Why I am getting the error message:Assignment has more non-singleton rhs dimensions than non-singleton subscripts

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Sandra Maria Cherian
评论： Sandra Maria Cherian 2018-7-9
syms tmv; tdcvv=[ 1.7691180562790619335090624118428, 3.7203521516231003235052331604841, 4.2549397326879975357442412372052; 1.8728284259647972559033795104949, 4.0675421128150192508477208506903, 5.0616967785172204893339442877428; 2.1175654748328456634231199033899, 3.1758086461927840904807328442577, 3.7653737399696897254758169563108]; S=40; C=60; a0=1.993; a1=0.002; a2=0.015; b0=38.086; b1=-0.176; b2=-0.633; c0=10.720; c1=1.256; c2=1.522; a=a0+a1.*S+a2.*C; b=b0+b1.*S+b2.*C; c=c0+c1.*S+c2.*C; eqs=a+b.*tmv+c.*tmv.^2; teqs=eqs==tdcvv; teqsd=vpa(teqs); tmvVV=sym('tmvVV',[3 3]); tic for i=1:3 for j=1:3 tmvVV(i,j)=vpasolve(teqsd(i,j),tmv); end end

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Dennis
Dennis 2018-7-9
tmvVV(i,j)=vpasolve(teqsd(i,j),tmv);
This line returns 2 values, but tmvVV(i,j) has only 1 field. Adding a third dimension tmvVV(i,j,:) or using cells should fix your problem.
syms tmv;
tdcvv=[ 1.7691180562790619335090624118428, 3.7203521516231003235052331604841, 4.2549397326879975357442412372052; 1.8728284259647972559033795104949, 4.0675421128150192508477208506903, 5.0616967785172204893339442877428; 2.1175654748328456634231199033899, 3.1758086461927840904807328442577, 3.7653737399696897254758169563108];
S=40; C=60; a0=1.993; a1=0.002; a2=0.015; b0=38.086; b1=-0.176; b2=-0.633; c0=10.720; c1=1.256; c2=1.522; a=a0+a1.*S+a2.*C; b=b0+b1.*S+b2.*C; c=c0+c1.*S+c2.*C; eqs=a+b.*tmv+c.*tmv.^2; teqs=eqs==tdcvv;
teqsd=vpa(teqs);
tmvVV=sym('tmvVV',[3 3 2]);
for i=1:3
for j=1:3
tmvVV(i,j,:)=vpasolve(teqsd(i,j),tmv);
end
end

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