How to solve "Subscripted assignment dimension mismatch"?
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I have two images; imgA and imgB with the size of 43*54*3
I wanted them to display on my screen (1366*768)
I was using the below code to do the same, however it shows me an error say dimention mismatch( sometimes for imgA or imgB)
Please find below code
PS: n is number is trials. img final size is 1366*768
I am using 2016a
if disp_position(n) == 1
LorR{n}= 'Left up';
Img_final(279:322,548:602,:)=imgA;
Img_final(279:322,818:872,:)=imgB;
Img_final(447:490,548:602,:)=imgB;
Img_final(447:490,818:872,:)=imgB;
elseif disp_position(n) == 2
LorR{n}= 'Right Up';
Img_final(279:322,548:602,:)=imgB;
Img_final(279:322,818:872,:)=imgA;
Img_final(447:490,548:602,:)=imgB;
Img_final(447:490,818:872,:)=imgB;
elseif disp_position(n) == 3
LorR{n}= 'Left down';
Img_final(279:322,548:602,:)=imgB;
Img_final(279:322,818:872,:)=imgB;
Img_final(447:490,548:602,:)=imgA;
Img_final(447:490,818:872,:)=imgB;
else
LorR{n}= 'Right down';
Img_final(279:322,548:602,:)=imgB;
Img_final(279:322,818:872,:)=imgB;
Img_final(447:490,548:602,:)=imgB;
Img_final(447:490,818:872,:)=imgA;
end
Please help me solving this problem. I was trying to tackle this last one day but did not successed
2 comentarios
Looky
el 2 de Ag. de 2019
You made a little error with your indexing. The way it works in matlab is, that if you have
279 : 322
this means that the 279th element up to the 322th element (included) is used. (this is different then e.g. in python). As a consequence the number of elements here is 322 -279 +1 = 44 which is one element more than your images.
Just change either 279 to 280 or 322 to 321 and similar for the other indexes and the error should disappear.
Kurni Eswar
el 2 de Ag. de 2019
Respuesta aceptada
Más respuestas (2)
KSSV
el 2 de Ag. de 2019
0 votos
What the necessity to save those images into a bigger image? USe montage. Read about it.
1 comentario
Kurni Eswar
el 2 de Ag. de 2019
Walter Roberson
el 2 de Ag. de 2019
0 votos
279:322 is 44 rows not 43
548:602 is 55 columns not 54
Consider that 2:6 would designate 2 3 4 5 6 for a total of 5 values, not (6-2=4) values. Count is last minus first plus 1, not last minus first.
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