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 > >
