Replacing Negative Values in Table with Previous Value in Column

1 Ansicht (letzte 30 Tage)
I have a table 17520x5, in the last 2 columns I would like to replace all negative values with the previous value in that column. This is what I have tried so far and it is not working I still get negative values shown.
My table (T) had 5 columns, variable labels are (A, B, C, D, E) for each column for example
D(D < 0) = NaN;
E(E < 0) = NaN;
T(:, {'D', 'E'}) = fillmissing(T(:,{'D', 'E'}), 'previous');
disp(T)

Akzeptierte Antwort

Matt J
Matt J am 22 Nov. 2021
Bearbeitet: Matt J am 22 Nov. 2021
T(:, {'D', 'E'}) = num2cell( fillmissing([D,E], 'previous') );
  3 Kommentare
Matt J
Matt J am 22 Nov. 2021
Bearbeitet: Matt J am 22 Nov. 2021
We can try an example to show that it works:
T=array2table( rand(4,5)-0.5 ,'Var',{'A','B','C','D','E'});
D=T{:,4}; E=T{:,5};
D(D < 0) = NaN;
E(E < 0) = NaN;
T,
T = 4×5 table
A B C D E _______ ________ ________ ________ ________ 0.44208 0.49326 0.12868 0.42218 0.43518 0.24954 -0.42086 -0.39169 -0.31886 -0.13442 0.38232 -0.3545 -0.49562 -0.47566 -0.15402 0.28615 0.3741 0.11841 -0.16468 0.48483
T(:, {'D', 'E'}) = num2cell( fillmissing([D,E], 'previous') )
T = 4×5 table
A B C D E _______ ________ ________ _______ _______ 0.44208 0.49326 0.12868 0.42218 0.43518 0.24954 -0.42086 -0.39169 0.42218 0.43518 0.38232 -0.3545 -0.49562 0.42218 0.43518 0.28615 0.3741 0.11841 0.42218 0.48483

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Peter Perkins
Peter Perkins am 23 Nov. 2021
Stef, as near as I can tell, the only thing wrong with your original solution is that D and E are in T, not in the workspace. The following works fine, including repeated negative values and negative values in the first row. There's no need to explicitly pull D and E out of the table. Using Matt's setup:
>> T = array2table(rand(4,5)-0.5 ,'Var',["A" "B" "C" "D" "E"])
T =
4×5 table
A B C D E
________ ________ _________ ________ _________
-0.33782 -0.33435 0.18921 -0.27102 0.038342
0.29428 0.10198 0.24815 0.41334 0.49613
-0.18878 -0.23703 -0.049458 -0.34762 -0.42182
0.028533 0.15408 -0.41618 0.32582 -0.057322
>> T.D(T.D < 0) = NaN;
>> T.E(T.E < 0) = NaN
T =
4×5 table
A B C D E
________ ________ _________ _______ ________
-0.33782 -0.33435 0.18921 NaN 0.038342
0.29428 0.10198 0.24815 0.41334 0.49613
-0.18878 -0.23703 -0.049458 NaN NaN
0.028533 0.15408 -0.41618 0.32582 NaN
>> T(:, ["D" "E"]) = fillmissing(T(:,["D" "E"]), 'previous')
T =
4×5 table
A B C D E
________ ________ _________ _______ ________
-0.33782 -0.33435 0.18921 NaN 0.038342
0.29428 0.10198 0.24815 0.41334 0.49613
-0.18878 -0.23703 -0.049458 0.41334 0.49613
0.028533 0.15408 -0.41618 0.32582 0.49613

Kategorien

Mehr zu Characters and Strings finden Sie in Help Center und File Exchange

Produkte


Version

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by