I'd like to abbreviate this long code to a shorter one. I think I can use for and while..
Does anybody know to make it effectively?? Thank you in advance~~
%data = 33 x 300 matrix
p1_1 = zeros(1,300);
p1_2 = zeros(1,300);
p1_3 = zeros(1,300);
p1_4 = zeros(1,300);
p1_5 = zeros(1,300);
p2_1 = zeros(1,300);
p2_2 = zeros(1,300);
p2_3 = zeros(1,300);
p2_4 = zeros(1,300);
p2_5 = zeros(1,300);
p3_1 = zeros(1,300);
p3_2 = zeros(1,300);
p3_3 = zeros(1,300);
p3_4 = zeros(1,300);
p3_5 = zeros(1,300);
for i = 1: 300
p1_1(i) = data(1,5*(i-1)+1);
p1_2(i) = data(1,5*(i-1)+2);
p1_3(i) = data(1,5*(i-1)+3);
p1_4(i) = data(1,5*(i-1)+4);
p1_5(i) = data(1,5*(i-1)+5);
end
for i = 1: 300
p2_1(i) = data(2,5*(i-1)+1);
p2_2(i) = data(2,5*(i-1)+2);
p2_3(i) = data(2,5*(i-1)+3);
p2_4(i) = data(2,5*(i-1)+4);
p2_5(i) = data(2,5*(i-1)+5);
end
for i = 1: 300
p3_1(i) = data(3,5*(i-1)+1);
p3_2(i) = data(3,5*(i-1)+2);
p3_3(i) = data(3,5*(i-1)+3);
p3_4(i) = data(3,5*(i-1)+4);
p3_5(i) = data(3,5*(i-1)+5);
end

답변 (1개)

Andrei Bobrov
Andrei Bobrov 2017년 6월 26일
편집: Andrei Bobrov 2017년 6월 26일

1 개 추천

% Let data -> double array [2500 x 3]
n = 5;
[k0,m] = size(data);
k = k0/m;
p = permute(reshape(data,n,[],m),[3,1,2]);
% here p(2,5,:) - your 'p2_5'

댓글 수: 1

Jan
Jan 2017년 6월 26일
+1. This is much better than the original: Clear, compact, fast, easy to expand, less prone to typos.
@Jiwon Song: Avoid hiding indices in the names of variables. You see how much easier the code is with arrays.

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