question about how to avoid using eval
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Baozai
am 22 Nov. 2016
Kommentiert: Baozai
am 22 Nov. 2016
Hello, all.
I got to know that using "eval" is not efficient and may cause several problems. I have used "eval" a lot in my previous codes.
If I want to replace the following expression without using "eval", would anyone help me one this? Thank you very much.
for jjjj=1:5
for iiii=1:40
for kkkk=1:3
x1(kkkk,:) = (1:10)*kkkk;
eval(['Resp.Phi',num2str(jjjj),'.Mu',num2str(iiii),'.Respx1(kkkk,:) = x1(kkkk,:);']);
end
end
end
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Akzeptierte Antwort
Walter Roberson
am 22 Nov. 2016
Resp.(sprintf('Phi%d',jjjj).(sprintf('Mu%d',iiii)),.Respx1(kkkk,:) = x1(kkkk,:);
5 Kommentare
Walter Roberson
am 22 Nov. 2016
You are right,
Resp.(sprintf('Phi%d',jjjj)).(sprintf('Mu%d',iiii)).Respx1(kkkk,:) = x1(kkkk,:);
Weitere Antworten (1)
KSSV
am 22 Nov. 2016
iwant = cell(5,40,3) ;
for jjjj=1:5
for iiii=1:40
for kkkk=1:3
x1(kkkk,:) = (1:10)*kkkk;
iwant{jjjj,iiii,kkkk}= ['Resp.Phi',num2str(jjjj),'.Mu',num2str(iiii),'.Respx1(kkkk,:) = x1(kkkk,:);'] ;
end
end
end
3 Kommentare
KSSV
am 22 Nov. 2016
Bearbeitet: KSSV
am 22 Nov. 2016
Resp = struct ;
for jjjj=1:5
for iiii=1:40
for kkkk=1:3
x1(kkkk,:) = (1:10)*kkkk;
% iwant{jjjj,iiii,kkkk}= ['Resp.Phi',num2str(jjjj),'.Mu',num2str(iiii),'.Respx1(kkkk,:) = x1(kkkk,:);'] ;
end
for llll = 1:3
Resp(jjjj).Phi1(iiii).Mu1(llll).Respx1 = x1 ;
end
end
end
Resp(1).Phi1(1).Mu1
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