Finding both row and column indexes of nearest value

Hi experts,

Please, could someone help me to find the 2 indexes, that is, the row and column indexes, corresponding to the nearest longitude and/or latitude? I could figure it out easily in a vector of 1D with find and dsearchn functions, but the problem is I cannot simply transform my matrix into a vector because longitude and latitudes are not constant all along the same row and column.

I have two matrices, one for the longitude, and another for the latitude. Both matrices are 224x164.

I need the row and column indexes of the nearest longitude of 25.983 on the longitude matrix; And idem for the latitude -84.64.

Thank you very much in advance!

Luis

Answer 1

[Image Analyst]

To find the nearest row and column, subtract the reference value from your matrix and then use find. To be super explicit, here are the steps:

m = magic(7)

m = 7x7

30 39 48 1 10 19 28 38 47 7 9 18 27 29 46 6 8 17 26 35 37 5 14 16 25 34 36 45 13 15 24 33 42 44 4 21 23 32 41 43 3 12 22 31 40 49 2 11 20

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refValue = 27;
diffMatrix = abs(m - refValue)

diffMatrix = 7x7

3 12 21 26 17 8 1 11 20 20 18 9 0 2 19 21 19 10 1 8 10 22 13 11 2 7 9 18 14 12 3 6 15 17 23 6 4 5 14 16 24 15 5 4 13 22 25 16 7

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minDifferenceValue = min(diffMatrix, [], 'all')

minDifferenceValue = 0

[row, column] = find(diffMatrix == minDifferenceValue)

row = 2

column = 6

[Luís Henrique Bordin]

Hey, thank you very much for assisting me on this. It worked good, however, I still have a problem. The pairs of indexes for longitude and latitude are different, something that I should have expected before, but didn't realize. Now I need a way to find the intersection between both. Following your logic, I did:

ind_lon=abs(lon-(-84.64))==min(abs(lon-(-84.64)));
ind_lat=abs(lat-(25.983))==min(abs(lat-(25.983)));
ind=ind_lon+ind_lat;
imagesc(ind)

The image shows the intersection between the two lines, however, it would be much more straightforward if I could get the indexes automatically, instead of zooming and annotating it manually. Do you have any clue how could I get there?

Thanks a lot!

[Image Analyst]

Then I suggest you look at computing all distances between your reference location and your actual data with pdist2 and then find the min value, which would be the closest point of your actual data to the reference data.

[Luís Henrique Bordin]

I tried your suggestion but I couldn't handle. I just needed the intersection indexes of those lines, so I took the indexes == 1, and then I got the intersection indexes with polyxpoly function, and it worked.

[xlon, ylon] = find(ind_lon == 1);
[xlat, ylat] = find(ind_lat == 1);

[xi,yi] = polyxpoly(xlon, ylon, xlat, ylat)

Thank you anyway, it is always good to learn new alternatives.

[Image Analyst]

OK. Next time, include your data so we can get you answer(s) right away with your own actual data.