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

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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 Kommentare
Adam Danz
Adam Danz am 10 Sep. 2019
Bearbeitet: Adam Danz am 10 Sep. 2019
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 am 11 Sep. 2019
I may have confused which came first, thank you!

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Sebastian Körner
Sebastian Körner am 10 Sep. 2019
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 Kommentare
Ana Bianco
Ana Bianco am 11 Sep. 2019
That is just perfect! Exactly what I wanted.
Thanks Sebastian!
Stephen23
Stephen23 am 11 Sep. 2019
Bearbeitet: Stephen23 am 11 Sep. 2019
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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