how can sort the numbers according to the numbners in the first column
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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
2018 年 11 月 22 日
編集済み: Stephen23
2018 年 11 月 22 日
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
5 件のコメント
Stephen23
2018 年 11 月 22 日
編集済み: Stephen23
2018 年 11 月 22 日
@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:
その他の回答 (1 件)
madhan ravi
2018 年 11 月 22 日
編集済み: madhan ravi
2018 年 11 月 22 日
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 件のコメント
madhan ravi
2018 年 11 月 22 日
result=cell(1,n) %before loop
result{i}=a(b==i); %inside loop
celldisp(result) %outside loop
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