complicated for loop with 2 requirements or constraints

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Grace
Grace 2014-6-4
评论： Grace 2014-6-4
Hi,I have
a=[1 2; 3 4; 5 6;7 8];
Suppose I want my result to have two sets of number, which set 1 is [1 2; 3 4; 5 6] and set 2=[3 4; 5 6;7 8].
result=cell(2,1);
for m=1:2
for i=0:1
k=1:3;
result{m}=a(k+i,:);
end
end
This output shows 2 similar set of numbers. What can I do? Do I make myself clear?

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采纳的回答

Roger Wohlwend
Roger Wohlwend 2014-6-4
Then do the following:
if m > 4
UT{m} = id(m-4:m-1,:)
end
  1 个评论
Grace
Grace 2014-6-4
Hi Roger, now I'm trying to modify my original id by sorting its column to get 2 new id as follows:
id = [ 1 3; 2 6; 3 2; 4 5; 5 1; 6 4; 7 7];
[r c]=size(id);
new_id=cell(c,1);
UT=cell(r*c,1);
for col=1:c
new_id{col}=sortrows(id,col);
for m=1:7
if m>4
UT{m} = new_id(m-4:m-1,:);
else
j=1:m-1;
a=new_id(j,:);
i=m:4;
b=new_id(i+3,:);
UT{m}=cat(1,a,b);
end
end
end
Then from that two new id, i want to run the if else statement to get UT, but now I have 2 new id, means that I will get 14 UTs. But if i run the code above, there is an error, I can't get what I wanted.
Can you help? Thank you.

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更多回答（1 个）

Roger Wohlwend
Roger Wohlwend 2014-6-4
In the second loop you first (when i = 0) save a matrix in result{m} and then you override it when (i = 1). So the inner loop has no effect.
I do not understand why you use loops at all. You could simply write:
result{1} = a(1:3,:);
result{2} = a(2:4,:);
Or if you want a loop:
for m = 1 : 2
result{m} = a(m:m+2,:);
end

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