Derivative of table between NaN values

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puccapearl 2024-5-21

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2121226-derivative-of-table-between-nan-values

评论： Star Strider 2024-5-21

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!

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Answer 1

Star Strider 2024-5-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2121226-derivative-of-table-between-nan-values#answer_1461201

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.

.

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Star Strider 2024-5-21

在 MATLAB Online 中打开

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 2024-5-21

编辑：puccapearl 2024-5-21

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 2024-5-21

在 MATLAB Online 中打开

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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<mw-icon class=""></mw-icon>
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
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
each_col{3}
ans = 4x2
    0.0756    0.3853
    0.0619   -0.1851
    0.0412   -0.2135
    0.0342    0.3286
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
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
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
dRangeRate_dTime1{3}
ans = 4x1
    5.0954
   -2.9883
   -5.1801
    9.6161
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
dRangeRate_dTime2{1}
ans = 4x1
   -1.4969
    5.2931
    0.7106
   -3.5124
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
dRangeRate_dTime2{2}
ans = 7x1
   13.7506
    5.1580
   -1.4555
  -11.9580
    0.2875
   -1.1509
  -12.8691
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
dRangeRate_dTime2{3}
ans = 4x1
    5.0954
   -2.9883
   -5.1801
    9.6161
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

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.

.

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Answer 2

Torsten 2024-5-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2121226-derivative-of-table-between-nan-values#answer_1461206

移动：Torsten 2024-5-21

在 MATLAB Online 中打开

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
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
section_start = positions(end)+1;
section_end = size(a,1);
section_a = a(section_start:section_end,:)
section_a = 1x2
     5    10
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
%Use gradient