Assignment has more non-singleton rhs dimensions than non-singleton subscripts

Question

0 个投票

Hi guys, trying my best to battle through some code but cannot get it to work for my data.

for e = length(WAV(:,1));
    k = find(tss==WAV(e,1));
    WAV(e,5) = dir(k);
end

WAV is currently 50x3 double, tss is 1048576x1 double, as is dir. e = 50.

Many thanks guys

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Jan 2017-8-22

Today I've selected your code and pressed the "{} Code" button, to make it readable. Please do this by your own in the future. Thanks.

Answer 1

Jan 2017-8-22

编辑：Jan 2017-8-22

0 个投票

Perhaps you mean:

for e = 1:size(WAV, 1)
  k        = (tss == WAV(e,1));
  WAV(e,5) = dir(k);
end

size(X,1) is more efficient than creating a vector only to measure its length by length(X(:,1).
The for loop requires a start points: "1:n". With only "for k = size(WAV,1)" the loop runs over one index only
I've omitted the find() to use the faster logical indexing.
"dir" is a bad name for a variable, because it shadows the builtin function with the same name. This is not an error, but it causes troubles frequently, when the user wants to access the command again.