Replace certain cell arrays by other cell arrays
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Ahmed Alsaadi
on 4 Jan 2019
Commented: Star Strider
on 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)
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Accepted Answer
Star Strider
on 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.
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