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John D'Errico wrote:
> Ben <bw.games_n0spam@gmail.nospam.com> wrote in message <mTWYi.
> 10525$ib1.5678@newsfe3-win.ntli.net>...
>> PB wrote:
>>> Den Thu, 08 Nov 2007 14:30:22 +0000 skrev Ben:
>>>
>>>> Hi there,
>>>>
>>>> I'm trying to perform some operations on a matrix - I have a 20x20
>>>> matrix that I've got to remove some rows from the top, and re-
> generate
>>>> lines at the bottom to leave the matrix the same size as it was.
>>>>
>>>> So, say I have a 10x10 matrix, I remove the top row. What I now want
> to
>>>> do is generate a row between the row at the bottom, and the one above
>>>> it, that I can put in to keep it 10x10.
>>>>
>>>> I've looked at interp2, and it sort of seems to do what I want, e.g
>>>> running interp2(Matrix) gives me a 19x19 matrix with an interpolated
> row
>>>> between each. Is there any way to point interp2 at a matrix, with row
>>>> vector 9, and get it to return a 1x10 matrix containing the interpolated
>>>> row between row vectors 9 and 10, that I can just add to the first
>>>> matrix?
>>>>
>>>> I believe interp2 can sort-of do this, using Xi and Yi, but that appears
>>>> to be more related to generating an interpolated value from two data
>>>> points, not two rows?
>>>>
>>>> Anyone have any advice?
>>>>
>>>> Thanks,
>>>>
>>>> Ben
>>> Does this help?
>>>
>>> a=magic(3);
>>> a(:,3)=[] % remove last row
>>> ix=1:0.5:2;
>>> [xi,yi]=meshgrid(ix)
>>> y=interp2(a,xi,yi)
>>>
>>>
>>> /PB
>> Thanks, but your a= line actually removes the last column and not the
>> last row...
>>
>> I changed it to remove the last column, but the final result bears no
>> resemblance to the original square I'm not sure how the meshgrid of ix
>> interacts with the interpolation, which is what I think is causing the
>> wrong results as I changed it from column to row...?
>>
>
> When you remove information by deleting a row
> from a magic square, an interpolant will not know
> that the square was a magic one. Also, PB made
> a mistake in his example.
>
> a=magic(3);
> a(3,:)=[]; % remove last row
> ix=1:0.5:2;
> iy = 1:3;
> [xi,yi]=meshgrid(ix,iy);
> y=interp2(a,yi,xi)';
>
>
> a =
> 8 1 6
> 3 5 7
>
> y =
> 8 1 6
> 5.5 3 6.5
> 3 5 7
>
> Note the transpose at the end!
>
> HOWEVER. dropping the last row of an array,
> then inserting a new one internally to make
> up for it by interpolation will not generally
> make sense to me. But its your data.
>
> HTH,
> John
>
>
Thanks for the feedback.
I think I was a bit unclear, sorry.
What I want to do, is remove the top row of an array.
Then I want to interpolate at the bottom of the array and create an
interpolated row between the last row, and the row previous to that, to
leave the array the same size as it was before, and the data is roughly
the same.
This has the effect of shifting the previous centre row of the array up
by one, without dramatically affecting my data, which is what I want.
Any ideas?
Thanks,
Ben
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