I worked on nested loop.

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seema niran
seema niran 2015-10-7
评论： seema niran 2015-10-7
y is a 5000 by 24 matrix. i need a new 46 by 24 matrix such that row 1 of y repeats 23 times and row 2 repeats 23 times. the code is given below
p=1;
k=1;
a=ones([1,24]);
while k<3
sub=y(k,:);
k=k+1;
while j<24
a(j,:)=sub;
j=j+1;
end
end
the answer i expect is a 46 by 24 matrix. but i get a 23 by 24 matrix.

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采纳的回答

per isakson
per isakson 2015-10-7
编辑：per isakson 2015-10-7
Assuming that your goal is to understand loops, try this
>> [ a01, a02 ] = cssm();
>> all(a01(:)==a02(:))
ans =
1
where
function [ a01, a02 ] = cssm()
y = randi( 100, [3,24] );
a01 = nan([46,24]);
a02 = nan([46,24]);
kk = 1;
while kk<3
sub = y(kk,:);
kk = kk+1;
jj = 1;
while jj<24
a01((kk-2)*23+jj,:)=sub;
jj = jj+1;
end
end
for kk = 1 : 2
for jj = 1 : 23
a02((kk-1)*23+jj,:) = y(kk,:);
end
end
end
&nbsp
Comment: &nbsp The codes based on while-loops and for-loops, respectively, produces identical answers. In this special case for-loops are easier to work with than while-loops.
"but i get a 23 by 24 matrix". That's because in the second turn of the outer loop the counter, j, has the value 24 and the code of the inner loop will not execute.
Stalin's answer provides a more efficient solution, which takes advantage of special features of Matlab.
  1 个评论
seema niran
seema niran 2015-10-7
yes sir , i got the correct matrix when i incremented j (j=j+23) in the first loop. thank you for the comment

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更多回答（1 个）

Stalin Samuel
Stalin Samuel 2015-10-7
A_new = ones(46,24)
A = rand(5000,24);
A_new(1:23,:) = A(1,:);
A_new(24:46,:) = A(2,:) ;

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