simply !
start=1
st=30000
k=6000001
v=[start:fld(k,st):k];push!(v,k)
Paul
W dniu wtorek, 3 lutego 2015 14:24:16 UTC+1 użytkownik paul analyst napisał:
>
>
>
> 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 <[email protected]> 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
>>>
>>>
