Possibly, but it doesn't look like the one I wrote and you did not provide its source code, so I cannot tell.
Best, Tamas On Tue, Feb 03 2015, paul analyst <[email protected]> wrote: > Thx Tamas, i see idea, but in this function is somthing wrong ? > > julia> range(1,30000,6000003) > 1:30000:180000060001 > > Paul > > W dniu wtorek, 3 lutego 2015 14:08:57 UTC+1 użytkownik Tamas Papp napisał: >> >> Eg >> >> function range_end(a,step,b) >> v = a:step:b >> if v[end] != b >> v = [v,b] >> end >> v >> end >> >> [1 4 ... ] is not a vector, but a matrix. You will need to transpose the >> result if you want that. Also, if the range does not need to be >> extended, it is left as is, should work fine in most applications, but >> you can of course convert it. >> >> Best, >> >> Tamas >> >> On Tue, Feb 03 2015, paul analyst <[email protected] <javascript:>> >> wrote: >> >> > How to build (automaticly) a vector with a range of indivisible ? >> > k=1 >> > l=11 >> > p=3 >> > I need vec : >> > [1 4 7 10 11 ] >> > Paul >> >>
