Is very fast, big thx!!
Paul
W dniu czwartek, 4 grudnia 2014 19:04:03 UTC+1 użytkownik tshort napisał:
>
> You could look to see if the PooledDataArray in the DataArrays package
> will do what you are after or at least give you ideas.
> On Dec 4, 2014 12:35 PM, "paul analyst" <paul.a...@mail.com <javascript:>>
> wrote:
>
>> I have :
>> Julia> J
>> 1
>> 5
>> 10000000
>> 2
>> 10000000
>> 1
>>
>>
>> i need recode to lowest values as posible
>> Julia> J
>> 1
>> 3
>> 4
>> 2
>> 4
>> 1
>>
>> Paul
>>
>>
>>
>>
>>
>> W dniu czwartek, 4 grudnia 2014 17:37:15 UTC+1 użytkownik Stefan
>> Karpinski napisał:
>>>
>>> What are you trying to do? Reduce each value in J mod 1000?
>>>
>>> On Thu, Dec 4, 2014 at 10:44 AM, paul analyst <paul.a...@mail.com>
>>> wrote:
>>>
>>>> I have a long vector J, which has little value. Unfortunately, these
>>>> values
>>>> are very different (small and large) J = [1 3 7 2 12548745 2
>>>> 125485458544363456345 125485458544363456345 ... 3 12,548,545,854].
>>>> The following is a very slow like:
>>>> julia> J
>>>> 1846623-element Array{Int64,1}:
>>>> 930070
>>>> 1475172
>>>> 21474836496435345
>>>> 21474836496435345
>>>> 4296445417
>>>> 641
>>>> 130
>>>> 130
>>>> 143
>>>> ...
>>>> 4296444827
>>>> 143
>>>> 148
>>>> 144
>>>> 145
>>>>
>>>> J1=zeros(J)
>>>> suJ=sort(unique(J))
>>>> l=length(suJ)
>>>>
>>>> for i =1:l
>>>> temp=findin(J,[suJ[i]])
>>>> J1[temp]=i;if mod(i,1000).==0 println(i) end;
>>>> end;
>>>>
>>>> How fast reduce the Vector range of values (Recode for smaler numbers)?
>>>> Paul
>>>>
>>>
>>>
