Skipping calculations using if

Tom
2013 12 3
1 个回答
更新时间：2013 12 3
3 次查看（30 天）

0 个投票

Hello,
I have the following code:
for j = 1:(length(acalc)-1)
for ii = (j+1):length(acalc)
dotp(count2,:) = dot(acalc(j,:),acalc(ii,:));
magp(count2,:) = mag(j)*mag(ii);
count2 = count2 + 1;
end
end
Which works fine if length(acalc) > 1. However when length(acalc) = 1 (as it sometimes is), an error is returned presumably because of the failure in accessing acalc(ii,:) = acalc(j+1,:) = acalc(2,:)
Could someone help me resolve this issue? i.e. by simply skipping the calculations if length(acalc) = 1?
I tried simply stating if length(acalc) > 1 before this section of code, but that didn't seem to work.
Many thanks,
Tom

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回答（1 个）

Matt J
Matt J 2013-12-3

0 个投票

No, if length(acalc)=1, the code you've shown would never be executed. For example, running the following will not result in anything displayed.
acalc=10;
for j = 1:(length(acalc)-1)
disp 'Executing'
end

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Tom
Tom 2013-12-3
Ok. Well the error being returned is:
"Attempted to access acalc(2,:); index out of bounds because size(acalc)=[1,3]."
Does that not seem to imply that the way I've written the code is asking it to access a row of acalc which doesn't exist?
Matt J
Matt J 2013-12-3
编辑：Matt J 2013-12-3
Yes. Since acalc is 1x3 it has one row, but you are trying to access a second row acalc(2,:).
Tom
Tom 2013-12-3
I thought that was the problem. But the problem is that most of the time (that section of the code is within a loop of its own) acalc consists of 4 or more rows and then everything is fine. What would be the way to skip those calculations for when acalc consists of only 1 row? I thought it would simply be if length(acalc) = 1 ... end. But that didn't work...
Matt J
Matt J 2013-12-3
编辑：Matt J 2013-12-3
It doesn't look like length(acalc) is the command you should be using if you really intend to measure the number of rows. You should be using size(acalc,1) instead.

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Tom
2013-12-3

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2013-12-3

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