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

1 回表示 (過去 30 日間)
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

採用された回答

Dennis
Dennis 2018 年 7 月 9 日
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

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeMATLAB についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by