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complicated calculation on the Arraylist of int[] !!

hi experts,

I got a Arraylist of int[] which contains data like belows:

321,371,361,305,112,88,109,89,115,97,101,91,922,468,586,4333,415047,1
317,370,357,294,113,87,110,89,110,93,107,92,924,468,591,4391,419170,1
306,362,348,265,114,88,110,90,110,91,103,81,908,475,581,4248,416936,2
.......
333,375,352,301,109,88,112,92,110,92,110,96,904,465,574,4374,409852,56
......
317,370,357,294,113,87,110,89,110,93,107,92,924,468,591,4391,419170,88
........
317,370,357,294,113,87,110,89,110,93,107,92,924,468,591,4391,419170,100
........................about 1000 rows and 17 columns !

the last column is used as a class identifier, writing a program for this ,
u may read in the dataset first, divide it into 10 subsets and store to ten 2D array. Use a For-loop to repeat the process 10 times, and do the classification experiment inside
this for loop. the classification experiment is using ED----http://www.nist.gov/dads/HTML/euclidndstnc.html

The ED method is used to measure the similarity between
two feature vectors(rows). For each sample in the testing set, compare it with
all the samples in the training set and find the most similar feature
vector(row) (by using ED, the shortest distance..), then
compare their class ID ( user ID), if match, it counts a correct
classified smaple, it counts a non-match classified sample otherwise.
Then, obtain the correct classification RATE for each fold, and
also the total classification rate.
Finally output the classification rate for each fold to a text file !

thanks !
0
mmccy
Asked:
mmccy
1 Solution
 
TimYatesCommented:
0
 
mmccyAuthor Commented:
Sorry that is the first part !! After having some experts's help I am tackling with it now and fully understand it (the first part)
However, I really have no idea of what I should do in this 2nd part !!
Can u give me some hints ?

thanks !
0
 
sciuriwareCommented:
And ... this is homework!!!!

;JOOP!
0
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Giant2Commented:
>divide it into 10 subsets and store to ten 2D array

      public ArrayList[] divide(ArrayList original, int numOfSubset){
            ArrayList[] ret=new ArrayList[numOfSubset];
            int subset=0;
            int nrlen=original.size()/numOfSubset;
            int pos=0;
            while (subset<numOfSubset && pos<original.size()){
                  ret[subset].add(original.get(pos));
                  pos++;
                  if (pos==nrlen)subset++;
            }
            if (pos<original.size()){
                  for (int i=pos;i<original.size();i++)
                  ret[numOfSubset-1].add(original.get(i));}
            return ret;
      }
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sciuriwareCommented:
Agreed; how do the moderators deal with this kind of questioners? Just curious.
;JOOP!
0
 
Giant2Commented:
>We can answer homework questions in Tutor mode but not to give full solutions.

Sorry Venabili,
but I see many (and many) Experts that giving the full solution have points; and I (giving only a Tutor solution) have NO points.
So I decide that if I have ready a solution, I'll post it (this is I have done in this thread).

Maybe I was not luckly than other because "mmccy" doesn't close the post where I gave the full solution, so now this thread is deleted (even if I gave a solution).

No problem, but I hope this control about the Tutor/FullSolution will be done even on the entire EE DB, not only on the thread not closed.

Bye, Giant.

P.S. I think this is one problem in EE, but, like I raised up in another thread I have this like answer: It's not possible to check the answer accepted/assisted.
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sciuriwareCommented:
Giant2,
I remark that the tone of some questions tells a story: literally from a book, paper or even examination.
You must not complain that you are not always stopped / rewarded or EE will turn into a police environment.
;JOOP!
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Giant2Commented:
Infact I told:
"It's not possible to check the answer accepted/assisted"

Bye, Giant.
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sciuriwareCommented:
Yes, indeed, now someone has the feeling he/she could get away with the answer without paying.
I agree with "Delete - no points refunded",
but annoying still to those who spent their time.
;JOOP!
0

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