how to read text file into formatted array
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Hi
I have a text file which contains many rows of this shape
'0862;1;10/09/2002;09:15:59;23.845089;38.018470;486253.80;4207588.10'
I need to read each line in different cells. each cell contains data between ";" and the third and forth one are date and time while others are double numbers.
would you please help me to do so
2 Kommentare
per isakson
am 15 Dez. 2014
Bearbeitet: per isakson
am 15 Dez. 2014
"I need to read each line in different cells"   Why is that? And what does it exactly mean?
Akzeptierte Antwort
per isakson
am 15 Dez. 2014
Bearbeitet: per isakson
am 15 Dez. 2014
Here is a function, which does what I believe is what you want. The third column of out is a serial date number.
>> out = cssm('cssm.txt')
out =
1.0e+06 *
0.0009 0.0000 0.7315 0.0000 0.0000 0.4863 4.2076
0.0009 0.0000 0.7315 0.0000 0.0000 0.4863 4.2076
0.0009 0.0000 0.7315 0.0000 0.0000 0.4863 4.2076
>> datestr( out(:,3), 31 )
ans =
2002-10-09 09:15:59
2002-10-09 09:15:59
2002-10-09 09:15:59
where
function out = cssm( filespec )
fid = fopen( filespec );
cac = textscan( fid, '%f%f%s%s%f%f%f%f' ...
, 'Delimiter',';', 'CollectOutput',true );
fclose( fid );
str = cell2mat(cac{2});
sdn = datenum( str, 'mm/dd/yyyyHH:MM:SS');
out = cat( 2, cac{1}, sdn, cac{3} );
end
and where cssm.txt contains
0862;1;10/09/2002;09:15:59;23.845089;38.018470;486253.80;4207588.10
0862;1;10/09/2002;09:15:59;23.845089;38.018470;486253.80;4207588.10
0862;1;10/09/2002;09:15:59;23.845089;38.018470;486253.80;4207588.10
4 Kommentare
per isakson
am 15 Dez. 2014
Bearbeitet: per isakson
am 15 Dez. 2014
cssm is the function defined after the word, where, in my answer.
Weitere Antworten (1)
Azzi Abdelmalek
am 15 Dez. 2014
Bearbeitet: Azzi Abdelmalek
am 15 Dez. 2014
fid=fopen('FileName.txt')
tline = fgetl(fid)
out=[];
while ischar(tline)
out{end+1,1} =tline
tline= fgetl(fid);
end
fclose(fid)
%You can aadd
A=regexp(out,';','split')
n=numel((A{1}))
B=reshape([A{:}],n,[])'
res=[cell
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