How to use for loop correctly and indexing my result?

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Eric Chua
Eric Chua 2020-5-31
评论： Eric Chua 2020-5-31
a = [1 1 1 0 0 0 0 1 0 1 0 1 1 0 0 1] ;
b = [1 1 1 0 1 0 0 0 1 1 1 0 0 0 0 1] ;
S = [a, a, a, a, a, a, a, a, a ,a;b, b, b, b, b, b, b, b, b, b];
x1 = S(1:1,:);
x2 = S(2:2,:);
L = 3;
for i = 3:160
x11 = [x1(i),x2(i)]
x12 = [x1(i-1),x2(i-1)]
x13 = [x1(i-L+1),x2(i-L+1)]
x14 = [1]
X1 = [x11 x12 x13 x14]
end
Hi this is my code. After I run it, it display the answer for x11, x12, x13, x14, and X1 directly with i = 3 to 160, a total of 158 sets of results. How to code such that i want to select my 3rd result to show only?

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回答（1 个）

madhan ravi
madhan ravi 2020-5-31
Use ; at the end of each line inside the loop.
X1(3,:) % after the loop
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madhan ravi
madhan ravi 2020-5-31
Please illustrate with a short example with an expected result, so it's easy to understand the main goal.
Eric Chua
Eric Chua 2020-5-31
I have training sequence a and b with length of 16 binary element each. The S I have defined is to concatenate the sequence by itself for 10 times. So my new 'a ' will be a 1x160 matrix/row vector, and similar to b, which i have defined as x1(for new 'a') and x2(for new 'b'). For example if I concatenate the a by itself for one time,
the new a = [1 1 1 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 1 0 0 1]
From the above posted question, X1 is made up of three 1x2 vector (x11, x12, and x13) and a 1x1 vector (a noise which will remain at 1 as the last element of X1)
For example, for i = 3,
x11 = [1 1] (the first element here is from the third element of the new 'a' and second element here is from
the third element of the new 'b')
x12 = [1 1]
x13 = [1 1]
x14 = [1]
so, my X3 = [1 1 1 1 1 1 1]
For i = 4,
X11 = [0 0]
X12 = [1 1]
X13 = [1 1]
X14 = [1]
so, my X4 = [0 0 1 1 1 1 1]
Im wondering how to code this and when I want my X150 for example, I can write it in the command window and get the vector of X150.
Hope you can understand my questions, much thanks!

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