How do i make this code as simple as possible? :(
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final_0 = 0;
final_1 = final(numel(final_0)+1:(floor(numel(final)*1/10))*1);
final_2 = final(floor(numel(final)*1/10)*1+1:floor(numel(final)*1/10)*2);
final_3 = final(floor(numel(final)*1/10)*2+1:(floor(numel(final)*1/10))*3);
final_4 = final(floor(numel(final)*1/10)*3+1:(floor(numel(final)*1/10))*4);
final_5 = final(floor(numel(final)*1/10)*4+1:(floor(numel(final)*1/10))*5);
final_6 = final(floor(numel(final)*1/10)*5+1:(floor(numel(final)*1/10))*6);
final_7 = final(floor(numel(final)*1/10)*6+1:(floor(numel(final)*1/10))*7);
final_8 = final(floor(numel(final)*1/10)*7+1:(floor(numel(final)*1/10))*8);
final_9 = final(floor(numel(final)*1/10)*8+1:(floor(numel(final)*1/10))*9);
final_10 = final(floor(numel(final)*1/10)*9+1:floor(numel(final)));
I think anyone who has studied a little bit can make it simple, but I'm not
For your information, the class of those codes is char.
The size of one data is so large that you want to split it into 10 pieces.
Please let me know, and if you have time, I would like to know if this code can be expressed in loop. :((((((((
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Walter Roberson
le 30 Déc 2020
1 vote
buffer(final, floor(numel(final)/10)).'
Needs the signal processing toolbox.
The final (11th) row will contain 110227 entries only the first two of which will be nonblank. You might want to consider using ceil instead of floor which will get you 10 rows the last of which has 8 padding positions.
1 commentaire
Jungwu Kim
le 31 Déc 2020
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