Indexing Multiple Vectors in MATLAB
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time = [ 1 2 3 4 5 6 7 8 9 10]
xvalues = [ 1 1.2 1.5 1.6 2 2.5 2.7 4.5 5.6 5.7 .... 9.9 10] etc
When one generates multiple vectors in a loop in MATLAB say
[index,~] = find(time)
for i = index(1):index(end)
[idx(i),~] = find( xvalues>=time(i));
idx(i+1) = idx(i)xvalues(idx(i))<time(i));
end
there seems to be a problem with idx(i+1) = idx(i)xvalues(idx(i))<time(i)); but i think the problem occurs earlier in the line above that because I can't recall idx(i)
The two lines in the loop are trying to find elements in a vector
So if A = [ 1 2 3 4 5 6 7 8 9 ]
I want to obtain, say, one vector as 1 2 3, another as 4 5 6 and the other as 7 8 9
Error: ()-indexing must appear last in an index expression.
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採用された回答
Matt J
2013 年 3 月 24 日
編集済み: Matt J
2013 年 3 月 24 日
The previous line potentially has does have a problem. In
[idx(i),~] = find( xvalues>=time(i));
you're going to get an error if find() returns empty or a non-scalar. idx(i) can only hold a scalar.
The expression
idx(i+1) = idx(i)xvalues(idx(i))<time(i))
is definitely a syntax error, but it is not clear what you want this to do, and therefore not clear how to fix it.
3 件のコメント
Matt J
2013 年 3 月 24 日
Sounds like HISTC might be what you're looking for
[~,bins]=histc(xvalues,time)
and now you can split up the data as you wish
xvalues(bins==1)
will be all the xvalues between time(1) and time(2), etc...
If that's not what you're trying to do, I'm afraid it's still very unclear. It's also clear what form you want the final data. A cell array, perhaps?
その他の回答 (1 件)
the cyclist
2013 年 3 月 24 日
I am not 100% sure what you were trying to do, but
idx(i)xvalues(idx(i))
is not valid MATLAB syntax. MATLAB is confused by this.
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