>From the help for `cat`:
cat(dims, A...)
Concatenate the input arrays along the specified dimensions in the
iterable dims....
And indeed, if
size(M1)=(3,3,3)
and
size(M2)=(3,3)
Then,
size(cat(3,M1,M2)) = (3,3,4)
This method may not be efficient (though in terms of memory layout it could
be).
On Saturday, February 13, 2016 at 1:06:02 PM UTC+2, Mauro wrote:
>
> I think this is not possible. Instead use a Vector, append! to that and
> then reshape in the end. The reshape should result in a view and not a
> copy, thus will be fast.
>
> Mauro
>
> On Sat, 2016-02-13 at 10:51, Vishnu Raj <[email protected]
> <javascript:>> wrote:
> > Hi,
> >
> > I have a three dimensional array (say 'z' ) like this :
> >
> > julia> z
> > 4x3x3 Array{Int64,3}:
> > [:, :, 1] =
> > 1 5 9
> > 2 6 10
> > 3 7 11
> > 4 8 12
> >
> > [:, :, 2] =
> > 13 17 21
> > 14 18 22
> > 15 19 23
> > 16 20 24
> >
> > [:, :, 3] =
> > 25 29 33
> > 26 30 34
> > 27 31 35
> > 28 32 36
> >
> > Now I have a matrix 'z1' as
> > julia> z1
> > 4x3 Array{Int64,2}:
> > 37 41 45
> > 38 42 46
> > 39 43 47
> > 40 44 48
> >
> > I want to put z1 at the end of third dimension of 'z'. So that the
> output
> > will be like
> > julia> z[*:,:,4*]
> > 4x3 Array{Int64,2}:
> > 37 41 45
> > 38 42 46
> > 39 43 47
> > 40 44 48
> >
> > I tried push!() and append!(), both gives me errors.
> >
> > Kindly suggest a way to do this. I want 'z' to grow in third dimension
> as
> > simulation progresses.
>
