# Derivative of table between NaN values

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puccapearl el 21 de Mayo de 2024
Comentada: Star Strider el 21 de Mayo de 2024
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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### Respuestas (2)

Star Strider el 21 de Mayo de 2024
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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puccapearl el 21 de Mayo de 2024
Editada: puccapearl el 21 de Mayo de 2024
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 el 21 de Mayo de 2024
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)
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)
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.
.

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Torsten el 21 de Mayo de 2024
Movida: Torsten el 21 de Mayo de 2024
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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for i = 2:numel(positions)
section_start = positions(i-1)+1;
section_end = positions(i)-1;
section_a = a(section_start:section_end,:)
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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