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changing a value in an array with matrix indices

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Metin
Metin 2013-8-27
hello all, I want to make such that
if true
i_i=zeros(48,1);
a_a=[3 9 11;25 2 4;5 7 1];
i_i(a_a(2,1) a_a(2,2) a_a(2,3))=1
end
it gives error of Unexpected MATLAB expression. if anyone knows please help thanks
  1 个评论
dpb
dpb 2013-8-27
It's a problem w/ the interface but please edit your question to remove the 'if true' and 'end' wrapper on your code. It shows up when one uses the CODE button w/ a blank area. It's simpler to just type two blanks in front of the first line of code instead...

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采纳的回答

Roger Stafford
Roger Stafford 2013-8-27
You need to replace
i_i(a_a(2,1) a_a(2,2) a_a(2,3))=1
with
i_i([a_a(2,1),a_a(2,2),a_a(2,3)]) = 1;
(I assume you want to set the 25th, the 2nd, and the 4th elements of i_i to 1.)

更多回答（1 个）

dpb
dpb 2013-8-27
Other than you preallocated a 1D (column) array and then wrote access to it w/ three subscripts, the expression
i_i(a_a(2,1) a_a(2,2) a_a(2,3))=1;
would expand to be
i_i(25 2 4)=1;
That clearly isn't proper array addressing syntax as it's missing the commas between the three locations.
Now it comes to me that perhaps you intended to address the three elements in the vector instead of a 3D expression is more likely -- in that case what you need is to turn the row of a_a into a vector itself.
Explicitly it could be written as
i_i([a_a(2,1) a_a(2,2) a_a(2,3)])=1;
but since it's the 2nd row, use the colon operator and write
i_i(a_a(2,:))=1;
instead.

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