grouping data by value

3 vues (au cours des 30 derniers jours)
wesso Dadoyan
wesso Dadoyan le 4 Août 2015
I have a vector of data X that I want to divide into 10 different groups depending on the values. i,e the highest 10% together, then the lower 10% together etc.... How can I create another vector that assigns numbers to such groups. For example highest 10% should have a value of 1, then the lower 10% a value of 2 , ... then the lowest 10% a value of 10. I used :
DecileThresholds= quantile(X,(1:9)/10);
to find the points that separates the categories but I don't want to write a loop to categorize them. I thought that matlab probably has a function that automatically and elegantly give me the category number. Thanks
  2 commentaires
Image Analyst
Image Analyst le 4 Août 2015
It's not clear to me. Let's say you have a normal distribution with 1000 elements with values going from 0 to 10. Do you want 10 arrays where each array has 100 (which is 10% of 1000) elements? Or do you want 10 arrays where each array can have a different number of elements, because the number of elements in the range 500-600 will be more than the number with values in the range 0-100?
wesso Dadoyan
wesso Dadoyan le 4 Août 2015
I want 10 arrays where each array has a different number of observations. Assume the value range is between 1 and 100 , then the 1st array (1st group) is of observations that fall between 1 and 10, 2nd group falls between 11 and 20 etc.... the number of values between 1 and 10 could be higher than those between 11 and 20, so the number of each array elements are not the same.

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Cedric
Cedric le 4 Août 2015
Modifié(e) : Cedric le 4 Août 2015
x = [1, 8, 2, 3, 3, 7] ; % Dummy example.
[~, binId] = histc( x, [0, 4, 10] ) ; % Bin edges: 0 --bin1-- 4 --bin2-- 10
With that you get:
binId =
1 2 1 1 1 2
Then you can group them per bin:
grouped = accumarray( binId', x, [], @(v){v} ) ;
  1 commentaire
Chanaka Keerthisinghe
Chanaka Keerthisinghe le 7 Juin 2018
Thank you,

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Azzi Abdelmalek
Azzi Abdelmalek le 4 Août 2015
A=rand(1,106);
n=numel(A);
[b,idx]=sort(A);
m=round(n/10);
p=n-m*9;
ii=repmat(1:9,m,1);
jj=[ii(:) ;10*ones(p,1)];
out=accumarray(jj,b',[],@(x) {x});
celldisp(out)
  1 commentaire
wesso Dadoyan
wesso Dadoyan le 4 Août 2015
Thanks Azzi, the groups shouldn't be of equal size. Look at my comment to the inquiry of image analyst. I hve the impression that your codes generated equal groups and the remainder went to the last group

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