Wrong way to index: Index exceeds matrix dimensions

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Fernando
Fernando 2013 年 3 月 20 日
編集済み: Hideto Koizumi 2018 年 8 月 28 日
Hi,
I'm trying to use indices to expand a vector but I keep getting an error message and I don't understand why. In particular, what I'm trying to do works in an example such as
x=[1 2 3]';
y=[.1 .5 .3]';
z=[1 1 2 2 3];
yz=y(z);
Here I have two vectors, x and z that have the same values for the indices. I want to expand the vector y to have the same dimensions as z but the values of z must be repeated according to how the index is repeated.
Now, in my application this is not working. I have two vectors X and Z, X of dimensions 2270x1 and Z of dimensions 446723x1 that contains the same indices as X but a lot of times (and each number is repeated a different number of times). To check that the indices are the same I used isequal(unique(X),unique(Z)) and the output was 1.
Finally, I have a vector Y of dimensions 2270x1 that plays the role of y in the example above. When doing YZ=Y(Z) I get ??? Index exceeds matrix dimensions.
Any ideas what is going on?
Thanks,

採用された回答

ChristianW
ChristianW 2013 年 3 月 20 日
max(Z) is greater then 2270?
  3 件のコメント
Fernando
Fernando 2013 年 3 月 21 日
This works, thanks!
Hideto Koizumi
Hideto Koizumi 2018 年 8 月 28 日
編集済み: Hideto Koizumi 2018 年 8 月 28 日
I think this method could cut off the largest element of z, if z is something like z = [1 1 2 4] and y=[.1 .5 .3 .7]';

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その他の回答 (2 件)

Azzi Abdelmalek
Azzi Abdelmalek 2013 年 3 月 20 日
編集済み: Azzi Abdelmalek 2013 年 3 月 20 日
Check this
size(Y)
max(Z)
What did you get?

Fernando
Fernando 2013 年 3 月 20 日
Thanks. Yes, the max of Z is greater than the number of elements in Y, which generates the message. Is there any way to overcome this? Following on the example, if this was
x=[1 2 5]';
y=[.1 .5 .3]';
z=[1 1 2 2 5];
yz=y(z);
I get the message because max(z)> size(y). How can I get yz to be yz=[.1 .1 .5 .5 .3]';?
Thanks,

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