Filtering doubles in a struct by criteria of one double

Hi all,

I'm newer to matlab and am trying to filter some data by a condition in one double of my struct. Here's my attempt:

id = site_data.wlH13 < 0.01;

site_data.wlH13(id) = NaN;

site_data.wlTmean(id) = NaN;

So the idea is that I'm creating a logical vector with the id of 1 and 0 which works correctly... and then the values of position '1' become NaN in the two doubles. But it does nothing and I don't understand why. My vectors are the same length so I don't think that could be the issue.

The desired out come would be filtering out values in wlH13 < 0.01 to be NaN and then the corresponding values in wlTmean would also be NaN.

I suppose I could write a loop but I'm doing this a lot with different parts of my data so I wanted to keep the loops to a minimum and was hoping to keep it vectorized.

I'd appreciate any help!

Cheers!

[Voss]

The code looks ok to me.

Is it possible that no element of site_data.wlH13 is less than 0.01? If you are seeing elements of site_data.wlH13 less than 0.01 that should be set to NaN but are not, please attach site_data in a mat-file so that we can try to reproduce the problem.

You could also write the loop and see whether that works or has the same problem, in the interest of debugging.

[Amy Bredes]

So when looking at my data I realized the doubles from one of my intrusments is in one row and many columns not on column with many rows. In matlab does direction of data matter? This data is definitely not "tidy" in that orientation, does matlab have issues looping through data that way? Transposing data would be an easy solution if that's the case

[Voss]

Whether the variable is a column vector or a row vector should not matter for logical indexing (like you have here) or for looping over elements. Try transposing and see if anything is different, and/or, share (some of) the data here so we can try to reproduce the problem.

Answer 1

[Star Strider]

It seems to work correctly when I run it with random data —

% site_data.wlH13 = randn(1,15);                          % Vector
% site_data.wlTmean = randn(1,15)                         % Vector
% 
%  id = site_data.wlH13 < 0.01
%     site_data.wlH13(id) = NaN;
%     site_data.wlTmean(id) = NaN

I have no idea what the problem could be otherwise.

EDIT — (29 Dec 2021 at 18:37)

‘In matlab does direction of data matter?’

No, not when dealilng with a logical matrix —

site_data.wlH13 = randn(10,5);
site_data.wlTmean = randn(10,5);

disp(site_data.wlH13)                                   % Original Matrix

    0.3003    0.9086   -1.2836    1.8215    1.7077
   -0.9888    0.6339    0.2637   -1.3677    1.6837
    0.9247   -0.9102   -0.2969    0.9729   -0.2158
   -0.1942    0.9928    1.8947    1.2380   -0.3695
   -0.3756   -0.3936    0.0046   -1.2277    0.0436
   -0.1705    0.0382    1.3941   -0.4806   -0.8174
   -0.6724    0.2722   -0.9517    1.0568    0.6946
   -1.8165    0.3566    0.2843    0.0853    2.5592
   -1.6213   -1.3947    1.0908   -0.6446    1.4818
    0.2327   -0.3986    0.6838    0.7668   -0.1432

disp(site_data.wlTmean)                                 % Original Matrix

    0.6265   -1.4207   -0.2086   -1.4861   -1.9043
   -0.7122    1.2252    1.8280   -0.5985    0.9328
    1.5839    1.3330   -1.3305    1.3311   -0.7515
    0.8710    0.5038   -1.1801    0.8015    1.1475
   -0.0660    1.4313   -0.5130   -1.6096    0.1137
   -1.3846    1.6282   -0.2993    1.9394    0.8071
    1.1329    0.4906    0.8091   -0.8215   -1.2889
    1.0612   -1.0081    1.6008    0.0384    1.3599
    0.2117    0.3809    0.0677    0.9157    0.5923
    2.3382    1.1527   -1.0338    1.3011   -0.4536

id = site_data.wlH13 < 0.01                             % Logical Selection Matrix

id = 10×5 logical array
   0   0   1   0   0
   1   0   0   1   0
   0   1   1   0   1
   1   0   0   0   1
   1   1   1   1   0
   1   0   0   1   1
   1   0   1   0   0
   1   0   0   0   0
   1   1   0   1   0
   0   1   0   0   1

site_data.wlH13(id) = NaN;                              % Substitute Value
site_data.wlTmean(id) = NaN;                            % Substitute Value

disp(site_data.wlH13)                                   % Display Result

    0.3003    0.9086       NaN    1.8215    1.7077
       NaN    0.6339    0.2637       NaN    1.6837
    0.9247       NaN       NaN    0.9729       NaN
       NaN    0.9928    1.8947    1.2380       NaN
       NaN       NaN       NaN       NaN    0.0436
       NaN    0.0382    1.3941       NaN       NaN
       NaN    0.2722       NaN    1.0568    0.6946
       NaN    0.3566    0.2843    0.0853    2.5592
       NaN       NaN    1.0908       NaN    1.4818
    0.2327       NaN    0.6838    0.7668       NaN

disp(site_data.wlTmean)                                 % Display Result

    0.6265   -1.4207       NaN   -1.4861   -1.9043
       NaN    1.2252    1.8280       NaN    0.9328
    1.5839       NaN       NaN    1.3311       NaN
       NaN    0.5038   -1.1801    0.8015       NaN
       NaN       NaN       NaN       NaN    0.1137
       NaN    1.6282   -0.2993       NaN       NaN
       NaN    0.4906       NaN   -0.8215   -1.2889
       NaN   -1.0081    1.6008    0.0384    1.3599
       NaN       NaN    0.0677       NaN    0.5923
    2.3382       NaN   -1.0338    1.3011       NaN

.