# how can a make a loop for a double value in Matlab?

17 views (last 30 days)
DulceEien on 10 Aug 2021
Commented: DulceEien on 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 CommentsShowHide -1 older comments

Sign in to comment.

### Accepted Answer

Simon Chan on 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
##### 1 CommentShowHide None
DulceEien on 10 Aug 2021
yes, this is working, thank yoou

Sign in to comment.

### More Answers (1)

Scott MacKenzie on 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
##### 1 CommentShowHide None
DulceEien on 10 Aug 2021
thank you for replying

Sign in to comment.

### Categories

Find more on Matrices and Arrays in Help Center and File Exchange

### Community Treasure Hunt

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

Start Hunting!