how can sort the numbers according to the numbners in the first column
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johnson saldanha
le 22 Nov 2018
Modifié(e) : Stephen23
le 22 Nov 2018
suppose i have a matrix where the first column is x(:,1)=[ 1 1 1 2 2 1 3 3 ] and second column is x(:,2)=[ 2 3 2 4 6 9 7 8 9];
i want different matrices which give me the numbers corresponding to 1, 2 and so on. in this case i need three different matrices where the first one will give me [2 3 2 7]
second will give [4 6] and third is [8 9]
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Stephen23
le 22 Nov 2018
Modifié(e) : Stephen23
le 22 Nov 2018
A general solution using accumarray:
>> x = [1,2;1,3;1,2;2,4;2,6;1,7;3,8;3,9]
x =
1 2
1 3
1 2
2 4
2 6
1 7
3 8
3 9
>> C = accumarray(x(:,1),x(:,2),[],@(v){v});
>> C{:}
ans =
2
3
2
7
ans =
4
6
ans =
8
9
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Stephen23
le 22 Nov 2018
Modifié(e) : Stephen23
le 22 Nov 2018
@johnson saldanha: you could simply use a loop. Here are some alternatives:
>> fun = @(v){[min(v),max(v),mean(v)]};
>> C = accumarray(x(:,1),x(:,2),[],fun);
>> C{1} % min,max,average
ans =
2.0000 7.0000 3.5000
>> C{2} % min,max,average
ans =
4 6 5
>> C{3} % min,max,average
ans =
8.0000 9.0000 8.5000
Or you could do something a bit different, and just put the values into one matrix:
>> x = [1,2;1,3;1,2;2,4;2,6;1,7;3,8;3,9];
>> mx = accumarray(x(:,1),x(:,2),[],@max);
>> mn = accumarray(x(:,1),x(:,2),[],@min);
>> av = accumarray(x(:,1),x(:,2),[],@mean);
>> out = [x,mx(x(:,1)),mn(x(:,1)),av(x(:,1))]
out =
1.0000 2.0000 7.0000 2.0000 3.5000
1.0000 3.0000 7.0000 2.0000 3.5000
1.0000 2.0000 7.0000 2.0000 3.5000
2.0000 4.0000 6.0000 4.0000 5.0000
2.0000 6.0000 6.0000 4.0000 5.0000
1.0000 7.0000 7.0000 2.0000 3.5000
3.0000 8.0000 9.0000 8.0000 8.5000
3.0000 9.0000 9.0000 8.0000 8.5000
% x(:,1) x(:,2) max min average
But really the best solution would be to use a table, which are designed exactly to do the kind of processing that you are trying to do:
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madhan ravi
le 22 Nov 2018
Modifié(e) : madhan ravi
le 22 Nov 2018
By logical indexing you can do it:
a=x(:,1)
b=x(:,2)
first=b(a==1)
second=b(a==2)
third=b(a==3)
Note: It is assumed that x(:,1) and x(:,2) are column vectors and should contain equal number of elements because it’s extracted from a matrix and the above produces the result you need.
5 commentaires
madhan ravi
le 22 Nov 2018
result=cell(1,n) %before loop
result{i}=a(b==i); %inside loop
celldisp(result) %outside loop
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