W dniu wtorek, 3 lutego 2015 14:16:49 UTC+1 użytkownik paul analyst napisał:
>
> Thx Tamas, i see idea, but in this function is somthing wrong ?
>
> julia> range(1,30000,6000003)
> 1:30000:180000060001
>
> Paul
>
This way:
julia> a=1
1
julia> st=30000
30000
julia> finish=6000001
6000001
julia> [a:fld(finish,30000):finish]
30001-element Array{Int64,1}:
1
201
401
601
801
1001
...
5999801
6000001
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 <paul.a...@mail.com> 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
>>
>>
