Is it possible to find corresponding row from other matrix's row??
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y = [ 90 90 -45 0 0 45 45 0 -45 15 15;
90 90 -45 0 0 45 45 0 -45 15 15;
90 90 -45 0 0 45 45 0 -45 15 0;
90 90 -45 0 0 45 45 0 -45 15 0;
90 90 -45 0 0 45 45 0 -45 15 15;
90 90 -45 0 0 45 45 0 -45 15 15;];
y0 = sort(y,2);
y = y(sum([ones(size(y,1),1),diff(y0,[],2)~=0],2) >= numel(unique(y)),:);
y = permute(y ,[3 2 1]);
for k = 1:size(y,3)
[x0(:,:,k), x00(:,:,k)] = hist(y(:,:,k), unique(y(:,:,k)));
end
x0 =permute(x0,[3 2 1]); x00 =permute(x00,[3 2 1]);
x0(any(x0<2,2),:) = [];
I have to find corresponding row of "x00"
2 Commenti
Matthew Eicholtz
il 8 Mar 2016
Your question is a little confusing. Can you try rewording for clarity?
Where do you actually need help? In the last part?
Stephen23
il 9 Mar 2016
Continuation of this question:
Risposta accettata
Stephen23
il 9 Mar 2016
Modificato: Stephen23
il 9 Mar 2016
It is easy, you just need to assign the logical conditions to an index variable. So instead of this:
x0(any(x0<2,2),:) = [];
you should allocate that logical condition to a variable:
idx = any(x0<2,2);
and then you can use idx for any of the matrices:
>> xdel = x0(idx,:)
xdel =
2 4 1 2 2
2 4 1 2 2
>> xnew = x0(~idx,:)
xnew =
2 3 2 2 2
2 3 2 2 2
2 3 2 2 2
2 3 2 2 2
and now you can try x00(idx,:) and x00(~idx,:) yourself!
Più risposte (1)
Ced
il 8 Mar 2016
Modificato: Ced
il 8 Mar 2016
I am not sure I understood the full question, but in short, you want to delete rows in which there is an element < 2 ? You basically already answered the question. Since you like one liners, I think this should do the trick:
x(any(x<2,2),:) = [];
This finds all rows (DIM 2) in which there is ANY element smaller than 2, selects these rows, and deletes them.
If you want to match your centers, you can do:
rows_del = any(x<2,2);
x(rows_del,:) = [];
z(rows_del,:) = [];
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