how can a make a loop for a double value in Matlab?
    16 Ansichten (letzte 30 Tage)
  
       Ältere Kommentare anzeigen
    
    DulceEien
 am 10 Aug. 2021
  
    
    
    
    
    Kommentiert: DulceEien
 am 10 Aug. 2021
            when I obtain the data for the array I'm having the ID in a 22x1 double data, and when I run the if condition I have a problem with the size
"Unable to perform assignment because the indices on the left side are not compatible with the size of the right side." 

for i = 1:length(T1.ID)
    if ID(i) == 0
        EI(i) = 'Very high';
    elseif ID(i) == 1
        EI(i) = 'High';
    elseif ID(i) == 2
        EI(i) = 'Medium';
    elseif ID(i) == 3
        EI(i) = 'Low';
    end
end
the other values are taking a colum of the table generated. Sorry, I'm kind of new with matlab, thanks in advance 
0 Kommentare
Akzeptierte Antwort
  Simon Chan
      
 am 10 Aug. 2021
        I think it should put into a cell array:
EI = cell(1,length(T1.ID));                         % Cell array
for i = 1:length(T1.ID)
    if ID(i) == 0
        EI{i} = 'Very high';
    elseif ID(i) == 1
        EI{i} = 'High';
    elseif ID(i) == 2
        EI{i} = 'Medium';
    elseif ID(i) == 3
        EI{i} = 'Low';
    end
end
Weitere Antworten (1)
  Scott MacKenzie
      
 am 10 Aug. 2021
        It's not clear what 19 variables you have in the table T1, but I suspect your error will disappear via...
for i = 1:length(ID) % change from T1.ID to just ID
    if ID(i) == 0
        EI(i) = 'Very high';
    elseif ID(i) == 1
        EI(i) = 'High';
    elseif ID(i) == 2
        EI(i) = 'Medium';
    elseif ID(i) == 3
        EI(i) = 'Low';
    end
end
Siehe auch
Kategorien
				Mehr zu Loops and Conditional Statements 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!


