Replace certain cell arrays by other cell arrays
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Ahmed Alsaadi
am 4 Jan. 2019
Kommentiert: Star Strider
am 4 Jan. 2019
I have this code, I want to replace 0,1,-1 in the first column by T1, T2, T3 respectively and 0,1,-1 by P1, P2, P3 respectively and the same for the third column I want to use S1, S2, and S3 instead of 0, 1, -1. Any ideas?
clc, clear
T = {'T1','T2','T3'}'
PS = {'P1','P2','P3'}'
DS = {'S1','S2','S3'}'
d = bbdesign (3)
dd = num2cell(d)
2 Kommentare
Akzeptierte Antwort
Star Strider
am 4 Jan. 2019
A simple loop seems to work:
idxv = [0; 1; -1];
T = {'T1','T2','T3'}';
PS = {'P1','P2','P3'}';
DS = {'S1','S2','S3'}';
labels = cat(3, T, PS, DS);
d = bbdesign (3);
dd = num2cell(d);
for k1 = 1:size(dd,2) % Columns
for k2 = 1:numel(idxv) % Elements
dd(d(:,k1)==idxv(k2), k1) = labels(k2,k1);
end
end
for k1 = 1:size(dd,1) % Print Results (Delete Later)
fprintf('%s\t%s\t%s\n', dd{k1,:})
end
producing:
T3 P3 S1
T3 P2 S1
T2 P3 S1
T2 P2 S1
T3 P1 S3
T3 P1 S2
T2 P1 S3
T2 P1 S2
T1 P3 S3
T1 P3 S2
T1 P2 S3
T1 P2 S2
T1 P1 S1
T1 P1 S1
T1 P1 S1
It uses simple logical indexing to compare the column entries of °d’ with successive elements of ‘idxv’, and do the replacements. It requires the interim creation of ‘idxv’ and ‘labels’ to make the code reasonably efficient.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Resizing and Reshaping Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!