Filter löschen
Filter löschen

Dimension error in matrix manipulation

1 Ansicht (letzte 30 Tage)
Amandeep
Amandeep am 31 Aug. 2011
Hi, trying to figure out why this doesn't work and how to make it work:-
looktime2= [1 16; 2 55; 2 61; 2 66; 2 68; 5 20; 6 60; 8 47; 8 83; 8 105; 9 72; 10 44; 10 57]
t2 = [true;diff(looktime2(:,1))~=0]; t2 = [t2,looktime2(:,2)];
t2 = [1 16; 1 55; 0 61; 0 66; 0 68; 1 20; 1 60; 1 47; 0 83; 0 105; 1 72; 1 44; 0 57]
Rmatrix is a 10 by 121 matrix
for i = 1:size(t2,1)
for j = 1
if t2(i,j)==1
sd(i,j) = Rmatrix(looktime2(i,j),1:t2(i,j+1));
end
if t2(i,j)==0
sd(i,j) = Rmatrix(looktime2(i,j),t2(i-1,j+1):t2(i,j+1));
end
end
end
When I press enter I get an Subscripted assignment dimension mismatch error , but I'm not sure how to get it to work.
Any help will be much appreciated. Thanks
  2 Kommentare
Andrei Bobrov
Andrei Bobrov am 31 Aug. 2011
that want to get?
Andrei Bobrov
Andrei Bobrov am 31 Aug. 2011
error in the code
element double array assign vector

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 31 Aug. 2011
variant 1 (use 'arrayfun')
t = [true;diff(looktime2(:,1))~=0];
idx = [0;looktime2(1:end-1,2)];
idx(t)=1;
idx2 = [idx looktime2(:,2)];
di = diff(idx2,1,2)+1;
sd = arrayfun(@(i1)[Rmatrix(looktime2(i1),idx2(i1,1):idx2(i1,2)) zeros(1,max(di)-di(i1))],1:numel(t),'un',0);
sd=cat(1,sd{:});
variant 2 (use loop)
t = [true;diff(looktime2(:,1))~=0];
idx = [0;looktime2(1:end-1,2)];
idx(t)=1;
idx2 = [idx looktime2(:,2)];
di = diff(idx2,1,2)+1;
nt = numel(t);
sd = zeros(nt,max(di));
for i1 = 1:nt
sd(i1,1:di(i1)) = Rmatrix(looktime2(i1),idx2(i1,1):idx2(i1,2));
end
variant 3 (vectorization with use 'bsxfun')
l=looktime2';
t = [true diff(l(1,:),1,2)~=0];
idx = [0,l(2,1:end-1)].*~t+t;
di = diff([idx; l(2,:)])+1;
mdi = max(di);
sdt = zeros(mdi,numel(t));
sz = size(Rmatrix);
lm = bsxfun(@and,bsxfun(@le,(1:sz(2))',l(2,:)),bsxfun(@ge,(1:sz(2))',idx));
i1=bsxfun(@times,lm,l(1,:));
j1 = bsxfun(@times,(1:sz(2))',lm);
sdt(bsxfun(@le,(1:mdi)',di)) = Rmatrix(sub2ind(sz,i1(lm), j1(lm)));
sd = sdt';
variant 4 (your variant)
looktime2= [1 16; 2 55; 2 61; 2 66; 2 68; 5 20; 6 60; 8 47; 8 83; 8 105; 9 72; 10 44; 10 57];
t2 = [true;diff(looktime2(:,1))~=0];
t2 = [t2,looktime2(:,2)];
%t2 = [1 16; 1 55; 0 61; 0 66; 0 68; 1 20; 1 60; 1 47; 0 83; 0 105; 1 72; 1 44; 0 57]
%Rmatrix is a 10 by 121 matrix
sd = nan(size(Rmatrix));
for i1 = 1:size(t2,1)
if t2(i1,1)
k = 1:t2(i1,2);
sd(i1,1:numel(k)) = Rmatrix(looktime2(i1,1),k);
else
k = t2(i1-1,2):t2(i1,2);
sd(i1,1:numel(k)) = Rmatrix(looktime2(i1,1),k);
end
end
sd = sd(:,~all(isnan(sd)));

Weitere Antworten (1)

zohar
zohar am 31 Aug. 2011
Hi Amandeep,
This is the problem
sd(i,j) = Rmatrix(looktime2(i,j),1:t2(i,j+1));
Subscripted assignment dimension mismatch.
Rmatrix(looktime2(i,j),1:t2(i,j+1)) is vector , and sd(i,j) is 1x1 you can not do that.
Have fun

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by