search for the position of a number inside a cell (cell with existence of empty rows)
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Alberto Acri
il 10 Feb 2024
Risposto: Star Strider
il 10 Feb 2024
I am trying to look for each row of 'N' inside a cell.
I am using 'find' but it doesn't seem to work.
I should identify, in the cell, the rows of N. In this case, row 3 and row 19.
parameters = importdata("parameters.mat");
N = [parameters{3,1}; parameters{19,1}];
tt = 1;
for t = 1:height(parameters)
[r,c] = find(parameters{t,1}(1,1) == N(tt,1) & ...
parameters{t,1}(1,2) == N(tt,2) & ...
parameters{t,1}(1,3) == N(tt,3) & ...
parameters{t,1}(1,4) == N(tt,4));
end
0 Commenti
Risposta accettata
Star Strider
il 10 Feb 2024
Use ismember (since you have the exact values, otherwise ismembertol) and any (here using the default column orientation, dimension = 1) to be inclusive, with cellfun (that loops internally) —
load('parameters')
% whos
parameters
N = [parameters{3,1}; parameters{19,1}]; % 'Target' Values
idx = cell2mat(cellfun(@(x)any(ismember(x,N)), parameters, 'Unif',0)); % Logical Index Vector
Result_Index = find(idx) % Numeric Index
Result = parameters(idx) % Result Values
There are likely other ways to do this as well.
.
0 Commenti
Più risposte (1)
VBBV
il 10 Feb 2024
Modificato: VBBV
il 10 Feb 2024
Use a | instead of &. OR in place of AND
2 Commenti
VBBV
il 10 Feb 2024
Using for loop it can be like this
p = load("parameters.mat");
N = [p.parameters{3,1}; p.parameters{19,1}]
tt = 1;
for t = 1:height(p.parameters)-1
K = p.parameters{t+1,1};
if ~isempty(K)
if tt == 1 % N(1,:) inside parameter
[r,c] = find((K(1) - N(tt,1)) < 1e-3 & ...
(K(2) - N(tt,2)) < 1e-3 & ...
(K(3) - N(tt,3)) < 1e-3 & ...
(K(4) - N(tt,4)) < 1e-3)
else % N(2,:) inside parameter
tt = tt + 1;
[r,c] = find((K(1) - N(tt,1)) < 1e-3 & ...
(K(2) - N(tt,2)) < 1e-3 & ...
(K(3) - N(tt,3)) < 1e-3 & ...
(K(4) - N(tt,4)) < 1e-3)
end
end
end
You can try using ismember function to vaoid for loop
Vedere anche
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!