Linear FillMissing producing incorrect values?

My Code involves filling in missing values using linear interpolation, but its plain to see that the values its giving are wrong. User error or matlab bug? Instead of interpolating the missing value for 76.82, its simply halving the previous value and calling it a day. Thanks in advance guys!

Baye2 =

0 0

14.1800 NaN

29.0625 7.7096

58.1250 0

74.5625 4.3605

76.8200 NaN

91.0000 0

Bayee2 =

0 0

14.1800 3.8548

29.0625 7.7096

58.1250 0

74.5625 4.3605

76.8200 2.1803

91.0000 0

%Inputs%
%example from bottom horizontal on D11.10z%
clear
clc

DP = 38.2; %Design pressure in psf
ML = 91; %Mullion length in inches
B1W = 28.36; %Bay 1 width in inches
B2W = 60.67; %Bay 2 width in inches
B1BP = []; %Bay 1 break points in inches seperated by spaces
B2BP = [58.125]; %Bay 2 break points in inches seperated by spaces
B1 = [0 0];
B2 = [0 0];

if isempty(B1BP) %if there is only a single loading on bay 1 (no breakpoints)
    if ML > B1W % Trapezoidal loading
        B1 = [0 0; B1W/2 B1W/2/144*DP; ML-B1W/2 B1W/2/144*DP; ML 0];

    else %Triangular loading
        B1 = [0 0; ML/2 ML/2/144*DP; ML 0];

    end

else %if there are multiple loadings for bay 1
    B1BP = [0 B1BP ML];
    for n = 1:length(B1BP)-1 %for each loading on bay 1
        if (B1BP(n+1)-B1BP(n)) > B1W % Trapezoidal loading
            Bn1s{n} = [B1BP(n) 0; B1BP(n)+B1W/2 B1W/2/144*DP; B1BP(n+1)-B1W/2 B1W/2/144*DP; B1BP(n+1) 0];
            Bn1 = [B1BP(n) 0; B1BP(n)+B1W/2 B1W/2/144*DP; B1BP(n+1)-B1W/2 B1W/2/144*DP; B1BP(n+1) 0];
            B1 = [B1; Bn1];
            n=n+1;

        else %Triangular loading
            Bn1s{n} = [B1BP(n) 0; B1BP(n)+(B1BP(n+1)-B1BP(n))/2 (B1BP(n+1)-B1BP(n))/2/144*DP; B1BP(n+1) 0];
            Bn1 = [B1BP(n) 0; B1BP(n)+(B1BP(n+1)-B1BP(n))/2 (B1BP(n+1)-B1BP(n))/2/144*DP; B1BP(n+1) 0];
            B1 = [B1; Bn1];
            n=n+1;

        end
    end
end

B1 = unique(B1,'rows');


if isempty(B2BP) %if there is only a single loading on bay 2 (no breakpoints)
    if ML > B2W % Trapezoidal loading
        B2 = [0 0; B2W/2 B2W/2/144*DP; ML-B2W/2 B2W/2/144*DP; ML 0];

    else %Triangular loading
        B2 = [0 0; ML/2 ML/2/144*DP; ML 0];

    end

else %if there are multiple loadings for bay 2
    B2BP = [0 B2BP ML];
    for j = 1:length(B2BP)-1 %for each loading on bay 1
        if (B2BP(j+1)-B2BP(j)) > B2W % Trapezoidal loading
            Bn2s{j} = [B2BP(j) 0; B2BP(j)+B2W/2 B2W/2/144*DP; B2BP(j+1)-B2W/2 B2W/2/144*DP; B2BP(j+1) 0];
            Bn2 = [B2BP(j) 0; B2BP(j)+B2W/2 B2W/2/144*DP; B2BP(j+1)-B2W/2 B2W/2/144*DP; B2BP(j+1) 0];
            B2 = [B2; Bn2];
            j=j+1;

        else %Triangular loading
            Bn2s{j} = [B2BP(j) 0; B2BP(j)+(B2BP(j+1)-B2BP(j))/2 (B2BP(j+1)-B2BP(j))/2/144*DP; B2BP(j+1) 0];
            Bn2 = [B2BP(j) 0; B2BP(j)+(B2BP(j+1)-B2BP(j))/2 (B2BP(j+1)-B2BP(j))/2/144*DP; B2BP(j+1) 0];
            B2 = [B2; Bn2];
            j=j+1;

        end
    end
end

B2 = unique(B2,'rows');


B1L = B1(:,1); %extracts just the locations from the Bay 1 loadings
B1z = NaN(size(B1L,1),1); %creates a column of unknowns the length of the locations
B1L = [B1L B1z]; %combines these two columns 

B2L = B2(:,1);
B2z = NaN(size(B2L,1),1);
B2L = [B2L B2z];
 
B1 = [B1;B2L];
B1 = sortrows(B1);
[Bay1, ia] = unique(B1(:,1),'rows');
Baye1 = B1(ia,:);
Bayee1 = fillmissing(Baye1,'linear');

B2 = [B2;B1L];
B2 = sortrows(B2);
[Bay2, ib] = unique(B2(:,1),'rows');
Baye2 = B2(ib,:)
Bayee2 = fillmissing(Baye2,'linear')

Loadings = [Bayee1(:,1) Bayee1(:, 2) + Bayee2(:, 2)];
Loadings = round(Loadings, 2);

Answer 1

[the cyclist]

User error (or, really, misinterpretation).

Although the result happens to be half, the process is not to just take half of the previous value. In the case of your first NaN, it's linearly interpolating between 0 and 7.7096, which gives 3.8548. It's only because one of your values is zero, that the result is half of the other one.

[Nathan Reed]

Im not sure I (or you) understand. Matlab is taking half of the Y value (irrespective of the x coordinate). This is not a linear interpolation and is giving the wrong answer. For example, the first interpolated value should be 3.7616 not 3.8548. The error is much more pronounce on the second interpolated value however.

[the cyclist]

Ah, you are misunderstanding how fillmissing operates. The algorithm only uses values from its own column. This spelled out in the documentation.

You'll want to use interp1 to linearly interpolate Y based on the position in X.

[Nathan Reed]

I previously determined that Interp1 wouldnt work for my sepcific case without significantly increasing the complexity of my code. Can you explain why I cant use it in the way its being used in the documentation with y dependant on x, specifically under the "Interpolate Missing Data" section?

https://www.mathworks.com/help/matlab/ref/fillmissing.html

[the cyclist]

You can:

Baye2 = [
         0         0
   14.1800       NaN
   29.0625    7.7096
   58.1250         0
   74.5625    4.3605
   76.8200       NaN
   91.0000         0];

% What you did (uses implicit even spacing)
fillmissing(Baye2,'linear')

ans = 7×2

0 0 14.1800 3.8548 29.0625 7.7096 58.1250 0 74.5625 4.3605 76.8200 2.1803 91.0000 0

% How to use the first column of Baye2 as the sample points
fillmissing(Baye2,'linear','SamplePoints',Baye2(:,1))

ans = 7×2

0 0 14.1800 3.7616 29.0625 7.7096 58.1250 0 74.5625 4.3605 76.8200 3.7616 91.0000 0

You will need to be careful to not interpolate every column of your matrix, if that is not what you want to do.

[Nathan Reed]

That seems to be working great, thank you.