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Renaming cells inside structure from variable in for loop

조회 수: 1 (최근 30일)
Mohammed Hammad
Mohammed Hammad 2019년 1월 24일
댓글: Stephen23 2019년 1월 24일
I have a for loop which it goes through mat files and copy results to put them inside one structure called (level0). This structure has many cells based on the number of the files (e.g 10) and I want to rename them based on variables (dates).
in other words that each cell or fields will have the name with the date and it has inside a structure with these values
dir = 'directory';
filePattern3 = fullfile(dir, 'x*.mat');
mat3 = dir(filePattern3);
name3= {mat3.name}.';
numfiles3 = length(name3);
results3 = cell(numfiles3,1);
dates = [cellfun( @(S) datetime(S([3:9]), 'InputFormat', 'yyMMMdd'), name3, 'uniformoutput',false )];
dates = [cellfun(@datestr,dates,'un',0)];
for K = 1 : numfiles3
thisfile3 = name3{K};
datastruct = load(thisfile3);
xval = [datastruct.x_.coorx.val].'; %extract variable
yval = [datastruct.x_.coory.val].';
zval = [datastruct.x_.coorz.val].';
mjd = [datastruct.x_.coorx.mjd].';
t = mjd / 365.2425;
level0{K}.xval = xval;
level0{K}.yval = yval;
level0{K}.zval = zval;
level0{K}.t = t;
end
Please help me to achieve that, thanks in advance
The dates are
dates =
'04-Apr-2002'
'11-Apr-2002'
'18-Apr-2002'
'25-Apr-2002'
'25-Apr-2002'
'01-Aug-2002'
'08-Aug-2002'
'15-Aug-2002'
'22-Aug-2002'
'29-Aug-2002'
  댓글 수: 5
Mohammed Hammad
Mohammed Hammad 2019년 1월 24일
Can you please share similar example for indexing? I am still beginner with these stuff.
Thank you

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답변 (1개)

Stephen23
Stephen23 2019년 1월 24일
편집: Stephen23 2019년 1월 24일
You can certainly rename fields dynamically:
But really, it would likely be simpler and more efficient to use indexing and store the dates as data in their own right. Forcing meta-data (such as dates) into things like variable names or fieldnames just makes code slow and fragile. Remember that meta-data is data, and so it deserves to be treated as data and stored in a variable, not forced awkwardly into names of variables or fields:
S(k).X = [...];
S(k).Y = [...];
S(k).date = '2019-01-24';

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