Derivative of table between NaN values

I have data in a table, in between there are NaN values.

I want to calculate the derivatives (using the 1st column as x's and 2nd column as y's ) between each NaN space and put these values into a new vector.

Thanks!

Answer 1

[Star Strider]

Depending on what you want to do, use either fillmissing or rmmissing to either interpolate the NaN values (fillmissing) or remove the rows with NaN values (rmmissing) in at least one column.

Use the gradient function to calculate the numerical derivatives.

.

[puccapearl]

I don't want to do this because I want to calculate the derivatives in between each section (between the NaN values) not the whole data set at once.

[Star Strider]

Try something like this —

raw_data = array2table(sortrows(rand(20,2),1));
raw_data{randi(size(raw_data,1),3,1),:} = NaN

raw_data = 20x2 table
      Var1       Var2  
    ________    _______

    0.024944    0.56483
     0.19161    0.36235
     0.35388    0.28517
      0.3627    0.73479
     0.36546    0.79793
     0.38041    0.53506
     0.42025    0.27599
     0.45254    0.86484
         NaN        NaN
     0.53107    0.98757
     0.53572    0.64278
     0.72116    0.81537
     0.75179     0.6441
         NaN        NaN
     0.79652    0.88687
     0.83232    0.93528

NaN_rows = [1; find(isnan(raw_data{:,1})); size(raw_data,1)]

NaN_rows = 4x1

1 9 14 20

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for k = 1:numel(NaN_rows)-1
    idxrng = NaN_rows(k):NaN_rows(k+1);
    raw_data_segment{k} = rmmissing(raw_data(idxrng,:));
end

raw_data_segment{1}

ans = 8x2 table
      Var1       Var2  
    ________    _______

    0.024944    0.56483
     0.19161    0.36235
     0.35388    0.28517
      0.3627    0.73479
     0.36546    0.79793
     0.38041    0.53506
     0.42025    0.27599
     0.45254    0.86484

raw_data_segment{2}

ans = 4x2 table
     Var1       Var2  
    _______    _______

    0.53107    0.98757
    0.53572    0.64278
    0.72116    0.81537
    0.75179     0.6441

raw_data_segment{3}

ans = 6x2 table
     Var1       Var2  
    _______    _______

    0.79652    0.88687
    0.83232    0.93528
    0.89369    0.97574
    0.89435    0.44376
    0.91501    0.61939
    0.98391    0.48512

Then use gradient to calculate the derivatives.

.

[puccapearl]

I went ahead and put all the sections into tables of a cell, do you know how I can calculate the graident of each column?

[Star Strider]

Do you want the gradient of each column, or the gradient of the second column as a function of the first column?

I would do something like this —

raw_data = array2table(sortrows(rand(20,2),1));
raw_data{randi(size(raw_data,1),3,1),:} = NaN

raw_data = 20x2 table
     Var1        Var2   
    _______    _________

     0.1037      0.12961
    0.18754    0.0041072
    0.21735      0.73119
    0.38451      0.14408
        NaN          NaN
    0.43713      0.29112
    0.47086      0.75494
    0.53886      0.81585
    0.56965      0.61116
    0.58919        0.214
    0.59318      0.61792
    0.62804      0.16929
        NaN          NaN
     0.6597      0.57782
    0.73532      0.96316
    0.78359       0.2076

NaN_rows = [1; find(isnan(raw_data{:,1})); size(raw_data,1)]

NaN_rows = 5x1

1 5 13 18 20

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for k = 1:numel(NaN_rows)-1
    idxrng = NaN_rows(k):NaN_rows(k+1);
    raw_data_segment{k} = rmmissing(raw_data(idxrng,:));
end

raw_data_segment{1}

ans = 4x2 table
     Var1        Var2   
    _______    _________

     0.1037      0.12961
    0.18754    0.0041072
    0.21735      0.73119
    0.38451      0.14408

raw_data_segment{2}

ans = 7x2 table
     Var1       Var2  
    _______    _______

    0.43713    0.29112
    0.47086    0.75494
    0.53886    0.81585
    0.56965    0.61116
    0.58919      0.214
    0.59318    0.61792
    0.62804    0.16929

raw_data_segment{3}

ans = 4x2 table
     Var1       Var2  
    _______    _______

     0.6597    0.57782
    0.73532    0.96316
    0.78359     0.2076
    0.81776    0.53616

for k = 1:numel(raw_data_segment)
    each_col{k} = [gradient(raw_data_segment{k}{:,1}) gradient(raw_data_segment{k}{:,2})];
end

each_col{1}

ans = 4x2

0.0838 -0.1255 0.0568 0.3008 0.0985 0.0700 0.1672 -0.5871

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each_col{2}

ans = 7x2

0.0337 0.4638 0.0509 0.2624 0.0494 -0.0719 0.0252 -0.3009 0.0118 0.0034 0.0194 -0.0224 0.0349 -0.4486

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each_col{3}

ans = 4x2

0.0756 0.3853 0.0619 -0.1851 0.0412 -0.2135 0.0342 0.3286

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for k = 1:numel(raw_data_segment)
    dRangeRate_dTime1{k} = gradient(raw_data_segment{k}{:,2}, raw_data_segment{k}{:,1});
    dRangeRate_dTime2{k} = gradient(raw_data_segment{k}{:,2}) ./ gradient(raw_data_segment{k}{:,1});
end

dRangeRate_dTime1{1}

ans = 4x1

-1.4969 5.2931 0.7106 -3.5124

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dRangeRate_dTime1{2}

ans = 7x1

13.7506 5.1580 -1.4555 -11.9580 0.2875 -1.1509 -12.8691

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dRangeRate_dTime1{3}

ans = 4x1

5.0954 -2.9883 -5.1801 9.6161

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dRangeRate_dTime2{1}

ans = 4x1

-1.4969 5.2931 0.7106 -3.5124

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dRangeRate_dTime2{2}

ans = 7x1

13.7506 5.1580 -1.4555 -11.9580 0.2875 -1.1509 -12.8691

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dRangeRate_dTime2{3}

ans = 4x1

5.0954 -2.9883 -5.1801 9.6161

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The gradient function has changed over time, and since I am not certain which version you have, I calculated the time derivative using both types of syntax. The first syntax may give the correct result (it will in the most recent versions), the second syntax always will, however it may be less efficient.

.

Answer 2

[Torsten]

a=[1 4; 3 6; NaN NaN; 5 7; 9 0.2; NaN NaN;5 10];
positions = find(isnan(a(:,1)))

positions = 2x1

3 6

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section_start = 1;
section_end = positions(1)-1;
section = a(section_start:section_end,:)

section = 2x2

1 4 3 6

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%Use gradient
for i = 2:numel(positions)
    section_start = positions(i-1)+1;
    section_end = positions(i)-1;
    section_a = a(section_start:section_end,:)
    %Use gradient
end

section_a = 2x2

5.0000 7.0000 9.0000 0.2000

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section_start = positions(end)+1;
section_end = size(a,1);
section_a = a(section_start:section_end,:)

section_a = 1x2

5 10

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%Use gradient