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How to get the absolute value of the a vector inside an array?

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Ana Bianco
Ana Bianco 2019 年 9 月 10 日
編集済み: Stephen23 2019 年 9 月 11 日
Hi there, I am trying to extract the absolute value of the column vectors inside an array, but it keep giving me the message: Array indices must be positive integers or logical values., for the following code:
absvec = [];
for a = 1:270
for b =1:10
abs(H{a,b}) = absvec (a,b),
end
end
can anyone help me?
  2 件のコメント
Adam Danz
Adam Danz 2019 年 9 月 10 日
編集済み: Adam Danz 2019 年 9 月 10 日
Either that's not the error message you're getting or you shared the wrong code with us.
The error message returned by your code is "Index in position 1 exceeds array bounds." (r2019a) Of course this is happeneng because absvec is empty and you're trying to index it.
do you mean
absvec (a,b) = abs(H{a,b}) % ???
Ana Bianco
Ana Bianco 2019 年 9 月 11 日
I may have confused which came first, thank you!

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Sebastian Körner
Sebastian Körner 2019 年 9 月 10 日
relly not sure if this is what your question is about but maybe it helps.
since you provided no data to your problem i created some example data myself.
H ={[5,-6],[5,-6];[7,-8],[7,-8];[5,-6],[5,-6];[7,-8],[7,-8]};
v=1;
for i=1:4
for k=1:2
h(:,v)=abs(H{i,k});
v=v+1;
end
end
input: cell array with cevtors in each cell
output: a vector with the absolute values of each vector of the cell array
  2 件のコメント
Ana Bianco
Ana Bianco 2019 年 9 月 11 日
That is just perfect! Exactly what I wanted.
Thanks Sebastian!
Stephen23
Stephen23 2019 年 9 月 11 日
編集済み: Stephen23 2019 年 9 月 11 日
Simpler and more efficient with a comma-separated list:
>> out = abs(cat(1,H{:}))
out =
5 6
7 8
5 6
7 8
5 6
7 8
5 6
7 8
Timing (1e5 iterations):
Elapsed time is 13.833 seconds. % nested loops
Elapsed time is 2.239 seconds. % comma-separated list

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