'm also not buying the sqrt(length(a))+1 bit --- floor(2*length(a)) is more like it.
(The generic answer is "with some care". In particular, avoid being trapped by column-major storage layout and by in/excluding the diagonal.)
-pd
>
> Berend
>
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com
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