indexing with isnan in multidimensional arrays

I have this code:

load lon_nonan_reshape_use;
idx = ~isnan(lon_nonan_reshape_use);
lon_nonan_adj = lon_nonan_reshape_use(idx);

In the above example, lon_nonan_reshape_use and idx both are of size n*2 so I expect lon_nonan_adj also to be of n*2. MATLAB gives the results in n*1 dimension because lon_nonan_adj(:, 1) is not equal to lon_nonan_adj(:, 2) which is sensible. But I still want to get lon_nonan_adj as a n*2 matrix with NaNs retained in the place where there were NaNs previously. Could you please help?

[DGM]

It's not clear what you're after. None of these are 2x1 or 1x1.

load lon_nonan_reshape_use; size(lon_nonan_reshape_use)

ans = 1×2

208278 2

idx = ~isnan(lon_nonan_reshape_use); size(idx)

ans = 1×2

208278 2

lon_nonan_adj = lon_nonan_reshape_use(idx); size(lon_nonan_adj)

ans = 1×2

412344 1

ans(1)/2 % not the same number of elements as before because nans are missing

ans = 206172

If you want lon_nonan_adj to be the same shape as lon_nonan_reshape_use, with the NaNs preserved, then there's no reason to have made it in the first place. You would have removed the NaNs and then put them back in the same spots. I don't get it.

[Sagar Parajuli]

I edited the question, hope it is clearer now.

[DGM]

"But I still want to get lon_nonan_adj as a n*2 matrix with NaNs retained in the place where there were NaNs previously."

If you take matrix A, remove all the NaNs and then fill all the empty spots left over with NaN such that the size is preserved, then you're back at A. I don't see how the operation isn't superfluous.

[Sagar Parajuli]

I think you are right, so it looks like I should get rid of all NaNs and deal with the no-nan data only as answered below. Thank you.

Answer 1

[Walter Roberson]

load lon_nonan_reshape_use;
idx = ~all(isnan(lon_nonan_reshape_use),2);
lon_nonan_adj = lon_nonan_reshape_use(idx,:);

This retains rows provided there is at least one non-nan in the row.

[Walter Roberson]

If you want to remove all rows with NaN anywhere in the row, then

rmmissing(lon_nonan_reshape_use)

Answer 2

[Chunru]

% A small matrix with nans
a=randn(6, 3);
a([2 11 13])=nan

a = 6×3

-0.5340 0.4891 NaN NaN 0.2906 0.4462 -1.2660 -1.1701 2.0372 -0.3616 -0.4775 0.3251 0.6966 NaN 0.1013 0.4787 1.2229 -0.6159

% idx
idx = ~isnan(a)

idx = 6×3 logical array
   1   1   0
   0   1   1
   1   1   1
   1   1   1
   1   0   1
   1   1   1

a(idx)

ans = 15×1

-0.5340 -1.2660 -0.3616 0.6966 0.4787 0.4891 0.2906 -1.1701 -0.4775 1.2229

So a(idx) is a colum matrix based on the original a by removing all nans and arranged in a column vector.

If you want to remove all rows with nans (it looks so for your problem), then you can do the following

idx1 = all(~isnan(a), 2)

idx1 = 6×1 logical array
   0
   0
   1
   1
   0
   1

a(idx1, :)

ans = 3×3

-1.2660 -1.1701 2.0372 -0.3616 -0.4775 0.3251 0.4787 1.2229 -0.6159

[Sagar Parajuli]

I like your last solution because ultimately that is what I need (data with all NaNs removed), to input to 'griddata'. Thank you for forseeing my problem.