Hi I want to store values in a [3x3] matrix but getting error. Any help would be appreciated
for i=1:3
answerA(i,1)=1*i
answerB(i,2)=2*i
answerC(i,3)=i
% tableA=[answerA answerB answer C]
end
table=[reshape(answerA,[],1) reshape(answerB,[],1) reshape(answerC,[],1)]
% table=[reshape(answerA,[],1)]
ERROR:
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
Error in AllCurvesfit (line 80)
table=[reshape(answerA,[],1) reshape(answerB,[],1)]

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KSSV
KSSV 2019년 3월 13일
편집: KSSV 2019년 3월 13일

1 개 추천

answerA = zeros(3,1) ;
answerB = zeros(3,1) ;
answerC = zeros(3,1) ;
for i=1:3
answerA(i)=1*i ;
answerB(i)=2*i ;
answerC(i)=i ;
end
table=[answerA answerB answerC]
May be you wanted:
T = table(answerA, answerB, answerC)
The above can be achieved without loops also:
i = (1:3)' ;
A = [i 2*i i]

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도움말 센터File Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

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2019년 3월 13일

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2019년 3월 13일

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