Proper Syntax for a Nested Loop

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Frank Lehmann
Frank Lehmann 2023-8-7
评论： Matt J 2023-8-7
Hi All,
Im having issues with following code where I believe I may need an additional nested loop.Basically I want the first 2 rows for the 2nd column to reduce by 1 and rows 3 onwards to stay the same value as columns 2. The result Im getting is [ 74 15 12 33] the first 2 elements are correct but the last two should read [24 44], the final result I wish [ 74 15 24 44] I know I need to insert an additional loop somwhere could anyone guide me or provide have a simpler solution on how to reduce column 2 from 1 to 0 from rows 3 onwards?
motData=[12,2;37,3;11,4;15,2];
motData=sortrows(motData,'descend')
i=zeros(size(motData));
p=size(motData,1)
for t =1:p
newmodVector1(t)=(motData(t,1).*(motData(t,2)-1))
end
Thanks,
Frank

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采纳的回答

Voss
Voss 2023-8-7
编辑：Voss 2023-8-7

No additional loop is necessary.

You can replace your existing loop with this one:

for t =1:p
    newmodVector1(t)=motData(t,1).*(motData(t,2)-(t<=2));
end

Or you can do it without a loop at all:

newmodVector1=(motData(:,1).*(motData(:,2)-((1:p).'<=2))).';

Omit the final transpose (.') if you want newmodVector1 to be a column vector.

  2 个评论
Matt J
Matt J 2023-8-7
@Frank Lehmann Since Voss provided a valid solution for you should should Accept-click the answer. It would also be good if you went back to do the same with your previous posts.

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