This difference should be explained in the documentation for diag
The current documentation is kind of short:
Base.diag(M[, k])
The "k"-th diagonal of a matrix, as a vector.
Ivar
kl. 21:54:43 UTC+2 søndag 27. april 2014 skrev John Code følgende:
>
> Thank you.
>
> On Sunday, April 27, 2014 11:49:12 PM UTC+4, Andreas Noack Jensen wrote:
>>
>> Hi John
>>
>> In julia, the function diag extract the diagonal of a matrix and if the
>> matrix is rectangular, it extracts the diagonal of the largest square sub
>> matrix. Note that in julia, [1 2 3 4] is not vector but a matrix. To
>> construct a matrix from a vector you can either use the function diagm,
>> which does what you expected diag did,
>>
>> julia> diagm([1,2,3,4])
>> 4x4 Array{Int64,2}:
>> 1 0 0 0
>> 0 2 0 0
>> 0 0 3 0
>> 0 0 0 4
>>
>> but it is often better to use Diagonal, which creates a special Diagonal
>> matrix,
>>
>> julia> Diagonal([1,2,3,4])
>>
>> 4x4 Diagonal{Int64}:
>> 1 0 0 0
>> 0 2 0 0
>> 0 0 3 0
>> 0 0 0 4
>>
>>
>> 2014-04-27 21:40 GMT+02:00 John Code <jcod...@gmail.com>:
>> >
>> > Hi all,
>> > I would like to ask why there is a difference between Octave diag
>> function
>> > and the function that julia provide. For example, in the following
>> Octave session I get:
>> >
>> > ============================
>> > octave:1> v = [1 2 3 4]
>> > v =
>> >
>> > 1 2 3 4
>> >
>> > octave:2> a = diag(v)
>> > a =
>> >
>> > Diagonal Matrix
>> >
>> > 1 0 0 0
>> > 0 2 0 0
>> > 0 0 3 0
>> > 0 0 0 4
>> > =============================
>> >
>> > But in Julia I get:
>> >
>> > =============================
>> > julia> v = [1 2 3 4]
>> > 1x4 Array{Int64,2}:
>> > 1 2 3 4
>> >
>> > julia> a = diag(v)
>> > 1-element Array{Int64,1}:
>> > 1
>> >
>> >
>> > =============================
>> >
>> >
>> > Why is this the case and how to get a similar effect of the octave code.
>> > Thank you.
>>
>>
>>
>>
>> --
>> Med venlig hilsen
>>
>> Andreas Noack Jensen
>>
>
